INDEX [] TECHNOLOGICAL UNIVERSITY CIVIL ENGINEERING (06) NUMERICAL AND STATISTICAL METHODS FOR CIVIL ENGINEERING SUBJECT CODE: 2140606 B.E. 4th Semester Type of course: Engineering

INDEX

N.S.M.-2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

GTU Syllabus

Tutorial No. Name Of Topic

Statistics

Unit 1 Basic Probability

Unit 2 Probability

Unit 3 Descriptive Statistics

Unit 4 Correlation and Regression

Unit 5 Curve Fitting

Numerical Methods

Unit 6 Finite Differences and Interpolation

Unit 7 Numerical Integration

Unit 8 Solution of a System of Linear Equations

Unit 9 Roots of Algebraic and Transcendental Equations

Unit 10 Numerical solution of Ordinary Differential Equations

GTU Papers

June-2015

December-2015

June-2016

December-2016

GUJARAT TECHNOLOGICAL UNIVERSITY

CIVIL ENGINEERING (06) NUMERICAL AND STATISTICAL METHODS FOR CIVIL ENGINEERING

SUBJECT CODE: 2140606

B.E. 4th Semester

Type of course: Engineering Mathematics

Prerequisites: The students are required to have a reasonable mastery over calculus,

Differential equations and Linear algebra and introductory knowledge of probability and statistics.

Rationale: Mathematics is a language of Science and Engineering.

Teaching and Examination Scheme:

Teaching Scheme Credits Examination Marks Total

Marks L T P C Theory Marks Practical Marks

ESE

(E)

PA (M) PA (V) PA

(I) PA ALA ESE OEP

3 2 0 5 70 20 10 30 0 20 150

L- Lectures; T- Tutorial/Teacher Guided Student Activity; P- Practical; C- Credit; ESE- End Semester

Examination; PA- Progressive Assessment

Content:

Sr. No. Topics Teaching

Hrs.

Module

Weightage

Probability

1 Reorientation: Definition of probability, Exhaustive

events, Pair wise independent events, Multiplicative law of

probability, Conditional probability, Baye’s theorem

03 07

2 Probability Distributions: Random variable,

Mathematical Expectation, Standard Deviation, Binomial,

Poisson and Normal distributions, Mean, Median, Mode

05 12

Statistics

3 Descriptive Statistics: Mean, Median, Mode, Standard

deviation, Skewness

03 08

4 Correlation and Regression: Bivariate distribution,

Correlation coefficients, Regression lines, Formulas for

Regression coefficients, Rank correlation

04 10

5 Curve Fitting: Fitting of Linear, Quadratic, Exponential

and Logarithmic curves, Least squares method

03 08

Numerical Methods

6 Finite Differences and Interpolation: Finite Differences,

Forward, Backward and Central operators, Interpolation by

polynomials: Newton’s forward ,Backward interpolation

formulae, Gauss & Stirling’s central difference formulae ,

Newton’s divided and Lagrange’s formulae for unequal

intervals

08 15

7 Numerical Integration: Newton-Cotes formula,

Trapezoidal and Simpson’s formulae, error formulae,

Gaussian quadrature formulae

03 08

8 Solution of a System of Linear Equations: Gauss

elimination, partial pivoting , Gauss-Jacobi and Gauss-

Seidel methods

03 07

9 Roots of Algebraic and Transcendental Equations:

Bisection, false position, Secant and Newton-Raphson

methods, Rate of convergence

04 10

10 Numerical solution of Ordinary Differential Equations:

Taylor series method, Euler method, Runge-Kutta method

of order four, Milne’s Predictor-Corrector method

06 15

Suggested Specification table with Marks (Theory):

Distribution of Theory Marks

R Level U Level A Level N Level E Level

10 15 20 20 35

Legends: R: Remembrance; U: Understanding; A: Application, N: Analyze and E: Evaluate

and above Levels (Revised Bloom’s Taxonomy)

Note: This specification table shall be treated as a general guideline for students and teachers. The

actual distribution of marks in the question paper may vary slightly from above table

Reference Books:

Reference Books:

1. E. Kreyszig, Advanced Engineering Mathematics(8th Edition), John Wiley (1999)

2. S. D. Conte and Carl de Boor, Elementary Numerical Analysis-An Algorithmic

Approach (3rd Edition), McGraw-Hill, 1980

3. C.E. Froberg, Introduction to Numerical Analysis (2nd Edition), Addison-

Wesley,1981

4. Gerald C. F. and Wheatley P.O. , Applied Numerical Analysis (5th Edition),

Addison-Wesley, Singapore, 1998

5. Johnson Richard A., Miller and Freund's - Probability and Statistics (8th Edition) ,

PHI.

6. S.C. Gupta and V. K. Kapoor, Fundamentals of Mathematical Statistics (11th

Edition), Sultan Chand & Sons.

Course Outcomes:

After learning the course the students should be able to :

o Understand and apply the basic concepts of probability, random variables, probability distribution.

o Use statistical methodology and tools in the engineering problem solving process.

o Compute and interpret descriptive statistics using numerical and graphical techniques

o Understand the basic concepts of regression and curve fitting

o Calculate finite differences of tabulated data.

o use numerical methods to find integration and differentiation

o find an approximate solution of algebraic equations using appropriate method.

http://www.amazon.in/s/ref=dp_byline_sr_book_1?ie=UTF8&field-author=Johnson+Richard+A.&search-alias=stripbooks

o find an approximate solution of ordinary differential equations using appropriate

iterative method.

List of Open Source Software/learning website:

http://nptel.ac.in/courses/111101003/

http://nptel.ac.in/syllabus/syllabus.php?subjectId=111101004





ACTIVE LEARNING ASSIGNMENTS: Preparation of power-point slides, which include videos,

animations, pictures, graphics for better understanding theory and practical work – The faculty will

allocate chapters/ parts of chapters to groups of students so that the entire syllabus to be covered. The

power-point slides should be put up on the web-site of the College/ Institute, along with the names of

the students of the group, the name of the faculty, Department and College on the first slide. The best

three works should submit to GTU.


http://nptel.ac.in/syllabus/syllabus.php?subjectId=111101004





S-15 W-15 S-16 W-16

Conditional Probability 3 2 1+1 3+7

Theory 3

Bay's Theorem 3 7 4 7

Mathematical Expectation 3

Binomial 7 2+7 1 1+4

poission 4 7 3 3

Normal 7+7

Mean,Median,Mode 4 7 1+4 1

Standard Deviation 4 1+1

Correlation 3 1+1 4

Karl Pearson Product 1

Rank correlation 4

Regression Line 4+3 7+7 1+7 1+7

Straight Line 4

Parabola 7 4

Exponential 4 4

Relations & Theory 3 2 3+3 1+3

Newton Forward/Backwad 4 7 3

Newton's Divided Difference Formula 4 1 7

Langrange Interpolation Formula 4 7 7 1+4

Theory 1+1

Trapezoidal Formula 7 3

Simposon's 1/3 Formula 4 4

Simposon's 3/8 Formula 4 4 1

Weddel's Formula

Gaussian Integration 2

G.E. with Partial Pivoting 7 7 7+1 1+7+1

Gauss Seidel Method 7 7 7

Bisection Method 3 7

NR Method 3+7 3 3+1 3+1

Secant Method 3+1 3

Regula Falsi Method 7 7 3 3

Power Method 1

Picard's Method 1 4

Taylor's Series 3 3 4+4 7

Eular's Method 3 7 1 4

RK 4th Method 7 7 7 7

Predictor corrector 7

Paper Analysis of N. S. M. (2140606)

7

1

8

9

10

Unit no. Topic nameYear & Marks

23

45

6

Sta

tist

ics

Nu

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l M

eth

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1. Basic Concept Of Probability PAGE | 1

N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

Introduction

The words PROBABLE and POSSIBLE CHANCES are quite familiar to us.

We use these words when we are sure of the result of certain events. These words convey the sense of uncertainty of occurrence of events.

PROBABILITY is the word we use to calculate the degree of the certainty of events.

There are two types of approaches in the theory of PROBABILITY.

(1) Classical Approach – By Blaise Pascal

(2) Axiomatic Approach – By A. Kolmogorov

An experiment, in which we know all the possible outcomes in advance but which of them will occur is known only after the experiment is performed, is called a Random Variable.

Examples:

Event Outcomes

Toss Coin Head or Tail

Roll Die 1, 2, 3, 4, 5, 6

Pick a Card Ace,2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King (Clubs, Diamonds, Hearts, Spades)

Throw a stone It comes back surely. ( Why do we think about possibility )

Random Experiments

An experiment is called random experiment if it satisfies conditions:

(1) It has more than one possible outcome.

(2) It is not possible to predict the outcome in advance.

Sample Space

The set of outcomes is called the sample space of the experiment.

It is denoted by “ U ”.

Each element of the sample space is called a SAMPLE POINT.

If a sample space is in one-one correspondence with a finite set, {𝑥 ∈ ℕ | 1 ≤ 𝑥 ≤ 𝑛 , 𝑛 ∈ ℕ},

then it is called a finite sample space. Otherwise it is knowing as an infinite sample space.



Examples:

(1) Finite Sample Space: An experiment of tossing a coin twice.

U = {H, T} × {H, T} = {HH, HT, TH, TT}

(2) Infinite Sample Space: An experiment of tossing a coin until a HEAD comes up for the first time.

U = {H, TH, TTH, TTTH, TTTTH, TTTTTH, … }

Event

A subset of sample space is known as Event. Each member is called Sample Point.

Example:

Experiment U : Tossing a coin twice. U = {HH, HT, TH, TT}

Event A : getting TAIL both times. A = {TT}

Event B : getting TAIL exactly once. A = {HT, TH}

Definitions

1. The subset ∅ of a sample space is called “Impossible Events”. 2. The subset U(itself) of a sample space is called “Sure/Certain Events”. 3. If Subset contains only one element, it is called “Elementary/Sample Events”. 4. If Subset contains more than one element, it is called “Compound/Decomposable Events”. 5. The set contains all elements other than event A is known as “Complementary Event” of A.

It is known as A’. 6. A Union of Events A and B is Union of sets A and B (As per set theory). 7. An Intersection of Events A and B is Intersection of sets A and B (As per set theory). 8. If A ∩ B = ϕ, Events are called Mutually Exclusive Events (Disjoint set).

Set Notation: A ∩ B = { 𝑥 | 𝑥 ∈ A AND 𝑥 ∈ B }

9. If A ∪ B = U, Events are called Mutually Exhaustive Events.

Set Notation: A ∪ B = { 𝑥 | 𝑥 ∈ A OR 𝑥 ∈ B }

10. If A ∩ B = ϕ and A ∪ B = U, Events are called Mutually Exclusive & Exhaustive Events. 11. Suppose A1, A2, A3,…, An are Events.

If ,

1. Ai ∩ Aj = ϕ for i = 1,2, … , n ; j = 1,2, … , n ; i ≠ j.

2. ⋃ Ai

n

i=1

= U

Events A1, A2, A3,…, An are said to be Mutually Exclusive and Exhaustive Events. 12. Difference Events

A − B = A − (A ∩ B) = {x |x ∈ U, x ∈ A AND x ∉ B }

B − A = B − (A ∩ B) = {x |x ∈ U, x ∈ B AND x ∉ A }

13. Symmetric Difference Set A Δ B = (A − B) ∪ (B − A)



Exercise-1

C Que. 1 Describe the sample space for the indicated random experiments.

1. A coin is tossed 3 times. 2. A coin and die is tossed simultaneously.

Ans. 1. 𝐒 = {𝐇𝐇𝐇, 𝐇𝐇𝐓, 𝐇𝐓𝐇, 𝐇𝐓𝐓, 𝐓𝐇𝐇, 𝐓𝐇𝐓, 𝐓𝐓𝐇, 𝐓𝐓𝐓} 2. 𝐒 = {𝐇𝟏, 𝐇𝟐, 𝐇𝟑, 𝐇𝟒, 𝐇𝟓, 𝐇𝟔, 𝐓𝟏, 𝐓𝟐, 𝐓𝟑, 𝐓𝟒, 𝐓𝟓, 𝐓𝟔}

H Que. 2 A balanced coin is tossed thrice. If three tails are obtained, a balance die is rolled. Otherwise the experiment is terminated. Write down the elements of the sample space.

Ans. 𝐒 = { 𝐇𝐇𝐇, 𝐇𝐇𝐓, 𝐇𝐓𝐇, 𝐇𝐓𝐓, 𝐓𝐇𝐇, 𝐓𝐇𝐓, 𝐓𝐓𝐇,

𝐓𝐓𝐓𝟏, 𝐓𝐓𝐓𝟐, 𝐓𝐓𝐓𝟑, 𝐓𝐓𝐓𝟒, 𝐓𝐓𝐓𝟓, 𝐓𝐓𝐓𝟔}

C Que. 3 Let a coin be tossed. If it shows head we draw a ball from a box containing 3 identical red and 4 identical green balls and if it shows a tail, we throw a die. What is the sample space of experiments?

Ans. 𝐒 = {𝐇𝐑𝟏, 𝐇𝐑𝟐, 𝐇𝐑𝟑, 𝐇𝐆𝟏, 𝐇𝐆𝟐, 𝐇𝐆𝟑, 𝐇𝐆𝟒, 𝐓𝟏, 𝐓𝟐, 𝐓𝟑, 𝐓𝟒, 𝐓𝟓, 𝐓𝟔}

H Que. 4 An experiment consists of rolling a die and then tossing a coin once if the number on the die is odd. If the number on the die is even the coin is tossed twice. Write sample space.

Ans. 𝐒 = 𝐒𝐎 ∪ 𝐒𝐄 Where, 𝐒𝐎 = {𝟏𝑯, 𝟑𝑯, 𝟓𝑯, 𝟏𝑻, 𝟑𝑻, 𝟓𝑻} 𝐒𝐄 = {𝟐𝐇𝐇, 𝟐𝐇𝐓, 𝟐𝐓𝐇, 𝟐𝐓𝐓, 𝟒𝐇𝐇, 𝟒𝐇𝐓, 𝟒𝐓𝐇, 𝟒𝐓𝐓, 𝟔𝐇𝐇, 𝟔𝐇𝐓, 𝟔𝐓𝐇, 𝟔𝐓𝐓}

C Que. 5 Four card are labeled with A, B, C and D. We select and two cards at random without replacement. Describe the sample space for the experiments.

Ans. 𝐒 = {𝐀𝐁, 𝐀𝐂, 𝐀𝐃, 𝐁𝐂, 𝐁𝐃, 𝐂𝐃}

C Que. 6

A coin is tossed 3 times. Give the elements of the following events: 1. Event A: Getting at least two heads 2. Event B: Getting exactly two tails 3. Event C: Getting at most one tail 4. Event D: Getting at least one tail

Find A ∩ B , C ∩ D′, A ∪ C , B ∩ C , A′ ∪ C′.

Ans.

𝐒 = {𝐇𝐇𝐇, 𝐇𝐇𝐓, 𝐇𝐓𝐇, 𝐇𝐓𝐓, 𝐓𝐇𝐇, 𝐓𝐇𝐓, 𝐓𝐓𝐇, 𝐓𝐓𝐓} EVENT 𝐀 = {𝐇𝐇𝐇, 𝐇𝐇𝐓, 𝐇𝐓𝐇, 𝐓𝐇𝐇} EVENT 𝐁 = {𝐇𝐓𝐓, 𝐓𝐇𝐓, 𝐓𝐓𝐇} EVENT 𝐂 = {𝐇𝐇𝐇, 𝐇𝐇𝐓, 𝐇𝐓𝐇, 𝐓𝐇𝐇} EVENT 𝐃 = {𝐇𝐇𝐓, 𝐇𝐓𝐇, 𝐇𝐓𝐓, 𝐓𝐇𝐇, 𝐓𝐇𝐓, 𝐓𝐓𝐇, 𝐓𝐓𝐓}



H Que. 7

Describe the sample space associated with the experiment of selecting a child at random from three families each with a boy and a girl. Also write the elements of the following events:

1. There is at most one boy in the selection. 2. The selection consists of only girls. 3. The selection has exactly two girls.

Ans.

𝐒 = {𝐆𝐆𝐆, 𝐆𝐆𝐁, 𝐆𝐁𝐆, 𝐆𝐁𝐁, 𝐁𝐆𝐆, 𝐁𝐆𝐁, 𝐁𝐁𝐆, 𝐁𝐁𝐁} EVENT 𝐀 = {𝐆𝐆𝐆, 𝐁𝐆𝐆, 𝐆𝐁𝐆, 𝐆𝐆𝐁} EVENT 𝐁 = {𝐆𝐁𝐁, 𝐁𝐆𝐁} EVENT 𝐂 = {𝐆𝐆𝐁, 𝐆𝐁𝐆, 𝐁𝐆𝐆}

C Que. 8

An integer from 1 to 50 is selected at random. Write the elements of the following events. Event A: Randomly selected integer is a multiple of 2. Event B: Randomly selected integer is a multiple of 10. Event C: Randomly selected integer is a multiple of 4.

Ans.

𝐒 = {𝟏, 𝟐, 𝟑, 𝟒, … , 𝟓𝟎} EVENT 𝐀 = {𝟐, 𝟒, 𝟔, … , 𝟓𝟎} EVENT 𝐁 = {𝟏𝟎, 𝟐𝟎, 𝟑𝟎, 𝟒𝟎, 𝟓𝟎} EVENT 𝐂 = {𝟒, 𝟖, 𝟏𝟐, 𝟏𝟔, … , 𝟒𝟖}

C Que. 9

There are three identical balls, marked with a, b, c in a box. One ball is picked up from the box at random. The letter on it is noted and the ball is put back in the box, then another ball is picked up from the box and the letter on it is noted. Write the sample space of the experiment. Write down the elements of the following events: Event A: Ball marked “ a ” is selected exactly once. Event B: Balls selected have same letters marked. Event C: Ball with mark “ c ” is selected at least once.

Ans.

𝐒 = {𝐚𝐚, 𝐚𝐛, 𝐚𝐜, 𝐛𝐚, 𝐛𝐛, 𝐛𝐜, 𝐜𝐚, 𝐜𝐛, 𝐜𝐜} EVENT 𝐀 = {𝐚𝐛, 𝐚𝐜, 𝐛𝐚, 𝐜𝐚} EVENT 𝐁 = {𝐚𝐚, 𝐛𝐛, 𝐜𝐜} EVENT 𝐂 = {𝐚𝐜, 𝐛𝐜, 𝐜𝐚, 𝐜𝐛, 𝐜𝐜}

Description of different events in words and set notation are given in the following table.

No. Description of Events in Words Set Notation of Event

1. A is an event. A

2. Event does not occur. A’

3. Event B is surely occurs when A occurs. B ⊂ A

4. Impossible Events. ϕ

5. Certain Events. U (sample space)

6. Events A and B are Mutually exclusive A ∩ B = ϕ

7. Events A and B are Exhaustive A ∪ B = U



8. Among A and B events, Only B occurs B − A or B ∩ A′

9. Among A and B events, Only one event occurs (A − B) ∪ (B − A)

10. Events A and B occur together A ∩ B

11. At least one of the events A and B occurs A ∪ B

12. Among the events A,B,C Only A occurs A ∩ B′ ∩ C′

13. Among the events A,B,C Only A and B occurs A ∩ B ∩ C′

Definition: Probability Function

Let U be a finite sample space and S be its power set. Suppose that a set function P: S ⟶ R satisfies following axioms.

Axiom 1: For every A ∈ S, P(A) ≥ 0

Axiom 2: P(U) = 1

Axiom 3: ∀A1 ∈ S, A2 ∈ S, ifA1 ∩ A2 = ϕ, then P(A1 ∪ A2) = P(A1) + P(A2).

Then P is called a Probability function. For event A ∈ S, P(A) is called the probability of

the event A . The triplet (U, S, P) is called probability space.

Meaning of Definition

Axiom 1: Probability of any event is non-negative real number.

Axiom 2: Probability of certain event U is 1.At most value of probability of any event is 1.

Axiom 3: If A1, A2, A3,…, An ∈ S are mutually exclusive (n ≥ 2),

P(A1 ∪ A2 ∪ A3 ∪ … ∪ An) = P(A1) + P(A2) + P(A3) + ⋯ + P(An)

Definition: Equiprobable Events

Let U = {x1, x2, . . , xn} be a finite sample space. If P({x1}) = P({x2}) = P({x3}) = ⋯ = P({xn}), then the elementary events {x1}, {x2}, {x3}, … , {xn} are called Equiprobable Events.

Definition: Probability of an Event

If a finite sample space associated with a random experiments has "n" equally likely (Equiprobable) outcomes (elements) and of these "r“ (0 ≤ r ≤ n) outcomes are favorable for the occurrence of an event A, then probability of A is defined as follow.

𝐏(𝐀) =𝐟𝐚𝐯𝐨𝐫𝐚𝐛𝐥𝐞 𝐨𝐮𝐭𝐜𝐨𝐦𝐞𝐬

𝐭𝐨𝐭𝐚𝐥 𝐨𝐮𝐭𝐜𝐨𝐦𝐞𝐬=

𝐫

𝐧



Theorems

Thm. 1. For the Impossible Event , P(𝜙) = 0.

Thm. 2. Complementation Rule: For every Event A, P(A′) = 1 − P(A).

Thm. 3. If A ⊂ B, than P(B − A) = P(B) − P(A) and P(A) ≤ P(B)

Corollary 1. For every event A, 0 ≤ P(A) ≤ 1.

Corollary 2. For any events A and B,

P(A ∩ B′) = P(A) − P(A ∩ B)

P(A′ ∩ B′) = P(A ∪ B)′ = 1 − P(A ∪ B) (De Morgan’s Rule)

P(A′ ∪ B′) = P(A ∩ B)′ = 1 − P(A ∩ B) (De Morgan’s Rule)

Thm. 4. Addition Rule for Arbitrary Events

Let S be sample space and A, B and C be any events in S, then

P(A ∪ B) = P(A) + P(B) − P(A ∩ B) P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(B ∩ C) − P(C ∩ A) + P(A ∩ B ∩ C)

Exercise - 2

C Que. 1 If A and B are two mutually exclusive events with P(A) = 0.30 , P(B) = 0.45. Find the probability of A′ , A ∩ B , A ∪ B , A′ ∩ B′ .

[𝟎. 𝟕 , 𝟎 , 𝟎. 𝟕𝟓, 𝟎. 𝟐𝟓]

C Que. 2 If probability of an event A id

9

10, what is the probability of the

event “ not A ”. [𝟎. 𝟏]

H Que.3

A fair coin is tossed twice. Find the probability of (1) getting H exactly once. (2) getting T at least once.

[𝟏

𝟐,𝟑

𝟒]

T Que. 4

One card is drawn at random from a well shuffled pack of 52 cards. Calculate the probability that the card will be

(1) An Ace (2) A card of black colour (2) A diamond (4) Not an ace

[𝟏

𝟏𝟑,𝟏

𝟐 ,

𝟏

𝟒 ,

𝟏𝟐

𝟏𝟑]

C Que. 5

Four cards are drawn from the pack of cards. Find the probability that (i) all are diamonds (ii) there is one card of each suit (iii) there are two spades and two hearts.

[𝟎. 𝟎𝟎𝟐𝟔, 𝟎. 𝟏𝟎𝟓𝟓, 𝟎. 𝟎𝟐𝟐𝟓]

Nov-16

C Que. 6

A box contains 5 red, 6 white and 2 black balls. The balls are identical in all respect other than colour.

1. One ball is drawn at random from the box. Find the probability that the selected ball is black.

2. Two balls are drawn at random from the box. Find the probability that one ball is white and one is red.

[𝟐

𝟏𝟑 ,

𝟓

𝟏𝟑]



Conditional Probability

Let S be a sample space and A and B be any two events in S. Then the probability of the

occurrence of event A when it is given that B has already occurred is expressed by the

symbol P(A B⁄ ) and is defined as

P(A B⁄ ) =P(A ∩ B)

P(B); P(B) > 0,

Which is known as conditional probability of the event A relative to eventB.

Similarly, the conditional probability of the event B relative to eventA is

P(B A⁄ ) =P(B ∩ A)

P(A); P(A) > 0.

Properties of Conditional Probability

Let A1, A2 and B be any three events of a sample space S,then

P(A1 ∪ A2 B⁄ ) = P(A1 B⁄ ) + P(A2 B⁄ ) − P(A1 ∩ A2 B⁄ ); P(B) > 0.

Let A and B be any two events of a sample space S, then

P(A′ B⁄ ) = 1 − P(A B⁄ ); P(B) > 0.

Theorem (Multiplication Rule)

Let S be a sample space and A and B be any two events in S, then

P(A ∩ B) = P(A) ⋅ P(B A⁄ ); P(A) > 0.

Also, P(A ∩ B) = P(B) ⋅ P(A B⁄ ); P(B) > 0.

Corollary

Let S be a sample space and A, B and C be three events in S,then

P(A ∩ B ∩ C) = P(A) ⋅ P(B A⁄ ) ⋅ P(C A ∩ B⁄ ).

Theorem (Total Probability)

If B1 and B2 are two mutually exclusive and exhaustive events and P(B1) , P(B2) ≠ 0 , then for any event A.

P(A) = P(B1) ⋅ P(A B1⁄ ) + P(B2) ⋅ P(A B2)⁄

Corollary

If B1, B2 and B3 are mutually exclusive and exhaustive events and P(B1) , P(B2), P(B3) ≠ 0 , then for any event A.

P(A) = P(B1) ⋅ P(A B1⁄ ) + P(B2) ⋅ P(A B2)⁄ + +P(B3) ⋅ P(A B3)⁄



Bayes’ Theorem

Let B1, B2, B3 … , Bn be an n-mutually exclusive and exhaustive events of a sample space S and let A be any event such that P(A) ≠ 0, then

P(Bi A⁄ ) =P(Bi) ⋅ P(A Bi)⁄

P(B1) ⋅ P(A B1⁄ ) + P(B2) ⋅ P(A B2)⁄ + P(B3) ⋅ P(A B3)⁄ + ⋯ + P(Bn) ⋅ P(A Bn)⁄

Independent events

Let A and B be any two events of a sample space S, then A and B are called independent events if P(A ∩ B) = P(A) ⋅ P(B) . It also mean that, P(A B⁄ ) = P(A) and P(B A⁄ ) = P(B). This means that the probability of A does not depend on the occurrence or nonoccurrence of B, and conversely.

Remarks:

Let A , B and C are said to be Mutually independent, if

P(A ∩ B) = P(A) ⋅ P(B) , P(B ∩ C) = P(B) ⋅ P(C) P(C ∩ A) = P(C) ⋅ P(A) , P(A ∩ B ∩ C) = P(A) ⋅ P(B) ⋅ P(C)

Let A , B and C are said to be Mutually independent, if

P(A ∩ B) = P(A) ⋅ P(B) P(B ∩ C) = P(B) ⋅ P(C) P(C ∩ A) = P(C) ⋅ P(A)

Exercise – 3

C Que. 1

For two independent events A and B if P(A) = 0.3 and P(A ∪ B) = 0.6, find P(B).

[ 𝟑

𝟕 ]

H Que. 2

If A and B are independent events, where P(A) =1

4, P(B) =

2

3.

Find P(A ∪ B).

[ 𝟑

𝟒 ]

H Que. 3 If P(A) =

1

3, P(B) =

3

4 and P(A ∪ B) =

11

12 find P(A B⁄ ).

[ 𝟐

𝟗 ]

H Que. 4

If A and B are independent events, with P(A) =3

8, P(B) =

7

8.

Find P(A ∪ B) Find P(A B⁄ )and P(B A⁄ ).

[𝟓𝟗

𝟔𝟒,𝟑

𝟖 ,

𝟕

𝟖 ]

C Que. 5

If P(A) =1

3, P(B′) =

1

4 and P(A ∩ B) =

1

6 , then find P(A ∪ B),

P(A′ ∩ B′) and P(A′ B′⁄ ).

[𝟏𝟏

𝟏𝟐,

𝟏

𝟏𝟐,𝟏

𝟑]



C Que. 6

In a group of 200 students 40 are taking English,50 are taking mathematics,and 12 are taking both (a) If a student is selected at random, what is the probability that the student is taking English? (b) A student is selected at random from those taking mathematics. What is the probability that the student is taking English? (c) A student is selected at random from those taking English, what is the probability that the student is taking mathematics?

[𝟎. 𝟐𝟎 , 𝟎. 𝟐𝟒 , 𝟎. 𝟑]

H Que. 7

In a box, 100 bulbs are supplied out of which 10 bulbs have defects of type A, 5 bulbs have defects of type B and 2 bulbs have defects of both the type. Find the probability that

(a) a bulb to be drawn at random has a B type defect under the condition that it has an A type defect, and

(b) a bulb to be drawn at random has no B type defect under the condition that it has no A type defect.

[𝟎. 𝟐] , [𝟎. 𝟗𝟔𝟔𝟕]

C Que. 8

A person is known to hit the target in 3 out of 4 shots, whereas another person is known to hit the target in 2 out of 3 shots. Find the probability of the target being hit at all when they both try. What is the probability that the target will be hit?

[𝟔

𝟏𝟐,𝟏𝟏

𝟏𝟐]

May -15

Dec-15

C Que. 9

An urn contains 10 white and 3 black balls, while another urn contains 3 white and 5 black balls. Two balls are drawn from the first urn and put into the second urn and then a ball is drawn from the later. What is the probability that it is a white ball?

[𝟓𝟗

𝟏𝟑𝟎]

Nov-16

H Que. 10

There are two boxes A and B containing 4 white, 3 red and 3 white,7 red balls respectively. A box is chosen at random and a ball is drawn from it, if the ball is white, find the probability that it is from box A.

[𝟒𝟎

𝟔𝟏]

C Que. 11

State Bayes’ theorem. A microchip company has two machines that produce the chips. Machine-I produces 65% of the chips, but 5% of its chips are defective. Machine-II produces 35% of the chips, but 15% of its chips are defective. A chip is selected at random and found to be defective. What is the probability that it came from Machine-I? [𝟎. 𝟑𝟖𝟐𝟒]

Nov-16

H Que. 12

State Bayes’ theorem. In a bolt factory, three machines A, B and C manufacture 25%, 35% and 40% of the total product respectively. Out Of these outputs 5%, 4% and 2% respectively, are defective bolts. A bolt is picked up at random and found to be defective. What are the Probabilities that it was manufactured by machine A, B and C? [𝟎. 𝟑𝟔𝟐𝟑, 𝟎. 𝟒𝟎𝟓𝟕, 𝟎. 𝟐𝟑𝟏𝟖]

Dec-15



H Que. 13

A company has two plants to manufacture hydraulic machine. Plat I manufacture 70% of the hydraulic machines and plant II manufactures 30%. At plant I, 80% of hydraulic machines are rated standard quality and at plant II, 90% of hydraulic machine are rated standard quality. A machine is picked up at random and is found to be of standard quality. What is the chance that it has come from plant I? [𝟎. 𝟔𝟕𝟒𝟕]

May-15

C Que. 14

In a certain assembly plant, three machines, B1, B2 and B3, make 30%, 45% and 25%, respectively, of the products. It is known form the past experience that 2%, 3% and 2% of the products made by each machine respectively, are defective. Now, suppose that a finished product is randomly selected. What is the probability that it is defective? [𝟎. 𝟐𝟒𝟓]

May-16

C Que. 15

If proposed medical screening on a population, the probability that the test correctly identifies someone with illness as positive is 0.99 and the probability that test correctly identifies someone without illness as negative is 0.95. The incident of illness in general population is 0.0001. You take the test, the result is positive then what is the probability that you have the illness ?

[𝟎. 𝟎𝟎𝟏𝟗]

2. Probability Distribution PAGE | 11


Random Variable

An experiment, in which we know all the possible outcomes in advance but which of

them will occur is known only after the experiment is performed, is called a Random

Variable.

Types of Random Variables 1. Discrete random variable

2. Continuous random variable

Discrete Random Variable

A random variable, which can take only finite, countable, or isolated values in a

given interval, is called discrete random variable.

i.e. A random variable is one which can assume any of a set of possible values which

can be counted or listed.

For example, the numbers of heads in tossing coins, the number of auto passengers

can take on only the values 1 , 2 , 3 and so on.

Note: Discrete random variables can be measured exactly.

Continuous Random Variable

A random variable, which can take all possible values that are infinite in a given

interval is called Continuous random variable.

i.e. a continuous random variable is one which can assume any of infinite spectrum

of different values across an interval which can not be counted or listed

For example, measuring the height of a student selected at random, finding the

average life of a brand X tyre etc.

Note: Continuous random variables can not be measured exactly.

Probability Function

If for random variable X, the real valued function f(x) is such that P(X = 𝑥) = 𝑓(𝑥),

then 𝑓(𝑥) is called Probability function of random variable X.

Probability function 𝑓(𝑥) gives the measures of probability for different values of X

say 𝑥1, 𝑥2, … . , 𝑥𝑛 .

Sometimes Probability density function is denoted by p(𝑥).



Probability Density Function

If X is a continuous random variable then its probability function 𝑓(𝑥) is called continuous probability function OR probability density function. It is defined as below

P(𝑎 < 𝑥 < 𝑏) = ∫ 𝑓(𝑥)𝑑𝑥𝑏

𝑎

Note:

1. 𝑓(𝑥) ≥ 0

2. ∫ 𝑓(𝑥)𝑑𝑥∞

−∞= 1

Probability Mass Function If X is a discrete random variable then its probability function 𝑓(𝑥) is discrete probability function. It is also called probability mass function, then P(X = 𝑥𝑖) = f(𝑥𝑖).

Note:

1. 𝑓(𝑥) ≥ 0 2. ∑ 𝑓(𝑥𝑖)

𝑛𝑖=1 = 1

Properties of Probability function If 𝑓(𝑥) is a probability function of random variable X then it possess the following Properties:

1. (𝑥) is positive for any value of x i.e. 𝑓(𝑥) ≥ 0 for all 𝑥. 2. Total of all values of 𝑓(𝑥) for different x is always 1. i.e. ∑ 𝑓(𝑥) = 1,

summation is taken for all values of 𝑥. 3. 𝑓(𝑥) measures the probability for any given value of 𝑥. 4. 𝑓(𝑥) cannot be negative for any value of 𝑥.

Mathematical Expectation If X is a discrete random variable having various possible values 𝑥1 , 𝑥2 , … . , 𝑥𝑛 and if 𝑓(𝑥) is the probability function, the mathematical Expectation or simply expectation of X is defined and denoted by E(X).

E(X) = ∑ 𝑥𝑖 ⋅ 𝑓(𝑥𝑖)

n

i=1

= ∑ 𝑥𝑖 ⋅ p(𝑥𝑖)

n

i=1

= ∑ 𝑥𝑖 ⋅ pi

n

i=1

Where, ∑ pi

n

i=1

= p1 + p2 + ⋯ pn = 1

If X is a continuous random variable having probability density function 𝑓(𝑥), expectation of X

is defined as E(X) = ∫ 𝑥 𝑓(𝑥) 𝑑𝑥∞

−∞

Note:

E(X) is also called the mean value of the probability distribution of x and is denoted by μ.



Properties of Mathematical Expectation 1. Expected value of constant term is constant. i.e. E(c) = c

2. If c is constant, then E(cX) = c ∙ E(X)

3. If a and b are constants, then E(aX ± b) = aE(X) ± b

4. If a , b and c are constants, then E (aX+b

c) =

1

c[aE(X) + b]

5. If X and Y are two random variables , then E(X + Y) = E(X) + E(Y)

6. If X and Y are two independent random variable, then E(X ∙ Y) = E(X) ∙ E(Y)

7. If 𝑔(𝑥) is any function of random variable X and 𝑓(𝑥) is probability density

function , then E{𝑔(𝑥)} = ∑ 𝑔(𝑥) ∙ 𝑓(𝑥)

Variance of a Random Variable

Variance is a characteristic of random variable X and it is used to measure dispersion

(or variation) of X.

If X is a discrete random variable with probability density function 𝑓(𝑥), then expected

value of [X − E(X)]2 is called the variance of X and it is denoted by V(X).

i.e. V(X) = E[X − E(X)]2 = E(X2) − [E(X)]2

Properties of Variance 1. V(c) = 0, Where c is a constant.

2. V(cX) = c2 V(X) , where c is a constant.

3. V(X + c) = V(X), Where c is a constant.

4. If a and b are constants , then V(aX + b) = a2V(X)

5. If X and Y are the independent random variables, then

V(X + Y) = V(X) + V(Y)

6. V(X) = E(X2) − μ2

Standard Deviation of Random Variable The positive square root of V(X) (Variance of X) is called standard deviation of random

variable X and is denoted by σ.

i.e. S.D. σ = √V(X).

Note : σ2 is called variance of V(X).



Exercise - 1

C Que.1

Which of the following functions are probability function?

(i) 𝑓(𝑥) = (1

2)

𝑥

(1

2)

1−𝑥

; 𝑥 = 0,1

(ii) 𝑓(𝑥) = (−1

2)

𝑥

, 𝑥 = 0,1,2

(iii) X −1 0 1

𝑓(𝑥) 0.5 0.8 −3 (iv)

X 1 2 3

𝑓(𝑥) 0.3 0.5 0.2 [YES, NO, NO , YES]

T Que.2

The probability function of a random variable X is

𝑝(𝑥) =2𝑥+1

48, 𝑥 = 1,2,3,4,5,6.

(a)Verify whether 𝑝(𝑥) is probability function ? (b)Also find E(X).

[𝐘𝐞𝐬, 𝟒. 𝟐𝟑]

H Que.3

Find the expected value of a random variable X having the following probability distribution.

X −5 −1 0 1 5 8

P(X = 𝑥) 0.12 0.16 0.28 0.22 0.12 0.1

[𝟎. 𝟖𝟔𝟎𝟎]

C Que.4

A random variable X has the following function.

X 0 1 2 3 4 5 6 7

P(X = 𝑥) 0 2k 3k k 2k k2 7k2 2k2 + k

Find the value of k and then evaluate (i)P(X < 6)(ii)P(X ≥ 6)(iii)P(0 < x < 5). [Hint: (i)P(X < 6) = P(0) + P(1) + P(2) + P(3) + P(4) + P(5) (ii)P(X ≥ 6) = 1 − P(X < 6) (iii) P(0 < X < 5) = P(1) + P(2) + P(3) + P(4) ]

[(𝐢)𝟎. 𝟖𝟏𝟎𝟎, (𝐢𝐢)𝟎. 𝟏𝟗𝟎𝟎, (𝐢𝐢𝐢)𝟎. 𝟖𝟎𝟎𝟎]

C Que.5

(a)A contestant tosses a coin and receives $5 if heads appears and $1 if tail appears. What is the expected value of a trial? (b)A contestant receives $4.00 if a coin turns up heads and pays $3.00 if

it turns tails. What is the expected value? [ Hint:P(H) = P(T) =1

2],

[(𝐚)$𝟑. 𝟎𝟎 , (𝐛)$𝟎. 𝟓𝟎]

C Que.6

A machine produces on average of 500 items during the first week of the month and average of 400 items during the last week of the month. The probability for these being 0.68 and 0.32. Determine the expected value of the production.

[𝟒𝟔𝟖]

May-15



H Que.7

In a business, the probability that a trader can get profit of Rs. 5000 is 0.4 and probability for loss of Rs. 2000 is 0.6. Find his expected gain or loss.

[𝟖𝟎𝟎]

C Que.8 Three balanced coins are tossed, then find the mathematical expectation of tails.

[𝟏. 𝟓]

C Que.9 There are 8 apples in a box, of which 2 are rotten. A person selects 3 Apples at random from it. Find the expected value of the rotten Apples.

[𝟎. 𝟕𝟓𝟎𝟎]

C Que.10

There are 3 red and 2 white balls in a box and 2 balls are taken at random from it. A person gets Rs. 20 for each red ball and Rs. 10 for each white ball. Find his expected gain.

[𝐄(𝐱) = 𝟑𝟐]

H Que.11

There are 10 bulbs in a box, out of which 4 are defectives. If 3 bulbs are taken at random, find the expected number of defective bulbs.

[𝟏. 𝟐𝟎]

C Que.12

The probability distribution of a random variable x is given below.

X −2 −1 0 1 2

P(x) 1

12

1

3 P

1

4

1

6

Find (i)p (ii)E(x) (iii) E(2x + 3) (iv)E(x2 + 2) (v) V(x)

[𝐩 =𝟏

𝟔, 𝐄(𝐱) =

𝟏

𝟏𝟐, 𝐄(𝟐𝐱 + 𝟑) =

𝟏𝟗

𝟔, 𝐄(𝐱𝟐 + 𝟐) =

𝟒𝟑

𝟏𝟐 , 𝐕(𝐱) =

𝟐𝟐𝟕

𝟏𝟒𝟒]

H Que.13

The probability distribution of a random variable 𝑥 is as follows. Find p, E(x).

X 0 1 2 3 4 5

p(𝑥) 𝑃 1

5

1

10 𝑃

1

20

1

20

[𝐩 = 𝟎. 𝟑𝟎, 𝐄(𝐱) = 𝟏. 𝟕𝟓𝟎𝟎]

C Que.14

For a random variable X, if E(2𝑥 − 10) = 20 and E(𝑥2) = 400. Find the standard deviation of x.

[𝐕(𝐱) = √𝟏𝟕𝟓]

C Que.15 If mean and standard deviation of a random variable x are 5 and 5 respectively. Find E(𝑥2) and E(2𝑥 + 5)2

[𝐄(𝐱𝟐) = 𝟓𝟎, 𝐄(𝟐𝐱 + 𝟓)𝟐 = 𝟑𝟐𝟓]



Binomial Probability Distribution

This distribution is associated with repeated trials of an experiment.

Bernoulli Trials Suppose a random experiment has two possible outcomes, which are complementary,

say success (S) and failure (F). If the probability 𝑝(0 < 𝑝 < 1) of getting success at each

of the 𝑛 trials of this experiment is constant, then the trials are called Bernoulli trials.

Binomial Distribution

𝑷(𝑿 = 𝒙) = 𝒑(𝒙) = 𝑪𝒙 𝒑𝒙 𝒒𝒏−𝒙𝒏 ; 𝒙 = 𝟎, 𝟏, 𝟐 … , 𝒏.

Where, 𝑪𝒙𝒏 =𝒏!

𝒙! (𝒏 − 𝒙)!

q = probability of failure in each trial=1 − 𝑝

Examples Of Binomial Distribution 1. Number of defective bolts in a box containing 𝑛 bolts.

2. Number of post-graduates in a group of 𝑛 people.

3. Number of oil wells yielding natural gas in a group of 𝑛 wells test drilled.

4. Number of machines lying idle in a factory having 𝑛 machines.

Properties Of Binomial Distribution The Binomial distribution holds under the following conditions.

1. The number of trials 𝑛 is finite.

2. There are only two possible outcomes, success or failure.

3. The trials are independent of each other.

4. The probability of success 𝑝 is constant for each trial.

Note

The mean and variance of binomial distribution with parameters 𝑛 and 𝑝 are

defined as follow,

𝐌𝐞𝐚𝐧 𝜇 = E(𝑋) = 𝑛𝑝 ; 𝐕𝐚𝐫𝐢𝐚𝐧𝐜𝐞 V(𝑋) = 𝜎2 = 𝑛𝑝𝑞

𝐒𝐭𝐚𝐧𝐝𝐚𝐫𝐝 𝐝𝐞𝐯𝐢𝐚𝐭𝐢𝐨𝐧 = 𝜎 = √𝑛𝑝𝑞

𝐶𝑥𝑛 = 𝐶𝑛−𝑥𝑛

𝐶0𝑛 = 𝐶𝑛𝑛 = 1

We use math expression " ≥ " for at least and " ≤ " for at most.



Exercise - 2

C Que 1. Find the binomial distribution for n = 4 and p = 0.3.

[𝐏(𝐗 = 𝟎) = 𝟎. 𝟐𝟒𝟎𝟏, 𝐏(𝐗 = 𝟏) = 𝟎. 𝟒𝟏𝟏𝟔, 𝐏(𝐗 = 𝟐) = 𝟎. 𝟐𝟔𝟒𝟔, 𝐏(𝐗 = 𝟑) = 𝟎. 𝟎𝟕𝟓𝟔, 𝐏(𝐗 = 𝟒) = 𝟎. 𝟎𝟎𝟖𝟏]

C Que 2. 20% of the bulbs produced are defective .Find probability that at most 2 bulbs out of 4 bulbs are defective.

[𝟎. 𝟗𝟕𝟐𝟖]

C Que 3.

12% of the tablets produced by a tablet machine are defective. What is the probability that out of a random sample of 20 tablets produced by the machine, 5 are defective?

[𝟎. 𝟎𝟓𝟔𝟕]

C Que 4.

The probability that India wins a cricket test match against Australia is

given to be 1

3. If India and Australia play 3 tests matches, what is the

probability that (I) India will lose all the three test matches? (II) India will win at least one test match?

[(𝐢) 𝟎. 𝟐𝟗𝟔𝟑, (𝐢𝐢) 𝟎. 𝟕𝟎𝟑𝟕]

H Que 5.

What are the properties of Binomial Distribution ? The average percentage of failure in a certain examination is 40. What is the probability that out of a group of 6 candidates, at least 4 passed in examination ? [𝟎. 𝟓𝟒𝟒𝟑]

Dec-15

H Que 6.

The probability that in a university, a student will be a post-graduate is 0.6. Determine the probability that out of 8 students

(i) None (ii) Two (iii) At least two will be post-graduate

[(𝐢)𝟎. 𝟎𝟎𝟎𝟕, (𝐢𝐢)𝟎. 𝟎𝟒𝟏𝟑, (𝐢𝐢𝐢)𝟎. 𝟗𝟗𝟏𝟒]

T Que 7.

The probability that an infection is cured by a particular antibiotic drug within 5 days is 0.75. Suppose 4 patients are treated by this antibiotic drug. What is the probability that (a) no patient is cured (b) exactly two patient are cured (c) At least two patients are cured.

[(𝐚) 𝟎. 𝟎𝟎𝟑𝟗, (𝐛)𝟎. 𝟐𝟏𝟎𝟗, (𝐜)𝟎. 𝟗𝟒𝟗𝟐]

C Que 8.

Assume that on the average one telephone number out of fifteen called between 1 p.m. and 2 p.m. on weekdays is busy. What is the probability that if 6 randomly selected telephone numbers were called (i) not more than three, (ii) at least three of them would be busy?

[𝟎. 𝟗𝟗𝟗𝟕] , [𝟎. 𝟎𝟎𝟓𝟏]

Nov-16

T Que 9.

A dice is thrown 6 times getting an odd number of success, Find probability

(a) Five success (b) At least five success (c) At most five success.

[(𝐚)𝟑

𝟑𝟐, (𝐛)

𝟕

𝟔𝟒, (𝐜)

𝟔𝟑

𝟔𝟒]



C Que 10.

A multiple choice test consist of 8 questions with 3 answer to each question (of which only one is correct). A student answers each question by rolling a balanced dice and checking the first answer if he gets 1 or 2, the second answer if he gets 3 or 4 and the third answer if he gets 5 or 6. To get a distinction, the student must secure at least 75% correct answers. If there is no negative marking, what is the probability that the student secures a distinction? [𝐏(𝐗 ≥ 𝟔) = 𝟎. 𝟎𝟏𝟗𝟕]

May-15

C Que 11. Obtain the binomial distribution for which mean is 10 and variance is 5.

[𝐏(𝐗 = 𝒙) = 𝐶𝑥20 (𝟎. 𝟓)𝐱(𝟎. 𝟓)𝟐𝟎−𝐱] Dec-15

H Que 12.

Determine binomial distribution whose mean is 4 and variance is 3 and hence evaluate P(X ≥ 2).

[𝐏(𝐗 = 𝐱) = { 𝑪𝒙𝟏𝟔 (𝟏

𝟒)

𝐱

(𝟑

𝟒)

𝟏𝟔−𝐱

, 𝐗 = 𝟎, 𝟏, … 𝟏𝟔

𝟎 , 𝐨𝐭𝐡𝐞𝐫𝐰𝐢𝐬𝐞

& 𝐏(𝐗 ≥ 𝟐) = 𝟎. 𝟗𝟑𝟔𝟓]

H Que 13.

For the binomial distribution with n = 20, p = 0.35. Find (a) Mean, (b) Variance and (c) Standard deviation.

[(𝐚)𝟕. 𝟎𝟎𝟎𝟎 (𝐛)𝟒. 𝟓𝟓, (𝐜)𝟐. 𝟏𝟑𝟑𝟏]

C Que 14. If the probability of a defective bolt is 0.1 Find mean and standard deviation of the distribution of defective bolts in a total of 400.

[𝛍 = 𝟒𝟎, 𝛔 = 𝟔]

Poisson Distribution

A discrete random variable 𝑋 is said to follow Poisson distribution if it assume only

nonnegative values and its probability mass function is given by

𝐏(𝐗) =𝒆−𝝀𝝀𝒙

𝒙!; 𝒙 = 𝟎, 𝟏, 𝟐, 𝟑, …

Examples Of Poisson Distribution 1. Number of defective bulbs produced by a reputed company.

2. Number of telephone calls per minute at a switchboard.

3. Number of cars passing a certain point in one minute.

4. Number of printing mistakes per page in a large text.

5. Number of persons born blind per year in a large city.

Properties Of Poisson Distribution The Poisson distribution holds under the following conditions.

1. The random variable 𝑥 should be discrete.

2. The number of trials 𝑛 is very large.

3. The probability of success 𝑝 is very small (very close to zero).

4. 𝜆 = 𝑛𝑝 Is finite.

5. The occurrences are rare.

Note

The mean and variance of the Poisson distribution with parameter 𝜆 are defined as

follows. 𝐌𝐞𝐚𝐧 𝛍 = 𝐄(𝐗) = 𝛌 = 𝐧 𝐩 ; 𝐕𝐚𝐫𝐢𝐚𝐧𝐜𝐞 𝐕(𝐗) = 𝛔𝟐 = 𝛌.



Exercise - 3

C Que 1.

In a company, there are 250 workers. The probability of a worker remain absent on any one day is 0.02. Find the probability that on a day seven workers are absent.

[𝐏(𝐗 = 𝟕) = 𝟎. 𝟏𝟎𝟒]

C Que 2.

A book contains 100 misprints distributed randomly throughout its 100 pages. What is the probability that a page observed at random contains at least two misprints. Assume Poisson Distribution.

[𝟎. 𝟐𝟔𝟒𝟐]

Dec-15

H Que 3. For Poisson variant X, if P(X = 3) = P(X = 4) then, Find P(X = 0).

[𝐏(𝐗 = 𝟎) = 𝐞−𝟒]

T Que 4.

In a bolt manufacturing company, it is found that there is a small chance

of 1

500 for any bolt to be defective. The bolts are supplied in a packed of 30

bolts. Use Poisson distribution to find approximate number of packs, (a) Containing no defective bolt. (b) Containing two defective bolt, in the consignment of 10000

packets. [(𝐚)𝟗𝟒𝟐𝟎 , (𝐛) 𝟏𝟕]

H Que 5.

In sampling a large number of parts manufactured by a machine, the mean number and of defectives in a sample of 20 is 2. Out of 1000 such samples, how many would be expected to contain exactly two defective parts?

[𝐏(𝐗 = 𝟐) = 𝟎. 𝟐𝟕𝟎]

May-15

C Que 6.

Potholes on a highway can be serious problems. The past experience suggests that there are, on the average, 2 potholes per mile after a certain amount of usage. It is assumed that the Poisson process applies to the random variable “no. of potholes”. What is the probability that no more than four potholes will occur in a given section of 5 miles?

[𝐏(𝐗 ≤ 𝟒) = 𝟎. 𝟎𝟑𝟏𝟓]

May-16

T Que 7.

A car hire firm has two cars, which it hires out day by day. The number of demands for a car on each day is distributed on a Poisson distribution with mean 1.5. Calculate the proportion of days on which neither car is used and proportion of days on which some demand is refused. (e−1.5 =0.2231).

[𝐏(𝐗 = 𝟎) = 𝟎. 𝟐𝟐𝟑𝟏 ; 𝟏 − 𝐏(𝐗 ≤ 𝟐) = 𝟎. 𝟏𝟗𝟏𝟐]

Nov-16

C Que 8.

100 Electric bulbs are found to be defective in a lot of 5000 bulbs. Find the probability that at the most 3 bulbs are defective in a box of 100 bulbs.

[𝐏(𝐗 ≤ 𝟑) = 𝟎. 𝟖𝟓𝟔𝟕]

H Que 9.

Certain mass produced articles of which 0.5% are defective ,are packed in cartons each containing 100. What proportion of cartons are free from defective articles, and what proportion contain 3 or more defectives?

[𝐏(𝐗 = 𝟎) = 𝟔𝟎. 𝟔𝟓, 𝐏(𝐗 ≥ 𝟑) = 𝟎. 𝟎𝟏𝟒𝟒]



T Que 10.

If a bank receives an average six back cheques per day what are the probability that bank will receive

(i) 4 back cheques on any given day (ii) 10 back cheques on any consecutive day.

[(𝐢)𝐏(𝐗 = 𝟒) = 𝟎. 𝟏𝟑𝟑𝟖 (𝐢𝐢)𝟐𝐏(𝐗 = 𝟏𝟎) = 𝟎. 𝟎𝟖𝟐𝟔]

H Que 11.

Suppose 1% of the items made by machine are defective. In a sample of 100 items find the probability that the sample contains

(i) All good (ii) 1 defective (iii) At least 3 defective

[𝐏(𝐗 = 𝟎) = 𝟎. 𝟑𝟔𝟕𝟗𝐏[𝐗 = 𝟏] = 𝟎. 𝟑𝟔𝟕𝟗

𝐏(𝐱 ≥ 𝟑) = 𝟎. 𝟎𝟖𝟎𝟑

]



Normal Distribution (or Gauss Distribution)

A continuous random variable 𝑿 is said to follow a normal distribution of its probability

density function is given by

𝒇(𝒙) =𝟏

𝝈√𝟐𝝅 𝒆𝐱𝐩 [−

𝟏

𝟐(

𝐱 − 𝛍

𝛔)

𝟐

] ; −∞ < 𝐱 < ∞, 𝛔 > 𝟎

μ = mean of the distribution

σ = Standard deviation of the distribution

Note: μ and σ2 (variance) are called parameters of the distribution.

If X is a normal random variable with mean 𝛍 and standard deviation 𝛔, and if we find

the random variable 𝐙 =𝐗−𝛍

𝛔 with mean 0 and standard deviation 1, then Z in called the

standard (standardized) normal variable.

The probability destiny function for the normal distribution in standard form is given by

𝒇(𝒛) =𝟏

√𝟐𝝅𝒆−

𝟏𝟐

𝒛𝟐

; −∞ < 𝒛 < ∞

For normal distribution

1. p(−∞ ≤ z ≤ ∞) = 1 (Total area)

2. p(−∞ ≤ z ≤ 0) = p (0 ≤ z ≤ ∞) = 0.5

3. p(−z1 ≤ z ≤ 0) = p(0 ≤ z ≤ z1) (∵ Symmetry)

Exercise - 4

C Que 1.

Compute the value of following: I. P(0 ≤ z ≤ 1.43)

II. p(−0.73 ≤ z ≤ 0) III. p(−1.37 ≤ z ≤ 2.02) IV. p(0.65 ≤ z ≤ 1.26) V. p(z ≥ 1.33)

VI. p(|z| ≤ 0.5)

[[𝟎. 𝟒𝟐𝟑𝟔], [𝟎. 𝟐𝟔𝟕𝟑], [𝟎. 𝟖𝟗𝟑𝟎], [𝟎. 𝟏𝟓𝟒], [𝟎. 𝟎𝟗𝟏𝟖], [𝟎. 𝟑𝟖𝟑]]

T Que 2.

The compressive strength of the sample of cement can be modelled by a normal distribution With a mean 6000 kg/cm2 and standard deviation of 100 kg/cm2. (i) What is the probability that a sample strength is less than 6250 kg/cm2 ? (ii)What is the probability if sample strength is between 5800 and 5900 kg/cm2 ?(iii)What strength is exceeded by 95% of the samples? [𝑷(𝒛 = 𝟐. 𝟓) = 𝟎. 𝟗𝟗𝟑𝟖, 𝑷(𝒛 = 𝟏) = 𝟎. 𝟖𝟒𝟏𝟑, 𝑷(𝒛 = 𝟐) =𝟎. 𝟗𝟕𝟕𝟐, 𝑷(𝒛 = 𝟏. 𝟔𝟓) = 𝟎. 𝟗𝟓]

[(𝟏)𝟎. 𝟗𝟗𝟑𝟖 , (𝟐)𝟎. 𝟏𝟖𝟏𝟓 , (𝟑) 𝟏. 𝟔𝟓 , 𝒙 = 𝟔𝟏𝟔𝟓]

May-16



C Que 3.

In a photographic process, the developing time of prints may be looked upon as a random variable having the normal distribution with a mean of 16.28 seconds and a standard deviation of 0.12 seconds. Find the probability that it will take (i)Anywhere from 16.00 to 16.50 sec to develop one of the prints; (ii)At least 16.20 sec to develop one of the prints; (iii)At most 16.35 sec to develop one of the prints. [𝑷(𝒛 = 𝟏. 𝟖𝟑) = 𝟎. 𝟗𝟔𝟔𝟒, 𝑷(𝒛 = 𝟎. 𝟔𝟔) = 𝟎. 𝟕𝟒𝟓𝟒, 𝑷(𝒛 = 𝟎. 𝟓𝟖) =𝟎. 𝟕𝟏𝟗𝟎]

[𝑷(𝟏𝟔 < 𝑿 < 𝟏𝟔. 𝟓) = 𝟎. 𝟗𝟓𝟔𝟓

𝑷[𝑿 ≥ 𝟏𝟔. 𝟐𝟎] = 𝟎. 𝟕𝟒𝟕𝟓𝑷(𝑿 ≤ 𝟏𝟔. 𝟑𝟓) = 𝟎. 𝟕𝟏𝟗𝟎

]

May-16

C Que 4.

A sample of 100 dry battery cell tested and found that average life of 12 hours and standard deviation 3 hours. Assuming the data to be normally distributed what percentage of battery cells are expected to have life

(i) More than 15 hour (ii) Less than 6 hour (iii) Between 10 & 14 hours

[𝟏𝟓. 𝟖𝟕%, 𝟐. 𝟐𝟖%, 𝟒𝟗. 𝟕𝟐%]

T Que 5.

In AEC company, the amount of light bills follows normal distribution with standard deviation 60. 11.31% of customers pay light-bill less than Rs.260. Find average amount of light bill.

[𝛍 = 𝟑𝟑𝟐. 𝟔𝟎]

C Que 6.

In examination, minimum 40 marks for passing and 75 marks for distribution are required. In this examination 45% student passed and 9% obtained distribution. Find average Marks and standard deviation of this distribution of marks.

[𝛍 = 𝟑𝟔. 𝟒, 𝛔 = 𝟐𝟖. 𝟖]

T Que 7.

Weights of 500 students of college is normally distributed with average weight 95 lbs & σ =7.5 Find how many students will have the weight between 100 and 110.

[𝟏𝟏𝟒]

C Que 8.

Distribution of height of 1000 soldiers is normal with Mean 165cms and standard deviation 15cms how many soldiers are of height

(i) Less than 138cms (ii) More than 198cms (iii) Between 138 and 198cms

[(𝐢) = 𝟑𝟔 , (𝐢𝐢) = 𝟏𝟒 , (𝐢𝐢𝐢) = 𝟗𝟓𝟎 ]

H Que 9. In a normal distribution 31% of the items are below 45 and 8% are above 64. Determine the mean and standard deviation of this distribution

[𝛍 = 𝟒𝟗. 𝟗𝟕𝟒, 𝛔 = 𝟗. 𝟗𝟓]

T Que 10.

The breaking strength of cotton fabric is normally distributed with E(x) =16 and σ(x) = 1. The fabric is said to be good if x ≥ 14 what is the probability that a fabric chosen at random is good ?

[𝟎. 𝟗𝟕𝟕𝟐]



Standard Normal ( Z ) Table , Area between 0 and Z

𝑧 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.0359

0.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.0753

0.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.1141

0.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.1517

0.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.1879

0.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.2224

0.6 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.2549

0.7 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.2852

0.8 0.2881 0.2910 0.2939 0.2967 0.2995 0.3023 0.3051 0.3078 0.3106 0.3133

0.9 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.3389

1.0 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.3621

1.1 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.3830

1.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3990 0.3997 0.4015

1.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4115 0.4131 0.4147 0.4162

1.4 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.4319

1.5 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 0.4429 0.4441

1.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.4545

1.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.4633

1.8 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.4706

1.9 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.4767

2.0 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.4817

2.1 0.4821 0.4826 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 0.4854 0.4857

2.2 0.4861 0.4864 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884 0.4887 0.4890

2.3 0.4893 0.4896 0.4898 0.4901 0.4904 0.4906 0.4909 0.4911 0.4913 0.4916

2.4 0.4918 0.4920 0.4922 0.4925 0.4927 0.4929 0.4931 0.4932 0.4934 0.4936

2.5 0.4938 0.4940 0.4941 0.4943 0.4945 0.4946 0.4948 0.4949 0.4951 0.4952

2.6 0.4953 0.4955 0.4956 0.4957 0.4959 0.4960 0.4961 0.4962 0.4963 0.4964

2.7 0.4965 0.4966 0.4967 0.4968 0.4969 0.4970 0.4971 0.4972 0.4973 0.4974

2.8 0.4974 0.4975 0.4976 0.4977 0.4977 0.4978 0.4979 0.4979 0.4980 0.4981

2.9 0.4981 0.4982 0.4982 0.4983 0.4984 0.4984 0.4985 0.4985 0.4986 0.4986

3.0 0.4987 0.4987 0.4987 0.4988 0.4988 0.4989 0.4989 0.4989 0.4990 0.4990

3. Descriptive Statistics PAGE | 25


Introduction

Descriptive statistics are used to describe the basic features of the data in a study. They provide simple summaries about the sample and the measures. Together with simple graphics analysis, they form the basis of virtually every quantitative analysis of data.

Descriptive Statistics are used to present quantitative descriptions in a manageable form. In a research study we may have lots of measures. Or we may measure a large number of people on any measure. Descriptive statistics help us to simplify large amounts of data in a sensible way. Each descriptive statistic reduces lots of data into a simpler summary. For instance, his single number describes a large number of discrete events. Or, consider the scourge of many students, the Grade Point Average (GPA). This single number describes the general performance of a student across a potentially wide range of course experiences.

Every time you try to describe a large set of observations with a single indicator you run the risk of distorting the original data or losing important detail. The GPA doesn't tell you whether the student was in difficult courses or easy ones, or whether they were courses in their major field or in other disciplines. Even given these limitations, descriptive statistics provide a powerful summary that may enable comparisons across people or other units.

Definition: Variable

A variable is any characteristics, number, or quantity that can be measured or counted. A variable may also be called a data. Age, sex, business income and expenses, country of birth, capital expenditure, class grades, eye color.

Type of Variable

A discrete variable is a variable whose value is obtained by counting.

E.g. Number of students present, number of red marbles in a jar, number of heads when flipping three coins, students’ grade level

A continuous variable is a variable whose value is obtained by measuring.

E.g. Height of students in class, weight of students in class, time it takes to get to school, distance traveled between classes.

Note

A random variable is a variable whose value is a numerical outcome of a random phenomenon. A random variable is denoted with a capital letter, the probability distribution of a random variable 𝑋 tells what the possible values of 𝑋 are and how probabilities are assigned to those values, a random variable can be discrete or continuous.



Definition: Data

Data is a set of values of qualitative or quantitative variables.

Types of data

On basis of Property

o Qualitative Data: Data includes property of object. E.g. Cast, residence area, etc.

o Numerical Data: Data includes numbers. E.g. height, weight, age, etc.

On the basis of Numbers

o Univariate Data: If one variable is required for observation, than it is called

Univariate data. e.g. {𝑥1 , 𝑥2 , 𝑥3 , … , 𝑥𝑛}. o Bivariate Data: If two variable is required for observation, than it is called

Univariate data. e.g. {(𝑥1, 𝑦1) , (𝑥2, 𝑦2) , (𝑥3, 𝑦3) , … , (𝑥n, 𝑦n)}. o Trivariate Data: If three variable is required for observation, than it is

called Univariate data. e.g. {(𝑥1, 𝑦1, 𝑧1) , (𝑥2, 𝑦2, 𝑧2) , (𝑥3, 𝑦3, 𝑧3) , … , (𝑥n, 𝑦n, 𝑧𝑛)}

o Multivariate Data: If more than one variable is required to describe the

data, than it is called Multivariate data.

In this chapter, we discuss about analysis of Univariate Data.

Univariate Analysis

Univariate analysis involves the examination across cases of one variable at a time. There are three major characteristics of a single variable that we tend to look at:

Distribution ( Data ) Central Tendency Dispersion

Distribution

Distribution of a statistical data set (or a population) is a listing or function showing all the possible values (or intervals) of the data and how often they occur.

Type of distribution (Data) :

Ungrouped data

E.g. Marks of AEM of 10 students are 10,25,26,35,03,08,19,29,30,18.

https://en.wikipedia.org/wiki/Set_%28mathematics%29

https://en.wikipedia.org/wiki/Value_%28computer_science%29

https://en.wikipedia.org/wiki/Qualitative_data

https://en.wikipedia.org/wiki/Quantitative_data



Grouped data

Discrete Frequency Distribution

E.g. Data of Students using Library during exam time.

No. reading hours(𝑥𝑖)

1 2 3 4

No. of hostel students(𝑓𝑖)

4 7 10 8

Range of Data

Difference of Maximum frequency and Minimum frequency is called range of

data. In above example, Range of data = 10 − 4 = 6 .

Continuous Frequency Distribution

E.g. Data of Students using Library during exam time.

No. reading hours (𝑥𝑖)

0 − 2 3 − 5 6 − 8 9 − 11

No. of hostel students (𝑓𝑖)

11 7 8 0

Class

I. Exclusive Class: If classes of frequency distributions are 0 − 2,2 − 4,4 − 6, … such classes are called Exclusive Classes.

II. Inclusive Class: If classes of frequency distributions are 0 − 2,3 − 5,6 − 8, … such classes are called Inclusive Classes.

Lower Boundary & Upper Boundary

In Class 𝑥𝑖 – 𝑥𝑖+1, Lower Boundary is 𝑥𝑖and Upper boundary is 𝑥𝑖+1 .

Mid-Point

It is defined as Lower Boundary + Upper Boundary

2.

For Continuous frequency distribution, Mid-Point of class is as 𝑥𝑖 of class.

Class Interval/ Class Length

It is defined as |Upper boundary – Lower Boundary|



Central Tendency

The central tendency of a distribution is an estimate of the "center" of a distribution of values. There are three major types of estimates of central tendency:

Arithmetic Mean Median Mode

Arithmetic Mean(�̅�)

The Mean or Average is probably the most commonly used method of describing central tendency. To compute the mean all you do is add up all the values and divide by the number of values.

Arithmetic Mean for Ungrouped data

If data is 𝑥1 , 𝑥2 , 𝑥3 , … , 𝑥𝑛 , than Arithmetic Mean �̅� is defined as below.

Direct Method:

�̅� =𝑥1 + 𝑥2 + 𝑥3 + ⋯ + 𝑥𝑛

𝑛=

1

𝑛⋅ ∑ 𝑥𝑖

𝑛

𝑖=1

Assumed Mean Method:

Suppose 𝑥𝑖 is the biggest data among data 𝑥1 , 𝑥2 , 𝑥3 , … , 𝑥𝑛.LetA = 𝑥𝑖 .

Arithmetic Mean is defined as below.

�̅� = 𝐴 +1

𝑛⋅ ∑ 𝑑𝑖

𝑛

𝑖=1

; 𝑊ℎ𝑒𝑟𝑒 𝑑𝑖 = A − 𝑥𝑖

Generally, highest 𝑥𝑖is the choice of A.

Arithmetic Mean for Grouped data

Arithmetic Mean for following grouped data is defined as below.

Direct Method:

�̅� =𝑥1𝑓1 + 𝑥2𝑓2 + 𝑥3𝑓3 + ⋯ + 𝑥𝑛𝑓𝑛

𝑛=

1

𝑛⋅ ∑ 𝑥𝑖𝑓𝑖

𝑛

𝑖=1

Assumed Mean Method:

Suppose 𝑥𝑖 is the biggest data among data 𝑥1 , 𝑥2 , 𝑥3 , … , 𝑥𝑛. Let A = 𝑥𝑖 . Arithmetic Mean is defined as below.

�̅� = A +1

𝑛⋅ ∑ 𝑑𝑖𝑓𝑖

𝑛

𝑖=1

; 𝑊ℎ𝑒𝑟𝑒 di = A − 𝑥𝑖

Generally, highest 𝑥𝑖is the choice of A.



Step Deviation Method:

Suppose 𝑥𝑖 is the biggest data among data 𝑥1 , 𝑥2 , 𝑥3 , … , 𝑥𝑛 with class length 𝑐.

Let 𝐴 = 𝑥𝑖 . Arithmetic Mean is defined as below.

�̅� = 𝐀 +𝒄

𝒏⋅ ∑ 𝒖𝒊𝒇𝒊

𝒏

𝒊=𝟏

; 𝑾𝒉𝒆𝒓𝒆 𝒖𝒊 =𝐀 − 𝒙𝒊

𝒄

Note:

Step deviation method is used for only continuous frequency distribution.

Exercise – 1

C Que.1

Find mean of temperature recorded in degree centigrade during a week in May, 2015, where the temperature recorded are 38.2, 40.9, 39, 44, 39.6, 40.5, 39.5.

[𝟒𝟎. 𝟐𝟒]

Nov-16

H Que.2

Find the arithmetic mean of the following frequency distribution.

𝑋 1 2 3 4

𝐹 4 5 2 1

[𝟐]

Jun-16

T Que.3

Find arithmetic mean from the following table.

𝑥 35 45 55 60 75 80

𝑓 12 18 10 6 3 11

[𝟓𝟒. 𝟎𝟖]

Nov-16

C Que.4

The marks obtained by 100 students of two classes in mathematics paper

consisting 100 marks are as follows :

xi 15 20 25 32 35 45 50 60 70 77 80 85 90 92 95 99

fi 2 3 7 4 10 12 9 8 6 8 11 9 4 2 3 2

xi = Marks obtained , fi = Number of students

Find the mean of the marks obtained by the students.

[𝟓𝟗. 𝟐𝟏]



C Que.5

Find the mean of the data given below by all the three methods :

Class 0

− 10

10

− 20

20

− 30

30

− 40

40

− 50

50

− 60

60

− 70

Frequency 4 8 3 20 3 4 8

[𝟑𝟓. 𝟖]

T Que.6

The mean of the following frequency distribution is 16 , find the missing

frequency. Where, X=class, F=frequency

X 0

− 4

4

− 8

8

− 12

12

− 16

16

− 20

20

− 24

24

− 28

28

− 32

32

− 36

F 6 8 17 23 16 15 ? 4 3

[𝟖]

Mode ( Z ) :

The Mode is the most frequently occurring value in the set. To determine the mode, you might again order the observations in numerical order, and then count each one. The most frequently occurring value is the mode.

Mode of Ungrouped data:

Most repeated observation among given data is called Mode of Ungrouped data.

Mode of Grouped data:

Discrete Frequency Distribution

The value of variate (variable) corresponding to maximum frequency.

Continuous Frequency Distribution

𝒁 = 𝒍 + (𝒇𝟏 − 𝒇𝟎

𝟐𝒇𝟏 − 𝒇𝟎 − 𝒇𝟐) × 𝒄

Note that, Modal Class is the class with highest frequency.

Where,

𝒍 = Lower boundary of Modal Class

𝒄 = class interval OR class length

𝒇𝟏 = Frequency of the modal class.

𝒇𝟎 = Frequency of the class preceding the modal class.

𝒇𝟐 = Frequency of the class succeeding the modal class.



Exercise-2

C Que 1. Find mode of following observation.

7,3,6,8,9,8,8,8,9,8,7,5,8. [𝟖]

H Que 2.

Find mode of following observation.

10,9,21,16,14,18,20,18,14,18,23,16,18,4. [𝟏𝟖]

H Que 3.

A survey conducted on 20 hostel students for their reading hours per day resulted in the following frequency table

Number of reading hours

1 − 3 3 − 5 5 − 7 7 − 9 9 − 11

Number of hostel students

7 2 8 2 1

[𝟔]

C Que 4.

The mark distribution of 30 students at mathematics examination in a class as below

Marks 10 − 25 25 − 40 40 − 55 55 − 70 70 − 85 85 − 100

No. of students

05 21 21 08 25 20

[𝟖𝟏. 𝟓𝟗𝟎𝟗]

T Que 5.

What is mode of the following frequency distribution?

Number of

reading hours 1 2 3 4

Number of hostel

students 4 7 10 8

[3]

Jun-16

T Que 6.

What is mode of the following frequency distribution?

No. of

children 0 1 2 3 4 5 6 7 8 9 10 11 12

No. of

family 97 110 254 95 67 43 21 13 7 6 5 4 2

[𝟐]



Median ( M )

The Median is the value found at the exact middle of the set of values. To compute the median is to list all observations in numerical order, and then locate the value in the center of the sample.

Median of Ungrouped Data

Let the total number of observation be𝑛.

𝐈𝐟 𝐧 𝐢𝐬 𝐨𝐝𝐝 𝐧𝐮𝐦𝐛𝐞𝐫, than M = (n + 1

2)

th

observation.

𝐈𝐟 𝐧 𝐢𝐬 𝐞𝐯𝐞𝐧 𝐧𝐮𝐦𝐛𝐞𝐫, 𝐭𝐡𝐚𝐧 𝐌 =(

𝒏

𝟐)

𝐭𝐡

𝐨𝐛𝐬𝐞𝐫𝐯𝐚𝐭𝐢𝐨𝐧 + (𝒏

𝟐+ 𝟏)

𝐭𝐡

𝐨𝐛𝐬𝐞𝐫𝐯𝐚𝐭𝐢𝐨𝐧

𝟐.

Median of Grouped Data:

Median of Discrete Grouped Data

Let the total number of observation be 𝑛.

𝐈𝐟 𝐧 𝐢𝐬 𝐨𝐝𝐝 𝐧𝐮𝐦𝐛𝐞𝐫, 𝐭𝐡𝐚𝐧 𝐌 = (𝐧 + 𝟏

𝟐)

𝐭𝐡

𝐨𝐛𝐬𝐞𝐫𝐯𝐚𝐭𝐢𝐨𝐧.

𝐈𝐟 𝐧 𝐢𝐬 𝐞𝐯𝐞𝐧 𝐧𝐮𝐦𝐛𝐞𝐫, 𝐭𝐡𝐚𝐧 𝐌 =(

𝒏

𝟐)

𝐭𝐡

𝐨𝐛𝐬𝐞𝐫𝐯𝐚𝐭𝐢𝐨𝐧 + (𝒏

𝟐+ 𝟏)

𝐭𝐡

𝐨𝐛𝐬𝐞𝐫𝐯𝐚𝐭𝐢𝐨𝐧

𝟐.

Note: Formulae for Ungrouped data and discrete grouped data are same.

Median Of Continuous Grouped Data

𝐌 = 𝒍 + (

𝒏

𝟐− 𝑭

𝒇 ) × 𝒄

Where,

𝑙 = lower boundary point of the Median class

𝑛 = total number of observation (sum of the frequencies)

𝐹 = cumulative frequency of the class preceding the median class.

𝑓 = the frequency of the median class

𝑐 = class interval OR class length

Note

Median class = Class whose commulative frequncy with property min {𝑐𝑓 | 𝑐𝑓 ≥𝑛

2} .



Exercise-3

C Que 1. Define Mode and also give the relationship between Mean, Median and

Mode. May-16

Nov-16

T Que 2. Find the median by the data 2,8,4,6,10,12,4,8,14,16.

[11] Nov-16

C Que 3.

Obtain the median for the following distribution:

X 1 2 3 4 5 6 7 8 9 10

Y 10 34 27 24 12 27 20 18 15 30

[𝟐𝟐]

T Que 4.

Find Median of following data

Marks obtained 18 22 30 35 39 42 45 47

Number of

students 4 5 8 8 16 4 2 3

[𝟑𝟕]

H Que 5.

Find Median of following data

Number of students 6 4 16 7 8 2

Marks obtained 20 9 25 50 40 80

[𝟐𝟓]

T Que 6.

A Survey regarding the weights (in kg) of 45 students of class X of a school

was conducted and the following data was obtained:

Find the median weight.

[𝟑𝟖. 𝟕𝟓]

Weight

(in kg) 20 − 25

25

− 30

30

− 35

35

− 40

40

− 45

45

− 50

50

− 55

No. of

students 2 5 8 10 7 10 3



C Que 7.

The following data represents the no. of foreign visitors in a multinational

company in every 10 days during last 2 months. Use the data to the

median.

X: 0

− 10 10 − 20

20

− 30

30

− 40

40

− 50

50

− 60

No. of

visitors f: 12 18 27 20 17 06

[27.4074]

Jun-16

C Que 8.

Obtain the median for the following distribution:

Mid value 15 20 25 30 35 40 45 50 55

Frequency 2 22 19 14 3 4 6 1 1

Cumulative 2 24 43 57 60 64 70 71 72

[𝟐𝟓. 𝟔𝟓𝟕𝟗]

Dec-13

H Que 9.

Obtain the median for the following information:

Marks < 0 < 10 < 20 < 30

students 50 38 20 5

[17.22]

T Que 10.

Obtain the median for the following information:

Daily

wage

50 < 50

− 100

100

− 150

150

− 200

200

− 250

250

− 300

300

− 350

frequency 2 4 7 21 25 20 21

[232]

C Que 11.

Calculate Mean, Median and Mode for the following data

[𝟔𝟑. 𝟖𝟐, 𝟔𝟑. 𝟔𝟔, 𝟔𝟑. 𝟐𝟖𝟓𝟕]

Class 50

− 53

53

− 56

56

− 59

59

− 62

62

− 65

65

− 68

68

− 71

71

− 74

74

− 77

F 3 8 14 30 36 28 16 10 5

Dec-15



Dispersion:

Dispersion refers to the spread of the values around the central tendency. There are two common measures of dispersion, the range and the standard deviation.

Range

It is simply the highest value minus the lowest value. In our example distribution, the high value is 36 and the low is 15, so the range is 36 − 15 = 21.

Standard Deviation(𝝈)

It is a measure that is used to quantify the amount of variation or dispersion of a set of data values.

Formula to find Standard Deviation

Method Ungrouped Data Grouped Data

Direct Method

σ = √∑ (xj − x̅)

2Nj=1

N

σ = √∑ fj(xj − x̅)

2Nj=1

N

Actual Mean Method σ = √∑ fixi

2ni=1

∑ fini=1

− (∑ fixi

ni=1

∑ fini=1

)

2

Assumed Mean Method σ = √∑ xi

2ni=1

N− (

∑ xini=1

N)

2

σ = √∑ fidi

2ni=1

∑ fini=1

− (∑ fidi

ni=1

∑ fini=1

)

2

Step Deviation Method σ = √∑ di

2ni=1

N− (

∑ dini=1

N)

2

σ = √∑ fiui

2ni=1

∑ fini=1

− (∑ fiui

ni=1

∑ fini=1

)

2

× c

Variance ( 𝑽(𝑿) )

Variance is the expectation of the squared deviation of a random variable from its mean, and it informally measures how far a set of (random) numbers are spread out from their mean.

Relation between Variance and Standard Deviation is as follow.

𝐕𝐚𝐫𝐢𝐚𝐧𝐜𝐞 = (𝐒𝐭𝐚𝐧𝐝𝐚𝐫𝐝 𝐃𝐞𝐯𝐢𝐚𝐭𝐢𝐨𝐧 )𝟐

https://en.wikipedia.org/wiki/Statistical_dispersion

https://en.wikipedia.org/wiki/Expected_value

https://en.wikipedia.org/wiki/Random_variable

https://en.wikipedia.org/wiki/Mean



Skewness

It is a measure of the asymmetry of the probability distribution of a real-valued random variable about its mean. The skewness value can be positive or negative, or even undefined.

Negative skew:

The left tail is longer; the mass of

the distribution is concentrated on

the right of the figure.

Positive skew:

The right tail is longer; the mass of

the distribution is concentrated on the left of the figure.

𝐒𝐤𝐞𝐰𝐧𝐞𝐬𝐬 =𝐌𝐞𝐚𝐧 − 𝐌𝐨𝐝𝐞

𝐒𝐭𝐚𝐧𝐝𝐚𝐫𝐝 𝐝𝐞𝐯𝐢𝐚𝐭𝐢𝐨𝐧 =

�̅� − 𝐙

𝛔

Exercise – 4

T Que 1.

The pH of a solution is measured 8 times by one operator using the same

instrument. She obtains the following data: 7.15, 7.20, 7.18, 7.19, 7.21, 7.16 and

7.18. Calculate the sample Mean, the sample variance and sample standard

deviation.

[𝟕. 𝟏𝟖𝟏𝟒𝟑, 𝟎. 𝟎𝟎𝟎𝟒, 𝟎. 𝟎𝟐]

May-16

C Que 2.

Find the Standard deviation and Skewness of the data given below by all the

three methods :

Class 0

− 10

10

− 20

20

− 30

30

− 40

40

− 50

50

− 60

60

− 70

Frequency 4 8 3 20 3 4 8

[𝟏𝟕. 𝟗𝟖, 𝟎. 𝟎𝟒𝟒𝟒𝟗𝟑]

H Que 3.

Find the Standard deviation and Skewness of the mark distribution of 30

students at mathematics examination in a class as below :

Marks 10 − 25 25 − 40 40 − 55 55 − 70 70 − 85 85 − 100

No of

students 05 21 21 08 25 20

[𝟐𝟑. 𝟖𝟕𝟑𝟔, −𝟎. 𝟖𝟖𝟏𝟑]

https://en.wikipedia.org/wiki/Probability_distribution

https://en.wikipedia.org/wiki/Real_number





T Que 4.

Find standard deviation from the following data.

Class 9 − 11 12 − 14 15 − 17 18 − 20

Frequency 2 3 4 1

[𝟐. 𝟕𝟒𝟗𝟓𝟓]

Jun-16

4. Correlation And Regression PAGE | 38


Coefficient Of Correlation

Correlation is the relationship that exists between two or more variables. Two variables are said to be correlated if a change in one variable affects a change in the other variable. Such a data connecting two variables is called bivariate data.

When two variables are correlated with each other, it is important to know the amount or extent of correlation between them. The numerical measure of correlation of degree of relationship existing between two variables is called the coefficient of correlation and is denoted by r and it is always lies between −1 and 1.

When 𝑟 = 1, it represents Perfect Direct or Positive Correlation. When 𝑟 = −1 it represents Perfect Inverse or Negative Correlation. When 𝑟 = 0, there is No Linear Correlation or it shows Absence Of Correlation. When the value of r is ±0.9 or ±0.8 etc. it shows high degree of relationship

between the variables and when r is small say ±0.2 or ±0.1 etc, it shows low degree of correlation.

Types Of Correlations

Correlation is classified into four types:

1. Positive and negative correlations 2. Simple and multiple correlations 3. Partial and total correlations 4. Linear and nonlinear correlations

Positive and Negative Correlations Depending on the variation in the variables, correlation may be positive or negative.

1. Positive Correlation If both the variables vary in the same direction, the correlation is said to be positive. In the other words, if the value of one variable increases, the value of the other variable also increases, or, if the value of one variable decreases, the value of the other variable also decreases, e.g., the correlation between heights and weights of group of persons is a positive correlation.

2. Negative Correlation If both the variables vary in the opposite direction, correlation is said to be negative. In other words, if the value of one variable increases, the value of the other variable also decreases, or, if the value of one variable decreases, the value of the other variable also increases, e.g., the correlation between the price and demand of a commodity is a negative correlation.

Simple and Multiple Correlation Depending upon the study of the number of variables, correlation may be simple or

multiple.

1. Simple Correlation When only two variables are studied, the relationship is described as simple correlation, e.g., the quantity of money and price level, demand and price, etc.



2. Multiple correlation When more than two variables are studied, the relationship is described as multiple correlation, e.g., relationship of price, demand, and supply of a commodity.

Partial and Total Correlation Multiple correlation may be either partial or total.

1. Partial Correlation

When more than two variables are studied excluding some other variables, the relationship is termed as partial correlation.

2. Total Correlation

When more than two variables are studied, without excluding any variables ,the relationship is termed as total correlation.

Linear and Nonlinear Correlation Depending upon the ratio of change between two variables, the correlation may be

linear or nonlinear.

1.Linear Correlation

If the ratio of change between two variables is constant, the correlation is said to be linear.

2. Nonlinear Correlation

If the ratio of change between two variables is not constant, the correlation is said to be nonlinear.

Methods of Studying Correlation

There are two different methods of studying correlation,

1. Graphic methods,

2. Mathematical methods.

Graphic methods are

1. scatter diagram

2. Simple graph

Mathematical methods are

1. Karl Pearson’s coefficient of correlation

2. Spearman’s rank coefficient of correlation

Karl Pearson’s Product Moment Method

This is most popular and widely used mathematical method. In this method the degree

of correlation between two variables can be measured by the coefficient of correlation

and it gives not only magnitude (degree) of correlation but also its direction.



Let (x1,y1), (x2,y2), . … … … (x𝑛,y𝑛)be n pairs of observations of two variables X and Y,

then coefficients of correlation (r) between X and Y is defined by

𝐫 =𝐜𝐨𝐯𝐚𝐫𝐢𝐚𝐧𝐜𝐞 𝐛𝐞𝐭𝐰𝐞𝐞𝐧 𝐗 𝐚𝐧𝐝 𝐘

(𝐬𝐭𝐚𝐧𝐝𝐚𝐫𝐝 𝐝𝐞𝐯𝐢𝐚𝐭𝐢𝐨𝐧 𝐨𝐟 𝐗)(𝐬𝐭𝐚𝐧𝐝𝐚𝐫𝐝 𝐝𝐞𝐯𝐢𝐚𝐭𝐢𝐨𝐧 𝐨𝐟 𝐘)=

𝑪𝒐𝒗(𝑿, 𝒀)

𝛔𝐱 . 𝛔𝐲… … (𝟏)

Where 𝑪𝒐𝒗 (𝑿, 𝒀) =𝚺(𝑿−�̅�)(𝒀−�̅�)

𝒏

Variance Of 𝒙 =𝚺(𝑿−�̅�)𝟐

𝒏

Variance Of 𝒚 =𝚺(𝒀−�̅�)𝟐

𝒏

�̅� = 𝐌𝐞𝐚𝐧 𝐨𝐟 𝐗 =∑ 𝑿

𝒏

�̅� = 𝐌𝐞𝐚𝐧 𝐨𝐟 𝐘 =∑ 𝒀

𝒏

𝐧 = 𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐗, 𝐘 𝐩𝐚𝐢𝐫𝐬

𝛔𝐱 = 𝐬𝐭𝐚𝐧𝐝𝐚𝐫𝐝 𝐝𝐞𝐯𝐢𝐚𝐭𝐢𝐨𝐧 𝐨𝐟 𝐗

= √𝚺(𝑿 − �̅�)𝟐

𝒏

𝛔𝐲 = 𝐬𝐭𝐚𝐧𝐝𝐚𝐫𝐝 𝐝𝐞𝐯𝐢𝐚𝐭𝐢𝐨𝐧 𝐨𝐟 𝐘

= √𝚺(𝒀 − �̅�)𝟐

𝒏

Then above formula is given by

𝒓 =𝚺(𝑿 − �̅�)(𝒀 − �̅�)

𝐧 𝛔𝐱 . 𝛔𝐲 … … (𝟐)

𝒓 =𝜮𝒙𝒚

𝒏 𝝈𝒙 . 𝝈𝒚 𝑾𝒉𝒆𝒓𝒆, 𝑿 − �̅� = 𝒙 𝒂𝒏𝒅 𝒀 − �̅� = 𝒚 … … (𝟑)

This formula is known as the Product Moment Formula of Coefficient Of Correlation.

From (2),

𝒓 =𝚺(𝑿 − �̅�)(𝒀 − �̅�)

𝒏 √𝚺(𝑿 − �̅�)𝟐

𝒏 √𝚺(𝒀 − �̅�)𝟐

𝒏

𝐎𝐑 𝒓 =𝚺(𝑿 − �̅�)(𝒀 − �̅�)

√𝚺(𝑿 − �̅�)𝟐 √𝚺(𝒀 − �̅�)𝟐 … … (𝟒)

𝒓 =𝚺𝐱𝐲

√𝚺𝐱𝟐√𝚺𝐲𝟐 … … (𝟓) 𝐖𝐡𝐞𝐫𝐞, 𝑿 − �̅� = 𝒙 𝐚𝐧𝐝 𝒀 − �̅� = 𝒚.

If there is a small number of items in two variables the correlation coefficient can

obtained by the formula,

𝒓 =𝒏𝚺𝐱𝐲 − (𝚺𝐱)(𝚺𝐲)

√𝐧𝚺𝐱𝟐 − (𝚺𝐱)𝟐√𝒏𝚺𝐲𝟐 − (𝚺𝒚)𝟐 … … (𝟔)

Since the correlation coefficient is independent of choice of origin, the formula for 𝑟,

when the deviation are taken from the assumed mean is given by

𝒓 =𝒏𝚺𝐝𝐱𝐝𝐲 − (𝚺𝐝𝐱)(𝚺𝐝𝐲)

√𝐧𝚺𝐝𝐱𝟐 − (𝚺𝐝𝐱)𝟐√𝒏𝚺𝐝𝐲𝟐 − (𝚺𝐝𝒚)𝟐 … … (𝟕)



Where 𝐝𝐱 = 𝐗 − 𝐀 and 𝐝𝐲 = 𝐘 − 𝐁. A and B being assumed means of X and Y

respectively, dx and dy are the deviations of X and Y respectively.

Short-Cut Method Of Computing Correlation Coefficient

Here, we take 𝒅𝒙 = 𝑿 − 𝑨 or 𝒅𝒙 =𝑿−𝑨

𝒊 𝒅𝒚 = 𝒀 − 𝑩 or 𝒅𝒚 =

𝒀−𝑩

𝒋

Where 𝐴 and 𝐵 are assumed mean (𝑖 and 𝑗 are used to reduce data in small figures).

Exercise-1

C Que 1.

Find the correlation coefficient between the serum diastolic B.P. and

serum cholesterol levels of 10 randomly selected data of 10 persons.

Person 1 2 3 4 5 6 7 8 9 10

Choles. 307 259 341 317 274 416 267 320 274 336

B.P. 80 75 90 74 75 110 70 85 88 78

[𝐫 = 𝟎. 𝟖𝟎𝟖𝟕]

T Que 2.

Find the correlation coefficient between the sales and expenses of the

following 10 firms.

Firms 1 2 3 4 5 6 7 8 9 10

Sales 50 50 55 60 65 65 65 60 60 50

Ex. 11 13 14 16 16 15 15 14 13 13

[𝐫 = 𝟎. 𝟕𝟖𝟔𝟓]

T Que 3.

Calculate correlation coefficient from following data.

x 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000

y 0.30 0.29 0.29 0.25 0.24 0.24 0.24 0.29 0.18 0.15

[𝐫 = −𝟎. 𝟕𝟗𝟎𝟔]

H Que 4.

Find correlation coefficient between x and y for the given values.

x 1 2 3 4 5 6 7 8 9 10

y 10 12 16 28 25 36 41 49 40 50

[𝐫 = 𝟎. 𝟗𝟓𝟖𝟐]



H Que 5.

Compute correlation coefficient for the following data and interpret it:

X 4 5 9 14 18 22 24

y 16 22 11 16 7 3 17

[𝐫 = −𝟎. 𝟓𝟐𝟓𝟓]

C Que 6. Find rxy from given data: n = 10, ∑(x − x̅)(y − y̅) = 1650, σx

2 = 196,

σy2 = 225. [𝐫𝐱𝐲 = 𝟎. 𝟕𝟖𝟓𝟕]

H Que 7.

Compute the correlation between 𝑥 and 𝑦 using following data.

[𝐫 = −𝟎. 𝟗𝟐𝟎𝟑]

𝑥 2 4 5 6 8 11

𝑦 18 12 10 8 7 5

Dec-12

T Que 8.

Find the correlation coefficient between Intelligence Ratio (I.R) and

Emotional Ratio (E.R) from the following data.

Student 1 2 3 4 5 6 7 8 9 10

I.R. 105 104 102 101 100 99 98 96 93 92

E.R. 101 103 100 98 95 96 104 92 97 94

[𝐫 = 𝟎. 𝟓𝟗𝟔𝟑]

C Que 9.

Calculate Karl-Pearson’s correlation coefficient between age and playing

habits from the data given below.

Age 20 21 22 23 24 25

No. of students 500 400 300 240 200 160

Regular players 400 300 180 96 60 24

[𝐫 = −𝟎. 𝟗𝟗𝟏𝟐]

C Que 10.

Dicofenac Sodium Sustained release tablets were analyzed in-vitro and

in-vivo. The results were summarized in the following table. Find out

whether both the methods of evaluation are correlated or not.

Time in minutes Amount of Drug released in %

In-vitro In-vivo

0 0 0



30 35.45 20.33

60 36.87 33.65

90 44.91 41.82

120 55.20 50.01

150 62.46 59.78

[𝐫 = 𝟎. 𝟗𝟕𝟎𝟓]

T Que 11.

Consider following 14 measurements of the concentration of sodium

chlorate produce in a chemical reactor operated at a pH of 7.0.

12 15 14.1 15.9 11.5 14.2 11.2

13.7 15.9 12.6 14.3 12.6 12.1 14.8

Calculate a) Mean b) Sample Standard Deviation c) Variance

[(𝒂)𝟏𝟑. 𝟓𝟔 (𝒃)𝟏. 𝟓𝟏(𝒄)𝟐. 𝟐𝟗]

C Que 12.

Determine the coefficient of correlation �̅� = 5.5; �̅� = 4; ∑ 𝑥2 = 385;

∑ 𝑦2 = 192 ; ∑(𝑥 + 𝑦)2 = 947.

[𝒓 = −𝟎. 𝟔𝟖𝟏]

Dec-16

Spearman’s Rank Correlation Method

Rank correlation is based on the rank or the order of the variables and not on the

magnitude of the variables. Here, the individuals are arranged in order of proficiency.

If the ranks are assigned to the individuals range from 1 to n, then the correlation

coefficient between two series of ranks is called rank correlation coefficient.

Edward Spearman’s formula for rank coefficient of correlation(R) is given by

𝑹 = 𝟏 −𝟔 ∑ 𝒅𝟐

𝒏(𝒏𝟐 − 𝟏)

Where, d = difference between the ranks 𝑅1 and 𝑅2 given by two judges

n = number of pairs

For example, we cannot measure beauty and intelligence quantitatively. It is possible to

rank the individuals in order.

The above formula is used when ranks are not repeated.



Tide Ranks

If there is a tie between two or more individuals ranks, the rank is divided among equal

individuals, e.g., if two items have fourth rank, the 4th and 5th rank is divided between

them equally and is given as 4+5

2= 4.5𝑡ℎ rank to each of them. If three items have the

same 4th rank, each of them is given 4+5+6

3= 5𝑡ℎ rank. As a result of this, the following

adjustment or correlation is made in the rank correlation formula. If m is the number of

item having equal ranks then the factor 1

12(𝑚3 − 𝑚) is added to ∑ 𝑑2. If there are more

than one cases of this types, this factor is added corresponding to each case.

𝒓 = 𝟏 −𝟔 [∑ 𝒅𝟐 +

𝟏𝟏𝟐

(𝒎𝟏𝟑 − 𝒎𝟏) +

𝟏𝟏𝟐

(𝒎𝟐𝟑 − 𝒎𝟐) + ⋯ ]

𝒏(𝒏𝟐 − 𝟏)

Exercise-2

H Que 1.

Two Judges in a beauty contest rank the 12 contestants as follows:

𝑋 1 2 3 4 5 6 7 8 9 10 11 12

𝑦 12 9 6 10 3 5 4 7 8 2 11 1

What degree of agreement is there between the Judges?

[𝝆 = −𝟎. 𝟒𝟓𝟒𝟓]

T Que 2.

The competitions in a beauty contest are ranked by three judges in the

following order:

1st judge 1 5 4 8 9 6 10 7 3 2

2nd judge 4 8 7 6 5 9 10 3 2 1

3rd judge 6 7 8 1 5 10 9 2 3 4

Use rank correlation coefficient method to discuss which pair of judges

has nearest approach to beauty.

[𝝆𝟏 = 𝟎. 𝟓𝟓𝟏𝟓 , 𝝆𝟐 = 𝟎. 𝟕𝟑𝟑𝟑 , 𝝆𝟑 = 𝟎. 𝟎𝟓𝟒𝟓𝟒]

C Que 3.

9 students secured the following percentage of marks in math and

chemistry. Find the rank correlation coefficient and comment on its value.

Roll no. 1 2 3 4 5 6 7 8 9

Marks in Math. 78 36 98 25 75 82 90 62 65

Marks in Chem. 84 51 91 60 68 62 86 58 53



[𝝆 = 𝟎. 𝟖𝟑𝟑𝟑]

C Que 4.

Calculate coefficient of correlation by spearman’s method from the

following.

Sales 45 56 39 54 45 40 56 60 30 36

cost 40 36 30 44 36 32 45 42 20 36

[𝝆 = 𝟎. 𝟕𝟔𝟑𝟔]

T Que 5.

(i) Ten competitors in a music competition are ranked by three judges in

the following order

1st judge 1 6 5 10 3 2 4 9 7 8

2nd judge 3 5 8 4 7 10 2 1 6 9

3rd judge 6 4 9 8 1 2 3 10 5 7

Use the rank correlation to discuss which pair of judges has the nearest

approach to common tastes in music.

(ii) What is correlation? Give methods to measure it.

[𝛒𝟏 = −𝟎. 𝟐𝟏𝟐𝟏, 𝛒𝟐 = −𝟎. 𝟐𝟗𝟔𝟕, 𝛒𝟑 = 𝟎. 𝟔𝟑𝟔𝟒]

1st and 3rd judge has nearest approach [𝛒 = 𝟏 −𝟔 ∑ 𝐝𝟐

𝐧𝟑−𝐧]

May-15

May-12

T Que 6.

Obtain the rank correlation coefficient for the following data:

X 68 64 75 50 64 80 75 40 55 64

Y 62 58 68 45 81 60 68 48 50 70

[𝝆 = 𝟎. 𝟓𝟒𝟓]



Regression

Regression is defined as a method of estimating the value of one variable when that of

the other is known and the variables are correlated. Regression analysis is used to

predict or estimate one variable in terms of the other variable. It is a highly valuable

tool for prediction purpose in economics and business.

Regression Equations The algebraic expressions of the regression lines are called Regression equations. Since

there are two regression lines, there are two regression equations. Using method of

least squares we have obtained the regression equation of 𝑦 on 𝑥 as 𝑦 = 𝑎 + 𝑏𝑥 and that

of 𝑥 on 𝑦 as 𝑥 = 𝑎 + 𝑏𝑦. The values of 𝑎 and 𝑏 depends on the means, the standard

deviations and coefficient of correlation between the two variables.

These equations are written as follows:

The Regression equation of 𝑦 on 𝑥 is

𝒚 − �̅� = 𝒓.𝝈𝒚

𝝈𝒙

(𝒙 − �̅�) … … … (𝟏)

And the regression equation of 𝑥 on 𝑦 is

𝒙 − �̅� = 𝒓.𝝈𝒙

𝝈𝒚

(𝒚 − �̅�) … … … (𝟐)

Where,

�̅� = Mean of 𝒙 =∑ 𝒙

𝒏

�̅� =Mean of 𝒚 =∑ 𝒚

𝒏

𝝈𝒙 =Standard deviation of 𝑥

𝝈𝒚 =Standard deviation of 𝑦

𝒓 =Correlation coefficient between 𝑥 & 𝑦

The coefficients of 𝑥 and 𝑦 in equations (1) and(2), namely 𝒓.𝝈𝒚

𝝈𝒙 and 𝒓.

𝝈𝒙

𝝈𝒚 are known as

regression coefficients and they are denoted by 𝑏𝑦𝑥 and 𝑏𝑥𝑦 respectively.

i.e. 𝒃𝒚𝒙 = 𝒓.𝝈𝒚

𝝈𝒙… … … (𝟑) & 𝒃𝒙𝒚 = 𝒓.

𝝈𝒙

𝝈𝒚… … … (𝟒)

Thus, the regression equations (1) and (2) can also be written as

𝒚 − �̅� = 𝒃𝒚𝒙(𝒙 − �̅�) … … … (𝟓)

𝒙 − 𝒙 = 𝒃𝒙𝒚(𝒚 − �̅�) … … … (𝟔)

Formula for Computations of Regressions Coefficients The following formulae are used for calculation of 𝑏𝑦𝑥 and 𝑏𝑥𝑦,

When 𝜎𝑥, 𝜎𝑦 𝑎𝑛𝑑 𝑟 are given 𝒃𝒚𝒙 = 𝒓 ∙𝝈𝒚

𝝈𝒙 & 𝒃𝒙𝒚 = 𝒓 ∙

𝝈𝒙

𝝈𝒚



When 𝑛, ∑ 𝑥 , ∑ 𝑦 , ∑ 𝑥2 , ∑ 𝑦2 𝑎𝑛𝑑 ∑ 𝑥𝑦 are given

𝒃𝒚𝒙 =𝒏 ∑ 𝒙𝒚 − (∑ 𝒙)(∑ 𝒚)

𝒏 ∑ 𝒙𝟐 − (∑ 𝒙)𝟐 & 𝒃𝒙𝒚 =

𝒏 ∑ 𝒙𝒚 − (∑ 𝒙)(∑ 𝒚)

𝒏 ∑ 𝒚𝟐 − (∑ 𝒚)𝟐

When the deviations are taken from their means

𝒃𝒚𝒙 =∑(𝑿 − �̅�)(𝒀 − �̅�)

∑(𝑿 − �̅�)𝟐=

∑ 𝒙𝒚

∑ 𝒙𝟐 & 𝒃𝒙𝒚 =

∑(𝑿 − �̅�)(𝒀 − �̅�)

∑(𝒀 − �̅�)𝟐=

∑ 𝒙𝒚

∑ 𝒚𝟐

Where 𝒙 = (𝑿 − �̅�), 𝒚 = (𝒀 − �̅�)

When the deviations are taken from assumed numbers

𝒃𝒚𝒙 =𝒏 ∑ 𝒅𝒙𝒅𝒚 − (∑ 𝒅𝒙)(∑ 𝒅𝒚)

𝒏 ∑ 𝒅𝒙𝟐 − (∑ 𝒅𝒙)𝟐 & 𝒃𝒙𝒚 =

𝒏 ∑ 𝒅𝒙𝒅𝒚 − (∑ 𝒅𝒙)(∑ 𝒅𝒚)

𝒏 ∑ 𝒅𝒚𝟐 − (∑ 𝒅𝒚)𝟐

Where = 𝒙 − 𝑨, 𝒅𝒚 = 𝒚 − 𝑩 . 𝐴 and 𝐵 are any assumed numbers.

When the values of covariance and variances are given

𝒃𝒚𝒙 =𝑪𝒐𝒗(𝑿,𝒀)

𝝈𝒙𝟐 =

∑(𝑿−�̅�)(𝒀−�̅�)

𝒏𝝈𝒙𝟐 =

∑ 𝒙𝒚

𝒏𝝈𝒙𝟐 & 𝒃𝒙𝒚 =

𝑪𝒐𝒗(𝑿,𝒀)

𝝈𝒚𝟐 =

∑(𝑿−�̅�)(𝒀−�̅�)

𝒏𝝈𝒚𝟐 =

∑ 𝒙𝒚

𝒏𝝈𝒚𝟐

Where 𝒙 = (𝑿 − �̅�), 𝒚 = (𝒀 − �̅�)

Properties of Regression Coefficients The geometric mean (𝑟) between two regression coefficients is given by i.e.

𝑟 = √𝑏𝑦𝑥 × 𝑏𝑥𝑦

Both the regression coefficients will have the same sign i.e. they are either both positive or both negative.

The values of both 𝑏𝑥𝑦 and 𝑏𝑦𝑥 individually cannot be more than one.

The Sign of 𝑟 is same as of regression coefficients , if 𝑟 < 0 then 𝑏𝑦𝑥 < 0 & 𝑏𝑥𝑦 < 0

Exercise-3

T Que 1.

From the following data calculate two equation of line of regression:

𝑋 𝑌

Mean 60 67.5

Standard deviation 15 13.5

Correlation coefficient between 𝑋 & 𝑌 is 0.50. Also estimate the value of 𝑌 for 𝑋 = 72 using the appropriate regression equation.

[𝒀 = 𝟎. 𝟒𝟓𝑿 + 𝟒𝟎. 𝟓, 𝑿 = 𝟎. 𝟓𝟓𝟔𝒀 + 𝟐𝟐. 𝟒𝟕, 𝒀(𝟕𝟐) = 𝟕𝟐. 𝟗]

Dec-14



H Que 2.

The following values are available for the variable 𝑥 & 𝑦.

𝑛 = 10 , ∑ 𝑥 = 30 ,∑ 𝑦 = 40 , ∑ 𝑥2 = 222 , ∑ 𝑦2 = 985 , ∑ 𝑥𝑦 = 384.

Obtain (a) two regression equation. (b) Correlation coefficient. [𝒀 = 𝟐𝑿 − 𝟐, 𝑿 = 𝟎. 𝟑𝟐𝒀 + 𝟏. 𝟕𝟐, 𝒓 = 𝟎. 𝟖]

C Que 3.

Obtain the two regression lines from the following data and hence find the

correlation coefficient.

X 6 2 10 4 8

Y 9 11 5 8 7

[𝒚 = 𝟏𝟏. 𝟗 − 𝟎. 𝟔𝟓𝒙 , 𝒙 = 𝟏𝟔. 𝟒 − 𝟏. 𝟑𝒚 , 𝒓 = −𝟎. 𝟗𝟏𝟗𝟐]

May-15

C Que 4.

For the following data find the two regression lines:

x 1 2 3 4 5 6 7 8 9 10

y 10 12 16 28 25 36 41 49 40 50

[𝒀 = 𝟒. 𝟔𝟕𝟐𝟕𝑿 + 𝟓, 𝑿 = 𝟎. 𝟏𝟗𝟔𝟓𝒀 − 𝟎. 𝟓𝟑𝟏𝟖]

May-14

H Que 5.

Following table gives the data on rain falls & discharge in a certain river. Obtain the line of regression of 𝑦 on 𝑥 :

Rain fall (inches) 𝑥 1.53 1.78 2.60 2.95 3.42

Discharge (1000 cc)

𝑦 33.5 36.3 40.0 45.8 53.5

[𝒀 = 𝟗. 𝟕𝟐𝟔𝟔𝑿 − 𝟏𝟕. 𝟗𝟏𝟑𝟏]

H Que 6.

Find the lines of regression of 𝑌 on 𝑋 if 𝑛 = 9 , ∑ 𝑥 = 30.3, ∑ 𝑦 = 91.1, ∑ 𝑥𝑦 = 345.09, & ∑ 𝑥2 = 115.11. Also find value of 𝑌(1.5) & 𝑌(5.0)

[𝒀 = 𝟐. 𝟗𝟑𝑿 + 𝟎. 𝟐𝟓𝟔𝟖, 𝒀(𝟏. 𝟓) = 𝟒. 𝟔𝟓𝟐𝟑, 𝒀(𝟓. 𝟎) = 𝟏𝟒. 𝟗𝟎𝟖𝟑]

C Que 7.

The amount of chemical compound , which were dissolved in 100 grams of water at various temperatures , 𝑥 were recorded as:

x(℃) 15 15 30 30 45 45 60 60

y(grams) 12 10 25 21 31 33 44 39

Find the line of regression of 𝑦 on 𝑥 & estimate the amount of chemical that will dissolve in 100 grams of water at50℃.

[𝒀 = 𝟎. 𝟔𝟕𝑿 + 𝟏. 𝟕𝟓, 𝒀(𝟓𝟎°) = 𝟑𝟓. 𝟐𝟓]



H Que 8.

The following table gives the age of tablet machine of certain make & annual maintenance costs. Obtain the regression equation for cost related to age.

Age of machine (years) 2 4 6 8

Maintenance cost (in thousand Rs.) 10 20 25 30

[𝒀 = 𝟑. 𝟐𝟓𝑿 + 𝟓]

T Que 9.

Obtain the line of regression of monthly sales(𝑌) on advertisement expenditure (𝑋) & estimate the monthly sales when the company will spend Rs. 50000 on advertisement, if the data on 𝑌 & 𝑋 are as follows:

Y (in lacs) 74 76 60 68 79 70 71 94

X (in thousand) 43 44 36 38 47 40 41 54

[𝒀 = 𝟏. 𝟕𝟐𝟓𝟒𝑿 + 𝟎. 𝟎𝟐𝟑𝟑, 𝒀(𝟓𝟎) = 𝟖𝟔. 𝟐𝟕𝟑𝟑]

T Que 10.

(i)The following data give the experience of machine operators and their performances rating as given by the number of good parts turned out per 100 piece

Operator 1 2 3 4 5 6

Performance rating(x)

23 43 53 63 73 83

Experience(Y) 5 6 7 8 9 10

Calculate the regression line of performing rating on experience and also estimate the probable performance if an operator has 11 years’ experience.

(ii)Define regression coefficients and give its properties.

[𝒙 = 𝟏𝟏. 𝟒𝟐𝟖𝒚 − 𝟐𝟗. 𝟑𝟖]

(performance rating on experience 𝒚 = 𝟏𝟏 years and 𝒙 = 𝟗𝟔. 𝟑𝟐 )

May-15

T Que 11.

A standard curve passing through the origin was prepared for colorimetric estimation of Salphadiazine. The concentration & absorbances are given below. Find the equation of line.

Concentrain 𝜇𝑔/𝑚𝑙

5 10 15 20 30 40

Absorbance 0.12 0.231 0.362 0.458 0.698 0.888

[𝒀 = 𝟎. 𝟎𝟐𝟐𝟏𝑿 + 𝟎. 𝟎𝟏𝟔𝟕]



H Que 12.

Obtain the two lines of regression for the following data:

Sales of Diclowin tablets(no. of

tablets) 190 240 250 300 310 335 300

Advertising expenditure (in Rs.)

5 10 15 20 20 30 30

[𝒀 = 𝟎. 𝟏𝟕𝟏𝟔𝑿 − 𝟐𝟖. 𝟔𝟐𝟑𝟑, 𝑿 = 𝟒. 𝟖𝟓𝟑𝟑𝒀 + 𝟏𝟖𝟒. 𝟖𝟔𝟕]

Dec-11

H Que 13.

Calculate the coefficient of correlation and obtain the lines of regression

for the following:

X 1 2 3 4 5 6 7 8 9

Y 9 8 10 12 11 13 14 16 15

[𝒓 = 𝟎. 𝟗𝟓, 𝒀 = 𝟎. 𝟗𝟓𝑿 + 𝟕. 𝟐𝟓, 𝑿 = 𝟎. 𝟗𝟓𝒀 − 𝟔. 𝟒]

Dec-15

C Que 14.

From the following regression equations8𝑥 − 10𝑦 = −66,40𝑥 − 18𝑦 =214 and variance of 𝑥 = 9 Find (i)Average values of 𝑥 and 𝑦.

(ii)correlation coefficient between the two variables.

(iii)Standard Deviation of 𝑦.

[�̅� = 𝟏𝟑, �̅� = 𝟏𝟕, 𝒓 = 𝟎. 𝟔, 𝝈𝒚 = 𝟒]

Dec-15

C Que 15.

A study of the amount of rainfall and the quantity of air pollution removed produced the following data:

Daily rainfall x (0.01 cm)

4.3 4.5 5.9 5.6 6.1 5.2 3.8 2.1 7.5

Particulate removed, y (𝜇𝑔/𝑚3)

126 121 116 118 114 118 132 141 108

(i)Find the equation of the regression line to predict the particulate removed from the amount of Daily rainfall.

(ii)Find the amount of particulate removed when daily rainfall is 𝑥 = 4.8 units.

[𝒚 = 𝟏𝟓𝟑. 𝟏𝟕𝟓 − 𝟔. 𝟑𝟐𝟒𝒙 ; 𝒚(𝟒. 𝟖) = 𝟏𝟐𝟐. 𝟖𝟏𝟗𝟖]

May-16

5. Curve Fitting PAGE | 51


Curve Fitting

Curve fitting is the process of finding the ‘best-fit’ curve for a given set of data. It is the

representation of the relationship between two variables by means of an algebraic

equation.

The Method Of Least Square

The method of least squares assumes that the best-fit curve of a given type is the curve

that has the minimum sum of the square of the deviation (least square error) from a

given set of data.

Suppose that the data points are (𝑥1, 𝑦1), (𝑥2, 𝑦2), … , (𝑥𝑛, 𝑦𝑛), where x is independent

and y is dependent variable. Let the fitting curve f(x) has the following deviations (or

errors or residuals) from each data points.

𝒅𝟏 = 𝒚𝟏 − 𝒇(𝒙𝟏), 𝒅𝟐 = 𝒚𝟐 − 𝒇(𝒙𝟐), … , 𝒅𝒏 = 𝒚𝒏 − 𝒇(𝒙𝒏).

Clearly, some of the deviations will be positive and others negative. Thus, to give equal

weightage to each error, we square each of these and form their sum; that is,

𝑫 = 𝒅𝟏𝟐 + 𝒅𝟐

𝟐 + ⋯ + 𝒅𝒏𝟐

Now, according to the method of least squares, the best fitting curve has the property

that

𝑫 = 𝒅𝟏𝟐 + 𝒅𝟐

𝟐 + ⋯ + 𝒅𝒏𝟐 = ∑ 𝒅𝒊

𝟐

𝒏

𝒊=𝟏

= ∑[𝒚𝒊 − 𝒇(𝒙𝒊)]𝟐 = 𝒂 𝒎𝒊𝒏𝒊𝒎𝒖𝒎.

𝒏

𝒊=𝟏

Fitting a Straight Line 𝐲 = 𝐚 + 𝐛𝐱 OR Linear Approximation.

Suppose the equation of a straight line of the form 𝑦 = 𝑎 + 𝑏𝑥 is to be fitted to the n-

data points

(𝒙𝟏, 𝒚𝟏), (𝒙𝟐, 𝒚𝟐), … , (𝒙𝒏, 𝒚𝒏), 𝒏 ≥ 𝟐,

Where a is y-intercept and b is its slope.

For the general point (𝑥𝑖, 𝑦𝑖), the vertical distance of this point from the line 𝑦 = 𝑎 + 𝑏𝑥

is the deviation 𝑑𝑖, then

𝒅𝒊 = 𝒚𝒊 − 𝒇(𝒙𝒊) = 𝒚𝒊 − 𝒂 − 𝒃𝒙𝒊.

Applying method of least squares, the values of a and b are so determined as to

minimize

𝑫 = ∑(𝒚𝒊 − 𝒂 − 𝒃𝒙𝒊)𝟐

𝒏

𝒊=𝟏



This will be so if

𝝏𝑫

𝝏𝒂= 𝟎 ⇒ −𝟐 ∑(𝒚𝒊 − 𝒂 − 𝒃𝒙𝒊) = 𝟎

𝒏

𝒊=𝟏

𝝏𝑫

𝝏𝒃= 𝟎 ⇒ −𝟐 ∑ 𝒙𝒊(𝒚𝒊 − 𝒂 − 𝒃𝒙𝒊) = 𝟎

𝒏

𝒊=𝟏

Simplifying and expanding the above equations, we have

∑ 𝒚𝒊 = 𝒂 ∑ 𝟏 + 𝒃 ∑ 𝒙𝒊

𝒏

𝒊=𝟏

,

𝒏

𝒊=𝟏

𝒏

𝒊=𝟏

∑ 𝒙𝒊𝒚𝒊 = 𝒂 ∑ 𝒙𝒊 + 𝒃 ∑ 𝒙𝒊𝟐

𝒏

𝒊=𝟏

𝒏

𝒊=𝟏

𝒏

𝒊=𝟏

Which implies

∑ 𝒚𝒊 = 𝒂𝒏 + 𝒃 ∑ 𝒙𝒊

𝒏

𝒊=𝟏

𝒏

𝒊=𝟏

… (𝟏)

∑ 𝒙𝒊𝒚𝒊 = 𝒂 ∑ 𝒙𝒊 + 𝒃 ∑ 𝒙𝒊𝟐

𝒏

𝒊=𝟏

𝒏

𝒊=𝟏

𝒏

𝒊=𝟏

… (𝟐)

According to the principle of least squares, these partial derivatives must be equal to

zero. Hence, equating zero and simplifying. we obtain following normal equations for

the best fitting straight line y = a + bx.

𝚺𝐱𝐲 = 𝐚𝚺𝐱 + 𝐛𝚺𝐱𝟐

𝚺𝐲 = 𝐧𝐚 + 𝐛𝚺𝐱}

“OR” We obtain following normal equations for the best fitting straight line y = ax + b.

𝚺𝐱𝐲 = 𝐚𝚺𝐱𝟐 + 𝐛𝚺𝐱

𝚺𝐲 = 𝐚𝚺𝐱 + 𝐧𝐛}



Exercise 1 Find the least square straight line to the following data or Find the equation 𝐲 = 𝐚𝐱 + 𝐛 of the

best fitting straight line for the following data.

C Que 1.

[𝐲 = 𝐱 + 𝟏]

X −1 0 1 2

Y 1 0 1 4

T Que 2.

[𝐲 = 𝟏𝟑. 𝟔𝐱]

X 1 2 3 4 5

Y 14 27 40 55 68 May-15

H Que 3.

[𝐲 = 𝟎. 𝟓𝟒𝟓𝟓 + 𝟎. 𝟔𝟑𝟔𝟒𝐱]

X 1 3 4 6 8 9 11 14

Y 1 2 4 4 5 7 8 9

T Que 4.

[𝐲 = 𝟎. 𝟏𝟖𝟕𝟗𝐱 + 𝟐. 𝟐𝟕𝟓𝟗]

X 50 70 100 120

Y 12 15 21 25 Dec-15

H Que 5.

[𝐲 = −𝟒. 𝟐 + 𝟖. 𝟖𝐱]

X 0 1 2 3

Y 0 1 8 27

H Que 6.

[𝐲 = 𝟎. 𝟗𝟒𝟑𝟏𝐱 − 𝟎. 𝟎𝟐𝟒𝟒]

X 2 5 6 9 11

Y 2 4 6 9 10

H Que 7.

[𝐲 = 𝟏. 𝟏𝟎𝟖𝟑 + 𝟎. 𝟕𝟓𝐱]

X 0 1 2

Y 0.5 3 2



T Que 8.

[𝐲 = 𝟕𝟎𝟐. 𝟏𝟕𝟐𝟏 + 𝟑. 𝟑𝟗𝟒𝟖𝐱]

X 20.5 32.7 51.0 73.2 95.7

Y 765 826 873 942 1032

C Que 9.

[𝐲 = 𝟎]

X −5 −3 −1 0 1 2 4

Y 0.4 −0.1 −0.2 −0.3 −0.3 0.1 0.4

C Que 10.

[𝐲 = 𝟎. 𝟒 − 𝟏. 𝟖𝐱]

X −1 0 1 2

Y 1 1 1 −5

T Que 11.

Find y when x = 1.8.

[𝐲 = 𝟎. 𝟕𝟖𝟐𝐱 − 𝟎. 𝟕𝟔𝟗𝟔, 𝐘(𝟏. 𝟖) = 𝟎. 𝟔𝟑𝟖]

X 1 1.2 1.4 1.6

y 0 0.182 0.336 0.47 May-15

Fitting a Parabola 𝐲 = 𝐚𝐱𝟐 + 𝐛𝐱 + 𝐜 by Least Square Approximation

Consider a set of n pairs of the given values (x, y) for fitting the curve 𝐲 = 𝐚𝐱𝟐 + 𝐛𝐱 + 𝐜.

The residual 𝐑 = 𝐲 − (𝐲 = 𝐚𝐱𝟐 + 𝐛𝐱 + 𝐜) is the difference between the observed and

estimated values of y. We have to find a, b, c such that the sum of the squares of the

residuals is minimum (least).

𝐋𝐞𝐭 𝐒 = ∑[𝐲 − (𝐚𝐱𝟐 + 𝐛𝐱 + 𝐜)]𝟐

𝐧

𝟏

… … (𝟏)

Differentiating S with respect to a, b, c and Equating zero

We obtain following normal Equations for the best fitting 𝑦 = 𝑎𝑥2 + 𝑏𝑥 + 𝑐

curve(parabola) of second degree.

𝚺𝐱𝟐𝐲 = 𝐚𝚺𝐱𝟒 + 𝐛𝚺𝐱𝟑 + 𝐜𝚺𝐱𝟐

𝚺𝐱𝐲 = 𝐚𝚺𝐱𝟑 + 𝐛𝚺𝐱𝟐 + 𝐜𝚺𝐱

𝚺𝐲 = 𝐚𝚺𝐱𝟐 + 𝐛𝚺𝐱 + 𝐧𝐜

“Or “ We obtain following normal Equations for the best fitting 𝒚 = 𝒂 + 𝒃𝒙 + 𝒄𝒙𝟐

curve(parabola) of second degree.



𝚺𝐱𝟐𝐲 = 𝐚𝚺𝐱𝟐 + 𝐛𝚺𝐱𝟑 + 𝐜𝚺𝐱𝟒

𝚺𝐱𝐲 = 𝐚𝚺𝐱 + 𝐛𝚺𝐱𝟐 + 𝐜𝚺𝐱𝟑

𝚺𝐲 = 𝐧𝐚 + 𝐛𝚺𝐱 + 𝐜𝚺𝐱𝟐

Exercise-2

Fit a parabola of second degree 𝐲 = 𝐚 + 𝐛𝐱 + 𝐜𝐱𝟐 in the least square sense for the data. Or Fit the

least square parabola to the following data.

C Que 1.

[𝐲 = 𝟎. 𝟓𝟓 + 𝟐. 𝟏𝟓𝐱 − 𝟎. 𝟐𝟓𝐱𝟐]

X −1 0 1 2

y −2 1 2 4

H Que 2.

[𝐲 = 𝟎. 𝟐𝟖𝟓𝟕𝐱𝟐 + 𝟎. 𝟒𝟖𝟓𝟕𝐱 + 𝟗. 𝟒]

X 1 2 3 4 5

Y 10 12 13 16 19

T Que 3.

[𝐲 = 𝟏. 𝟎𝟕𝟕𝟏 + 𝟎. 𝟒𝟏𝟓𝟕𝐱 − 𝟎. 𝟎𝟐𝟏𝟒𝐱𝟐]

X 0 1 2 3 4

Y 1 1.8 1.3 2.5 2.3 May-15

T Que 4.

[𝐲 = 𝟎. 𝟎𝟖𝟑𝐱𝟐 + 𝟒. 𝟗𝟔𝟒𝟐𝐱 + 𝟏𝟑. 𝟒𝟓𝟐𝟑]

X 0 1 2 3 4 5 6

Y 14 18 23 29 36 40 46

H Que 5.

[𝐲 = 𝟐. 𝟐𝟏𝟒𝟐𝐱𝟐 + 𝟎. 𝟐𝟒𝟐𝟖𝐱 + 𝟏. 𝟒𝟐𝟖𝟓]

X 0 1 2 3 4

y 1 5 10 22 38

T Que 6.

[𝐲 = 𝟎. 𝟗𝟗𝟏𝟗𝐱𝟐 − 𝟎. 𝟖𝟓𝟓𝟎𝐱 + 𝟎. 𝟔𝟗𝟔]

X 2 4 6 8 10

y 3.07 12.85 31.47 57.38 91.29



C Que 7.

[𝐲 = 𝟏. 𝟏𝟓 − 𝟏. 𝟎𝟓𝐱 − 𝟎. 𝟐𝟓𝐱𝟐]

X −1 0 1 2

Y 2 1 0 −2

C Que 8.

[𝐲 = 𝟏. 𝟑𝐱 + 𝟎. 𝟓𝐱𝟐]

X −2 −1 0 1 2

Y −1 0 0 1 5

H Que 9.

[𝐲 = −𝟎. 𝟎𝟖𝟗𝟐𝐱𝟐 + 𝟎. 𝟎𝟑𝟕𝐱 + 𝟎. 𝟔𝟏𝟒𝟓]

X −2 −1 0 1 2

Y 0.17 0.53 0.57 0.58 0.33

T Que 10.

Fit a second degree parabola y = ax2 + bx + c in least square sense for the

following data:

X 1 2 3 4 5

Y 10 12 13 16 19

[𝐲 = 𝟎. 𝟐𝟖𝟓𝟕𝐱𝟐 + 𝟎. 𝟒𝟖𝟓𝟕𝐱 + 𝟗. 𝟒]

Dec-15

T Que 11.

Find the least squares approximations of second degree for the following

data:

x -2 -1 0 1 2

Y=f(x) 15 1 1 3 19

[𝐲 =𝟏

𝟑𝟓(−𝟑. 𝟕 + 𝟑𝟓𝐱 + 𝟏𝟓𝟓𝐱𝟐)]

Dec-15



Non-polynomial Approximation OR Non-linear Regression

𝐲 = 𝐚𝐞𝐛𝐱 Yields

Taking Logarithm on both sides log y = loga + bx

Denoting logy = Y and loga = A , the above equation becomes

Y = A + bx Which is a straight line.

From above equation A, b can be found & consequently a=Antilog A can be

calculated.

𝐲 = 𝐚𝐱𝐛 Yields

Taking Logarithm on both sides log y = log a + blog x

Denoting log y = Y , loga = A and log x = X , we obtain

Y = A + bX, which is a straight line.

From above equation A, b can be found & consequently a=Antilog A can be

calculated.

𝐲 = 𝐚𝐛𝐱 Yields

Taking Logarithm on both sides log y = log a + x log b

Denoting logy = Y , log a = A , and log b = B , we obtain

Y = A + Bx , which is a straight line.

From above equation A, B can be found & consequently a=Antilog A and

b=Antilog B can be calculated.

Exercise-3

H Que 1.

Determine a and b so that y = aebx fits the data :

X 1 2 3 4

y 7 11 17 27

[𝐚 = 𝟒. 𝟒𝟔𝟖𝟎, 𝐛 = 𝟎. 𝟏𝟗𝟒𝟖]

T Que 2.

Fit a curve of the form y = abx for the data and hence find the

estimation for y when x = 8.

x 1 2 3 4 5 6 7

y 87 97 113 129 202 195 193

[𝐚 = 𝟕𝟑. 𝟕𝟒𝟏𝟔, 𝐛 = 𝟏. 𝟏𝟔𝟖𝟖, 𝐲(𝟖) = 𝟐𝟓𝟔. 𝟖𝟐𝟑𝟏]



C Que 3.

Fit a curve of the form 𝑦 = 𝑎𝑥𝑏 to the following data

X 20 16 10 11 14

y 22 41 120 89 56

[𝐚 = 𝟐𝟕𝟏𝟖𝟏. 𝟒𝟖𝟏𝟏, 𝐛 = −𝟐. 𝟑𝟔𝟎𝟖 ]

T Que 4.

By the method of least square fit a curve of the form 𝑦 = 𝑎𝑥𝑏 to the

following data:

𝑥 2 3 4 5

𝑦 27.8 62.1 110 161

[𝐚 = 𝟕. 𝟑𝟖𝟖𝟑, 𝐛 = 𝟏. 𝟗𝟐𝟗𝟑 ]

May-16

6. Interpolation PAGE | 59


SR. NO. TOPIC NAME

1. Definition of Operators

2. Relation Between Operators

3. Newton’s Forward Difference Formula

4. Newton’s Backward Difference Formula

5. Gauss’s Forward Difference Formula

6. Gauss’s Backward Difference Formula

7. Stirling’s Formula

8. Newton’s Divided Difference

9. Newton’s Divided Difference Formula

10. Lagrange’s interpolation Formula

Operator

Forward Difference

Backward Difference

Shift Operator

Average Operator

Differencial Operator



Definition of Operator

Relation between Operators

1. 𝐄 = 𝟏 + ∆ (Jun-13, Dec-14,May-16)

Proof

(1 + ∆)f(x) = f(x) + ∆f(x) = f(x) + f(x + h) − f(x) = Ef(x)

2. 𝐄𝛁 = ∆= 𝛁𝐄 = 𝜹𝑬𝟏

𝟐 (Jun-16)

Proof

E∇(f(x)) = E(∇f(x)) = E(f(x) − f(x − h)) = Ef(x) − Ef(x − h)

= f(x + h) − f(x) = ∆f(x)

⟹ E∇(f(x)) = ∆f(x); ∀ f(x)

⟹ E∇= ∆

Forward difference[∆] ∆𝐟(𝐱) = 𝐟(𝐱 + 𝐡) − 𝐟(𝐱)

Backward difference [𝛁] 𝛁𝐟(𝐱) = 𝐟(𝐱) − 𝐟(𝐱 − 𝐡)

Interrelation betn ∆ & 𝛁 ∆𝒚𝒏−𝟏 = 𝛁𝒚𝒏

Shift Operator [𝐄] 𝐄𝐧𝐟(𝐱) = 𝐟(𝐱 + 𝐧𝐡)

Central difference [𝛅]

𝛅𝐟(𝐱) = 𝐟 (𝐱 +𝐡

𝟐) − 𝐟 (𝐱 −

𝐡

𝟐)

𝛅 = 𝐄𝟏𝟐 − 𝐄−

𝟏𝟐

Average Operator [𝛍]

𝛍𝐟(𝐱) =𝟏

𝟐[𝐟 (𝐱 +

𝐡

𝟐) + 𝐟 (𝐱 −

𝐡

𝟐)]

𝛍 =𝐄

𝟏𝟐 + 𝐄−

𝟏𝟐

𝟐

Differential Operator [𝐃] 𝐃𝐟(𝐱) =𝐝

𝐝𝐱𝐟(𝐱) = 𝐟′(𝐱)



Now , 𝛿𝐸1

2 = (𝐸1

2 − 𝐸−1

2) 𝐸1

2 = (𝐸 − 1) = ∆(∵ E = 1 + ∆)

⟹ E∇= ∆= ∇E = 𝛿𝐸1

2

3. ∆𝛁 = ∆ − 𝛁 (Jun-16)

Proof

∆∇(f(x)) = ∆(∇f(x)) = ∆(f(x) − f(x − h)) = ∆f(x) − ∆f(x − h)

= [f(x + h) − f(x)] − [f(x) − f(x − h)]

= ∆f(x) − ∇f(x) = (∆ − ∇)f(x)

⟹ ∆∇(f(x)) = (∆ − ∇)f(x); ∀ f(x)

⟹ ∆∇= ∆ − ∇

4. ∆

𝛁−

𝛁

∆= ∆ + 𝛁

Proof

∆

∇−

∇

∆=

∆2 − ∇2

∆ ∙ ∇=

(∆ − ∇) ∙ (∆ + ∇)

∆ − ∇ (∵ ∆∇= ∆ − ∇)

= ∆ + ∇

5. (𝟏 + ∆)(𝟏 − 𝛁) = 𝟏 (Jun-16)

Proof

(1 + ∆)(1 − ∇)

= 1 − ∇ + ∆ − ∆. ∇

= 1 − ∇ + ∆ − (∆ − ∇) = 1

6. 𝛁 = 𝟏 − 𝐄−𝟏 (Dec-13,Dec-14)

Proof

1 − E−1 = 1 − (1 + ∆)−1(∵ E = 1 + ∆)

= 1 −1

1 + ∆=

1 + ∆ − 1

1 + ∆ =

∆

1 + ∆=

E∇

E= ∇ (∵ ∆= ∇E)

7. 𝐄 = 𝐞𝐡𝐃 (Dec-15)

Proof

Ef(x) = f(x + h) = f(x) + hf ′(X) +h2

2!f ′′(x) + ⋯ (By Taylor’s expansion)

= f(x) + hDf(X) +h2

2!D2f(x) + ⋯



= [1 + hD +h2

2!D2 + ⋯ ] f(x)

⟹ Ef(x) = ehDf(x) ⟹ E = ehD

8. ∆= 𝐞𝐡𝐃 − 𝟏 OR hD = log (1 + ∆) (Dec-14, Dec-15)

Proof

We Know that, E = ehD.

Taking E = 1 + ∆⟹ ∆= ehD − 1 ⟹ hD = log (1 + ∆)

9. 𝛍𝜹 =𝟏

𝟐(∆ + 𝛁) (Jun-16)

Proof

μ𝛿(f(x)) = μ(δf(x)) = μ [f (x +h

2) − f (x −

h

2)] (∵ δf(x) = f (x +

h

2) − f (x −

h

2))

= μf (x +h

2) − μf (x −

h

2)

=1

2[(f(x + h) + f(x)) − (f(x) + f(x − h))] (∵ μf(x) =

1

2[f (x +

h

2) + f (x −

h

2)])

=1

2[(f(x + h) − f(x)) + (f(x) − f(x − h))]

=1

2(∆ + ∇)

⟹ μ𝛿 =1

2(∆ + ∇)



Interpolation

Let the function 𝑦 = 𝑓(𝑥) take the values 𝑦0, 𝑦1, 𝑦2, … , 𝑦𝑛 corresponding to the values

𝑥0, 𝑥1, 𝑥2, … , 𝑥𝑛 of 𝑥. The process of finding the value of 𝑦 corresponding to any value of

𝑥 = 𝑥𝑖 between 𝑥0 and 𝑥𝑛 is called interpolation. Thus, interpolation is a technique of

finding the value of a function for any intermediate value of the independent variable. The

process of computing the value of the function outside the range of given values of the

variable is called extrapolation.

Newton’s Forward Difference Formula

If data are (x0, y0), (x1, y1)(x2, y2), … , (xn, yn)

x0, x1, x2, … , xn are equally spaced then.

𝐟(𝐱) = 𝐲 = 𝐲𝟎 + 𝐩∆𝐲𝟎 +𝐩(𝐩 − 𝟏)

𝟐!∆𝟐𝐲𝟎 +

𝐩(𝐩 − 𝟏)(𝐩 − 𝟐)

𝟑!∆𝟑𝐲𝟎 + ⋯ ; 𝐩 =

𝐱 − 𝐱𝟎

𝐡

Where 𝑥0 is initial point.

Difference Table

𝐱 𝐟(𝐱) = 𝐲 ∆𝐟(𝐱) ∆𝟐𝐟(𝐱) ∆𝟑𝐟(𝐱) ∆𝟒𝐟(𝐱)

𝐱𝟎 y0

∆y0 = y1 − y0

𝐱𝟏 y1 ∆2y0

∆y1 = y2 − y1 ∆3y0

𝐱𝟐 y2 ∆2y1 ∆4y0

∆y2 = y3 − y2 ∆3y1

𝐱𝟑 y3 ∆2y2

∆y3 = y4 − y3

𝐱𝟒 y4



Exercise-1

C Que 1.

Construct forward difference table for following data.

(1,1.1), (2,4.2), (3,9.3), (4,16.4). Nov-16

C Que 2.

Construct Newton’s forward interpolation polynomial for the

following data.

X 4 6 8 10

Y 1 3 8 16

Use it to find the value of y for x = 5.

[𝐲(𝐱) =𝟑𝐱𝟐−𝟐𝟐𝐱+𝟒𝟖

𝟖 , 𝐲(𝟓) =

𝟏𝟑

𝟖]

C Que 3.

Find sin 520 using the following values.

x sin 45° sin 50° sin 55° sin 60°

f(x) 0.7071 0.7660 0.8192 0.8660

[𝟎. 𝟕𝟖𝟖𝟔]

Nov-11

H Que 4.

Use Newton’s forward difference method to find the approximate

value of f(2.3) from the following data.

X 2 4 6 8

f(x) 4.2 8.2 12.2 16.2

[𝟒. 𝟖]

Dec-13

H Que 5.

Use Newton’s forward difference method to find the approximate value

of f(1.3) from the following data.

x 1 2 3 4

f(x) 1.1 4.2 9.3 16.4

[𝟏. 𝟖𝟐]

Jun-13

T Que 6.

Using Newton’s forward formula , find the value of f(1.6),if

x 1 1.4 1.8 2.2

f(x) 3.49 4.82 5.96 6.5

[𝟓. 𝟒𝟔]

Jun-11



H Que 7.

Determine the polynomial by Newton’s forward difference formula

from the following table.

x 0 1 2 3 4 5

y −10 −8 −8 −4 10 40

[𝐱𝟑 − 𝟒𝐱𝟐 + 𝟓𝐱 − 𝟏𝟎]

Jun-12

T Que 8.

Using Newton’s forward interpolation formula ,find the value of

f(218),if

x 100 150 200 250 300 350 400

f(x) 10.63 13.03 15.04 16.81 18.42 19.90 21.27

[𝟏𝟓. 𝟔𝟗𝟗]

Jun -14



Newton’s Backward Difference Formula

If data are (x0, y0), (x1, y1)(x2, y2), … , (xn, yn).


𝐟(𝐱) = 𝐲 = 𝐲𝐧 + 𝐩𝛁𝐲𝐧 +𝐩(𝐩 + 𝟏)

𝟐!𝛁𝟐𝐲𝐧 +

𝐩(𝐩 + 𝟏)(𝐩 + 𝟐)

𝟑!𝛁𝟑𝐲𝐧 + ⋯ ;

Where, p =x − x𝑛

h

Where 𝑥𝑛 is ending point.

Difference Table

𝐱 𝐟(𝐱) = 𝐲 𝛁𝐟(𝐱) 𝛁𝟐𝐟(𝐱) 𝛁𝟑𝐟(𝐱) 𝛁𝟒𝐟(𝐱)

𝐱𝟎 y0

∇y1 = y1 − y0

𝐱𝟏 y1 ∇2y2

∇y2 = y2 − y1 ∇3y3

𝐱𝟐 y2 ∇2y3 ∇4y4

∇y3 = y3 − y2 ∇3y4

𝐱𝟑 y3 ∇2y4

∇y4 = y4 − y3

𝐱𝟒 y4



Exercise-2

C Que 1.

Construct backward difference table from the following data

X 4 6 8 10

Y 1 3 8 16

Nov-15

C Que 2.

The area of circle of diameter d is given by

d 80 85 90 95 100

A 5026 5674 6361 7088 7854

Use suitable interpolation to find area of circle of diameter 98. Also

Calculate the error.

[𝐀 = 𝟕𝟓𝟒𝟑. 𝟎𝟔𝟕𝟐, 𝐄𝐱𝐚𝐜𝐭𝐀 = 𝟕𝟓𝟒𝟐. 𝟗𝟔𝟒𝟎, 𝐄𝐫𝐫𝐨𝐫 = 𝟎. 𝟏𝟎𝟑𝟐]

H Que 3.

Find the cubic polynomial which takes the following values :

y(0) = 1, y(1) = 0, y(2) = 1,and y(3) = 10.Hence, obtain y(4).

[𝐲(𝐱) = 𝐱𝟑 − 𝟐𝐱𝟐 + 𝟏, 𝐲(𝟒) = 𝟑𝟑]

H Que 4.

Consider following tabular values

X 50 100 150 200 250

Y 618 724 805 906 1032

Determine y(300).

[𝟏, 𝟏𝟒𝟖]

Jun-12

Dec-15

T Que 5.

The population of the town is given below. Estimate the population for

the year 1925.

year 1891 1901 1911 1921 1931

Population in

thousand 46 66 81 93 101

[𝟗𝟔. 𝟖𝟑𝟔𝟖]

May-15



T Que 6.

The following table gives distance (in nautical miles) of the visible

horizon for the heights (in feet) above earth’s surface. Find the values of

y when x = 390 feet.

Height(x) 100 150 200 250 300 350 400

Distance(y) 10.63 13.03 15.04 16.81 18.42 19.90 21.47

[𝟐𝟏. 𝟏𝟐𝟎𝟖]

Dec-15

H Que 7.

From the following table of half-yearly premium for policies maturing at

different ages, estimate the premium at the age of 63.

Age 45 50 55 60 65

Premium(in $)

114.84 96.16 83.32 74.48 68.48

[𝟕𝟎. 𝟓𝟖𝟓𝟏]

May-15

H Que 8.

Compute f(x) = ex at x = 0.02 and x = 0.38 Using suitable interpolation

formula for the data given below.

X 0.0 0.1 0.2 0.3 0.4

f(x) 1.0000 1.1052 1.2214 1.3499 1.4918

[𝟏. 𝟎𝟐𝟎𝟐, 𝟏. 𝟒𝟔𝟐𝟑]

T Que 9.

From the following table, find P when t = 1420 c and 1750 c using

appropriate Newton’s interpolation formula.

Temp. t0 C 140 150 160 170 180

Pressure P 3685 4854 6302 8076 10225

[𝐏𝟏𝟒𝟐° = 𝟑𝟖𝟗𝟖. 𝟔𝟔𝟖𝟖 ; 𝐏𝟏𝟕𝟓° = 𝟗𝟏𝟎𝟎. 𝟒𝟖𝟒𝟒]

Dec-14

C Que 10.

The population of the town is given below. Estimate the population for

the year 1895 and 1930 using suitable interpolation.

year 1891 1901 1911 1921 1931

Population

in

thousand

46 66 81 93 101

[𝐏𝐨𝐩𝐮𝐥𝐚𝐭𝐢𝐨𝐧 𝐨𝐟 𝐲𝐞𝐚𝐫 𝟏𝟖𝟗𝟓 = 𝟓𝟒. 𝟖𝟓𝟐𝟖

𝐩𝐨𝐩𝐮𝐥𝐚𝐭𝐢𝐨𝐧 𝐨𝐟 𝐲𝐞𝐚𝐫 𝟏𝟗𝟑𝟎 = 𝟏𝟎𝟎. 𝟒𝟕𝟎𝟓]

May-15



H Que 11.

Compute values of f(0.12) and f(0.40) using suitable interpolation

formula for the following data:

x 0.10 0.15 0.20 0.25 0.30

f(x) 0.1003 0.1511 0.2027 0.2553 0.3093

[𝐟(𝟎. 𝟏𝟐) = 𝟎. 𝟏𝟐𝟎𝟓, 𝐟(𝟎. 𝟒𝟎) = 𝟎. 𝟒𝟐𝟒𝟏]

Dec-15

May-16

T Que 12.

Compute cosh(0.56) & cosh(0.76) from the following table.

x 0.5 0.6 0.7 0.8

cosh x 1.127626 1.185465 1.255169 1.337435

[𝟏. 𝟏𝟔𝟎𝟗𝟒𝟓, 𝟏. 𝟑𝟎𝟐𝟗𝟕𝟕]

Nov-10

Jun-10

C Que 13.

Using Newton’s suitable formula , find the value of f(1.6) & f(2),if

x 1 1.4 1.8 2.2

f(x) 3.49 4.82 5.96 6.5

[𝟓. 𝟒𝟑𝟗𝟒, 𝟔. 𝟑𝟑𝟎𝟔]

Jun-11



Gauss’s Forward Difference Formula

If data are (𝑥−𝑛, 𝑦−𝑛), … , (𝑥−2, 𝑦−2), (𝑥−1, 𝑦−1), (𝑥0, 𝑦0), (𝑥1, 𝑦1), … , (𝑥𝑛, 𝑦𝑛).

𝑥0, 𝑥1, 𝑥2, … , 𝑥𝑛 are equally spaced then.

𝒇(𝒙) = 𝒚 = 𝒚𝟎 + 𝒑∆𝒚𝟎 +𝒑(𝒑 − 𝟏)

𝟐!∆𝟐𝒚−𝟏 +

(𝒑 + 𝟏)𝒑(𝒑 − 𝟏)

𝟑!∆𝟑𝒚−𝟏 + ⋯ ;

Where, p =x − x0

h

Where 𝑥0 is central point.

Difference Table

𝒙 𝒇(𝒙) = 𝒚 ∆𝒇(𝒙) ∆𝟐𝒇(𝒙) ∆𝟑𝒇(𝒙) ∆𝟒𝒇(𝒙)

𝒙−𝟐 𝑦−2

∆𝑦−2

𝒙−𝟏 𝑦−1 ∆2𝑦−2

∆𝑦−1 ∆3𝑦−2

𝒙𝟎 𝑦0 ∆2𝑦−1 ∆4𝑦−2

∆𝑦0 ∆3𝑦−1

𝒙𝟏 𝑦1 ∆2𝑦0

∆𝑦1

𝒙𝟐 𝑦2



Gauss’s Backward Difference Formula

If data are (𝑥−𝑛, 𝑦−𝑛), … , (𝑥−2, 𝑦−2), (𝑥−1, 𝑦−1), (𝑥0, 𝑦0), (𝑥1, 𝑦1), … , (𝑥𝑛, 𝑦𝑛).

𝑥0, 𝑥1, 𝑥2, … , 𝑥𝑛 are equally spaced then.

𝒇(𝒙) = 𝒚 = 𝒚𝟎 + 𝒑∆𝒚−𝟏 +(𝒑 + 𝟏)𝒑

𝟐!∆𝟐𝒚−𝟏 +

(𝒑 + 𝟏)𝒑(𝒑 − 𝟏)

𝟑!∆𝟑𝒚−𝟐 + ⋯ ;

Where, p =x − x0

h


Difference Table

𝒙 𝒇(𝒙) = 𝒚 ∆𝒇(𝒙) ∆𝟐𝒇(𝒙) ∆𝟑𝒇(𝒙) ∆𝟒𝒇(𝒙)

𝒙−𝟐 𝑦−2

∆𝑦−2

𝒙−𝟏 𝑦−1 ∆2𝑦−2

∆𝑦−1 ∆3𝑦−2

𝒙𝟎 𝑦0 ∆2𝑦−1 ∆4𝑦−2

∆𝑦0 ∆3𝑦−1

𝒙𝟏 𝑦1 ∆2𝑦0

∆𝑦1

𝒙𝟐 𝑦2



Stirling’s Formula

If data are (x−n, y−n), … , (x−2, y−2), (x−1, y−1), (x0, y0), (x1, y1), … , (xn, yn).


𝐟(𝐱) = 𝐲 = 𝐲𝟎 + 𝐩 [∆𝐲𝟎 + ∆𝐲−𝟏

𝟐] +

𝐩𝟐

𝟐!∆𝟐𝐲−𝟏 +

𝐩(𝐩𝟐 − 𝟏𝟐)

𝟑![∆𝟑𝐲−𝟏 + ∆𝟑𝐲−𝟐

𝟐]

+𝐩𝟐(𝐩𝟐 − 𝟏𝟐)

𝟒!∆𝟒𝐲−𝟐 +

𝐩(𝐩𝟐 − 𝟏𝟐)(𝐩𝟐 − 𝟐𝟐)

𝟓![∆𝟓𝐲−𝟐 + ∆𝟓𝐲−𝟑

𝟐] + ⋯

Where, p =x − x0

h


Difference Table

𝐱 𝐟(𝐱) = 𝐲 ∆𝐟(𝐱) ∆𝟐𝐟(𝐱) ∆𝟑𝐟(𝐱) ∆𝟒𝐟(𝐱)

𝐱−𝟐 y−2

∆y−2

𝐱−𝟏 y−1 ∆2y−2

∆y−1 ∆3y−2

𝐱𝟎 y0 ∆2y−1 ∆4y−2

∆y0 ∆3y−1

𝐱𝟏 y1 ∆2y0

∆y1

𝐱𝟐 y2



Exercise-3

C Que 1.

Apply Stirling’s formula to compute y(35) from the following table.

X 20 30 40 50

Y 512 439 346 243

[𝟑𝟗𝟒. 𝟔𝟖𝟕𝟓]

Jun-11

H Que 2.

Let f(40) = 836, f(50) = 682, f(60) = 436, f(70) = 272 use

Stirling’s Method to findf(55).

[𝟓𝟔𝟓. 𝟎𝟔𝟐𝟓]

Jun-12

T Que 3.

Using Stirling’s formula find 𝑦35 by using given data.

𝑋 10 20 30 40 50

𝑌 600 512 439 346 243

[𝟑𝟗𝟓. 𝟒𝟐𝟗𝟕]

Method Suitable Value of 𝑝

Gauss Forward Method 0 < p < 1

Gauss Backward Method −1 < p < 0

Stirling’s Method −0.25 < p < 0.2

Newton’s Divided Difference

[𝐱𝟎, 𝐱𝟏] = ⍋𝐲𝟎 = 𝐟(𝐱𝟏)−𝐟(𝐱𝟎)

𝐱𝟏−𝐱𝟎

[𝐱𝟎, 𝐱𝟏, 𝐱𝟐] = ⍋𝟐𝐲𝟎 = [𝐱𝟏,𝐱𝟐]−[𝐱𝟎,𝐱𝟏]

𝐱𝟐−𝐱𝟎

[𝐱𝟎, 𝐱𝟏, … , 𝐱𝐧] = ⍋𝐧𝐲𝟎 =[𝐱𝟏,𝐱𝟐,…,𝐱𝐧]−[𝐱𝟎,𝐱𝟏,…,𝐱𝐧−𝟏]

𝐱𝐧−𝐱𝟎

Newton’s Divided Difference Formula

𝐟(𝐱) = 𝐲 = 𝐲𝟎 + (𝐱 − 𝐱𝟎)[𝐱𝟎, 𝐱𝟏] + (𝐱 − 𝐱𝟎)(𝐱 − 𝐱𝟏)[𝐱𝟎, 𝐱𝟏, 𝐱𝟐] + ⋯



Divided Difference Table

𝐱 𝐟(𝐱) = 𝐲 ⍋𝐟(𝐱) ⍋𝟐𝐟(𝐱) ⍋𝟑𝐟(𝐱) ⍋𝟒𝐟(𝐱)

𝐱𝟎 y0

⍋y0

=y1 − y0

x1 − x0

𝐱𝟏 y1

⍋2y0

=⍋y1 − ⍋y0

x2 − x0

⍋y1

=y2 − y1

x2 − x1

⍋3y0

=⍋2y1 − ⍋2y0

x3 − x0

𝐱𝟐 y2

⍋2y1

=⍋y2 − ⍋y1

x3 − x1

⍋4y0

=⍋3y1 − ⍋3y0

x4 − x0

⍋y2

=y3 − y2

x3 − x2

⍋3y1

=⍋2y2 − ⍋2y1

x4 − x1

𝐱𝟑 y3

⍋2y2

=⍋y3 − ⍋y2

x4 − x2

⍋y3

=y4 − y3

x4 − x3

𝐱𝟒 y4



Exercise-4

C Que 1. Define Divided diff- interpolation formula. Nov-10

C Que 2.

If 𝑓(𝑥) =1

𝑥, find the divided differences [𝑎, 𝑏] and [𝑎, 𝑏, 𝑐].

[−𝟏

𝒂𝒃 𝒂𝒏𝒅

𝟏

𝒂𝒃𝒄]

Jun-10

H Que 3.

From the following table, find Newton’s divided difference formula

𝑥 1 2 7 8

𝑓(𝑥) 1 5 5 4

[𝒇(𝒙) = 𝟏 + 𝟒(𝒙 − 𝟏) −𝟐

𝟑(𝒙𝟐 − 𝟑𝒙 + 𝟐) +

𝟏

𝟏𝟒(𝒙𝟑 − 𝟏𝟎𝒙𝟐 + 𝟐𝟑𝒙 − 𝟏𝟒]

Jun-11

C Que 4.

The Shear stress in kips, per square foot for 5 specimens in a clay

stratum are:

Depth(m) 1.9 3.1 4.2 5.1 5.8

Stress(ksf) 0.3 0.6 0.4 0.9 0.7

Use Newton’s divided difference formula to compute the stress at 4.5m

depth.

[𝟎. 𝟓𝟒𝟏𝟓]

Dec-12

T Que 5.

Write a formula for divided difference [𝑥0 ,𝑥1] & [𝑥0 ,𝑥1,𝑥2] .Using

Newton’s divided difference formula find 𝑓(9.5) from following data

𝑥 8 9 9.2 11

𝑓(𝑥) 2.079442 2.197225 2.219203 2.397895

[𝟐. 𝟐𝟓𝟏𝟐𝟕𝟖]

Dec-13

T Que 6.

Write a formula for divided difference [𝑥0 ,𝑥1] & [𝑥0 ,𝑥1,𝑥2] .Using

Newton’s divided difference formula find 𝑓(10.5) from following data

𝑥 10 11 13 17

𝑓(𝑥) 2.3026 2.3979 2.5649 2.8332

[𝟐. 𝟑𝟓𝟏𝟒]

Jun-13



C Que 7.

Find 𝑓(8) from following data using Newton’s Divided difference formula.

𝑥 4 5 7 10 11 13

𝑓(𝑥) 48 100 294 900 1210 2028

[𝟒𝟒𝟖]

Nov-11

Dec-11

H Que 8.

Compute 𝑓(9.2) from the following value using Newton’s divided

difference.

𝑥 8 9 9.5 11.0

𝑓(𝑥) 2.079442 2.197225 2.251292 2.397895

[𝟐. 𝟐𝟏𝟗𝟐𝟎𝟖]

Jun-10

Nov-10

Dec-15

H Que 9.

Given following data for the polynomial 𝑓(𝑥) = 3𝑥3 − 5𝑥2 + 4𝑥 + 1.

Compute 𝑓(0.3) using Newton‘s divided difference formula.

𝑥 0 1 3 4 7

𝑓(𝑥) 1 3 49 129 813

[𝟏. 𝟖𝟑𝟏]

C Que 10.

Evaluate 𝑓(9), using Newton’s divided difference from the following data

𝑥 5 7 11 13 17

𝑓(𝑥) 150 392 1452 2366 5202

[𝟖𝟏𝟎]

Jun-14

T Que 11.

Construct Divided difference table for the data given below

x -4 -1 0 2 5

f(x) 1245 33 5 9 1335

[⍋𝟒𝐟(𝐱) = 𝟑 𝐨𝐫 [𝐱𝟎, 𝐱𝟏, 𝐱𝟐, 𝐱𝟑, 𝐱𝟒] = 𝟑]

May-15

May-16



H Que 12.

Using Newton’s divided difference formula find f(3) from the following

table.

x -1 2 4 5

F(x) -5 13 255 625

[𝟕𝟏]

Dec-15

Lagrange’s Interpolation Formula


x0, x1, x2, … , xn are unequally spaced then.

𝐲 = 𝐟(𝐱) =(𝐱 − 𝐱𝟏)(𝐱 − 𝐱𝟐) … (𝐱 − 𝐱𝐧)

(𝐱𝟎 − 𝐱𝟏)(𝐱𝟎 − 𝐱𝟐) … (𝐱𝟎 − 𝐱𝐧)𝐲𝟎 +

(𝐱 − 𝐱𝟎)(𝐱 − 𝐱𝟐) … (𝐱 − 𝐱𝐧)

(𝐱𝟏 − 𝐱𝟎)(𝐱𝟏 − 𝐱𝟐) … (𝐱𝟏 − 𝐱𝐧)𝐲𝟏

+ ⋯ +(𝐱 − 𝐱𝟎)(𝐱 − 𝐱𝟏) … (𝐱 − 𝐱𝐧−𝟏)

(𝐱𝐧 − 𝐱𝟎)(𝐱𝐧 − 𝐱𝟏) … (𝐱𝐧 − 𝐱𝐧−𝟏)𝐲𝐧

Proof : Let 𝑦0, 𝑦1, 𝑦2, … , 𝑦𝑛 be the value of 𝑓(𝑥) corresponding to the non-equally spaced

arguments 𝑥0, 𝑥1, 𝑥2, … , 𝑥𝑛.

Constructing a polynomial 𝑝(𝑥) of degree not exceeding 𝑛, satisfying 𝑦𝑘 = 𝑝(𝑥𝑘); 𝑘 =

0, 1,2, … , 𝑛

Let 𝒑(𝒙) = 𝑨𝟎(𝒙 − 𝒙𝟏)(𝒙 − 𝒙𝟐) … (𝒙 − 𝒙𝒏) + 𝑨𝟏(𝒙 − 𝒙𝟎)(𝒙 − 𝒙𝟐) … (𝒙 − 𝒙𝒏) +

𝑨𝟐(𝒙 − 𝒙𝟎)(𝒙 − 𝒙𝟏)(𝒙 − 𝒙𝟑) … (𝒙 − 𝒙𝒏) + ⋯ + 𝑨𝒏(𝒙 − 𝒙𝟎)(𝒙 − 𝒙𝟏)(𝒙 − 𝒙𝟐) … (𝒙 −

𝒙𝒏−𝟏) … (𝟏)

Substituting 𝑥 = 𝑥0 in (1), we have

𝒚𝟎 = 𝒑(𝒙𝟎) = 𝑨𝟎(𝒙𝟎 − 𝒙𝟏)(𝒙𝟎 − 𝒙𝟐) … (𝒙𝟎 − 𝒙𝒏)

∴ 𝑨𝟎 =𝒚𝟎

(𝒙𝟎 − 𝒙𝟏)(𝒙𝟎 − 𝒙𝟐) … (𝒙𝟎 − 𝒙𝒏)

Substituting 𝑥 = 𝑥1 in (1), we have

𝒚𝟏 = 𝒑(𝒙𝟏) = 𝑨𝟏(𝒙𝟏 − 𝒙𝟎)(𝒙𝟏 − 𝒙𝟐) … (𝒙𝟏 − 𝒙𝒏)

∴ 𝑨𝟏 =𝒚𝟏

(𝒙𝟏 − 𝒙𝟎)(𝒙𝟏 − 𝒙𝟐) … (𝒙𝟏 − 𝒙𝒏)

In similar way substituting 𝑥 = 𝑥𝑛 in (1), we have

𝒚𝒏 = 𝒑(𝒙𝒏) = 𝑨𝒏(𝒙𝒏 − 𝒙𝟎)(𝒙𝒏 − 𝒙𝟏) … (𝒙𝟏 − 𝒙𝒏−𝟏)



∴ 𝑨𝒏 =𝒚𝒏

(𝒙𝒏 − 𝒙𝟎)(𝒙𝒏 − 𝒙𝟏) … (𝒙𝟏 − 𝒙𝒏−𝟏)

Substituting all the values of 𝐴𝑘; 𝑘 = 0, 1, 2, … , 𝑛

Equation (1) becomes

𝒑(𝒙) =(𝒙 − 𝒙𝟏)(𝒙 − 𝒙𝟐) … (𝒙 − 𝒙𝒏)

(𝒙𝟎 − 𝒙𝟏)(𝒙𝟎 − 𝒙𝟐) … (𝒙𝟎 − 𝒙𝒏)𝒚𝟎 +

(𝒙 − 𝒙𝟎)(𝒙 − 𝒙𝟐) … (𝒙 − 𝒙𝒏)

(𝒙𝟏 − 𝒙𝟎)(𝒙𝟏 − 𝒙𝟐) … (𝒙𝟏 − 𝒙𝒏)𝒚𝟏

+ ⋯ +(𝒙 − 𝒙𝟎)(𝒙 − 𝒙𝟏)(𝒙 − 𝒙𝟐) … (𝒙 − 𝒙𝒏−𝟏)

(𝒙𝒏 − 𝒙𝟎)(𝒙𝒏 − 𝒙𝟏) … (𝒙𝟏 − 𝒙𝒏−𝟏)𝒚𝒏 … (𝟐)

Which is known as Lagrange’s interpolation formula for unequal interval.

Lagrange’s Inverse Interpolation Formula


x0, x1, x2, … , xn are unequally spaced then.

𝐱 =(𝐲 − 𝐲𝟏)(𝐲 − 𝐲𝟐) … (𝐲 − 𝐲𝐧)

(𝐲𝟎 − 𝐲𝟏)(𝐲𝟎 − 𝐲𝟐) … (𝐲𝟎 − 𝐲𝐧)𝐱𝟎 +

(𝐲 − 𝐲𝟎)(𝐲 − 𝐲𝟐) … (𝐲 − 𝐲𝐧)

(𝐲𝟏 − 𝐲𝟎)(𝐲𝟏 − 𝐲𝟐) … (𝐲𝟏 − 𝐲𝐧)𝐱𝟏 + ⋯

+(𝐲 − 𝐲𝟎)(𝐲 − 𝐲𝟏) … (𝐲 − 𝐲𝐧−𝟏)

(𝐲𝐧 − 𝐲𝟎)(𝐲𝐧 − 𝐲𝟏) … (𝐲𝐧 − 𝐲𝐧−𝟏)𝐱𝐧

Exercise-5

C Que 1.

Explain quadratic Lagrange’s interpolation. Determine the

interpolating polynomial of degree three and Compute f(2) by using

Lagrange’s interpolation method from the following data

x −1 0 1 3

f(x) 2 1 0 −1

[𝟎. 𝟐𝟕𝟐𝟖𝒙𝟑 − 𝟎. 𝟗𝟐𝟒𝟕𝒙𝟐 − 𝟎. 𝟑𝟒𝟖𝒙 + 𝟎. 𝟗𝟗𝟗𝟗, 𝑭(𝟐) = −𝟎. 𝟕𝟓]

Jun-13 Dec-15 Dec-16 Jun-10

H Que 2.

Find the Lagrange’s interpolating polynomial from the following

data. Also, Compute𝑓(4).

x 2 3 5 7

f(x) 0.1506 0.3001 0.4517 0.6259

[𝟎. 𝟎𝟎𝟓𝟓𝐱𝟑 − 𝟎. 𝟎𝟕𝟗𝟖𝒙𝟐 + 𝟎. 𝟒𝟒𝟒𝟓𝒙 − 𝟎. 𝟒𝟔𝟑𝟖, 𝑭(𝟒) = 𝟎. 𝟑𝟖𝟗𝟔]

Dec-12



C Que 3.

Explain quadratic Lagrange’s interpolation .Compute f(9.2) by using

Lagrange’s interpolation method from the following data

x 9 9.5 11

f(x) 2.1972 2.2513 2.3979

[𝟐. 𝟐𝟏𝟗𝟐]

Dec-13

T Que 4.

Using Lagrange’s formula of fit a polynomial to the data.

x −1 0 2 3

y 8 3 1 12

And hence find y(2).

𝐲(𝐱) =𝟏

𝟑[𝟐𝐱𝟑 + 𝟐𝐱𝟐 − 𝟏𝟓𝐱 + 𝟗], 𝐲(𝟐) = 𝟏

DEC-11

T Que 5.

Find the Lagrange’s interpolating polynomial from following data.

x 0 1 4 5

y 1 3 24 39

[𝐏(𝐱) =𝟏

𝟐𝟎[𝟑𝐱𝟑 + 𝟏𝟎𝐱𝟐 + 𝟐𝟕𝐱 + 𝟐𝟎]]

Nov-10

C Que 6.

From the following data find value of x when y = f(x) = 0.39.

x 20 25 30

Y=f(x) 0.342 0.423 0.500

[𝐱 = 𝟐𝟐. 𝟗𝟑𝟎𝟖]

Dec-15

C Que 7.

Apply Lagrange’s formula to find a root of the equationf(x) = 13.00.

x 44 45 46 47

f(x) 13.40 13.16 12.93 12.68

[𝟒𝟓. 𝟔𝟗]

T Que 8.

Apply Lagrange’s formula to find a root of the equationf(x) = 0.

x 30 34 38 42

f(x) −30 −13 3 18

[𝟑𝟕. 𝟐𝟑]



H Que 9.

Find y(9) by Lagrange’s Interpolation formula from following values.

x 5 7 11 13 17

f(x) 150 392 1452 2366 5202

[𝟖𝟏𝟎]

Jun -14

T Que 10.

Find y(12) by Lagrange’s Interpolation formula from following

values.

x 11 13 14 18 20 23

y 25 47 68 82 102 124

[𝐲(𝟏𝟐) = 𝟐𝟔. 𝟒𝟏]

Dec-14

C Que 11.

By Lagrange’s interpolation formula, Obtain the value of 𝑓(x) = 85.

[𝟒. 𝟔𝟏𝟒𝟏]

x 2 5 8 14

y 94.8 97.9 81.3 68.7 Dec-15

H Que 12.

Find the Lagrange’s interpolation polynomial from the following

data. Also find f(2).

[𝒇(𝒙) =𝟏

𝟐𝟎(𝟑𝒙𝟐 + 𝟏𝟎𝒙𝟐 + 𝟐𝟕𝒙 + 𝟐𝟎), 𝒇(𝟐) = 𝟔. 𝟗]

x 0 1 4 5

f(x) 1 3 24 39

Dec-15

T Que 13.

Find the value of tan 330 by Lagrange’s formula If, tan 300 = 0.5774, tan 320 = 0.6249 , tan 350 = 0.7002, tan 380 = 0.7813.

[𝟎. 𝟔𝟒𝟗𝟒]

May-16

C Que 14.

Express the function 3𝑥2−12𝑥+11

(𝑥−1)(𝑥−2)(𝑥−3) as a sum of partial fractions,

using Lagrange’s Formula.

[𝟏

𝒙 − 𝟏+

𝟏

𝒙 − 𝟐+

𝟏

𝒙 − 𝟑]

Jun-11



H Que 15.

Express the function 2𝑥2+3𝑥+5

(𝑥−1)(𝑥+2)(𝑥−2) as a sum of partial fractions,

using Lagrange’s formula.

[−𝟏𝟎

𝟑(𝒙 − 𝟏)+

𝟕

𝟏𝟐(𝒙 + 𝟐)+

𝟏𝟗

𝟒(𝒙 − 𝟐)]

7. Numerical Integration PAGE | 82

N.S.M. -2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

SR. NO. TOPIC NAME

1 Newton-Cotes Formula

1 Trapezoidal Rule

2 Simpson’s 1/3 – Rule

3 Simpson’s 3/8 – Rule

4 Weddle’s Rule

5 Gaussian Integration(Gaussian Quadrature)

Methods to find

Integraton

Trapezoidal Rule

Weddles Rule

Simpson's 3/8 Rule

Simpson's 1/3 Rule

Gaussian Integration



Newton-Cotes Formula

∫ 𝐟(𝐱)𝐝𝐱𝐛

𝐚

= 𝐡 [𝐲𝟎 +𝐧

𝟐 𝚫𝐲𝟎 +

𝐧(𝟐𝐧 − 𝟑)

𝟏𝟐𝚫𝟐𝐲𝟎 +

𝐧(𝐧 − 𝟐)𝟐

𝟐𝟒𝚫𝟑𝐲𝟎 + ⋯ ]

𝐖𝐡𝐞𝐫𝐞, 𝐡 =𝐛 − 𝐚

𝐧

Trapezoidal Rule (If 𝐧 is a multiple of 𝟏)


𝐚

=𝐡

𝟐[(𝐲𝟎 + 𝐲𝐧) + 𝟐(𝐲𝟏 + 𝐲𝟐 + ⋯ + 𝐲𝐧−𝟏)]

=𝐡

𝟐[(𝐅𝐢𝐫𝐬𝐭 𝐭𝐞𝐫𝐦 + 𝐋𝐚𝐬𝐭 𝐭𝐞𝐫𝐦) + 𝟐(𝐑𝐞𝐦𝐚𝐢𝐧𝐢𝐧𝐠 𝐚𝐥𝐥 𝐓𝐞𝐫𝐦𝐬)]

Where n is number of subinterval & h =b − a

n

Error bounds:

If 𝑓′′(𝑥) is continuous in [𝑎, 𝑏],then the error in the trapezoidal rule is no longer then

|𝒆𝒓𝒓𝒐𝒓| ≤(𝒃 − 𝒂)

𝟏𝟐𝒉𝟐|𝒇′′(𝑴)|

Where |𝑓′′(𝑀)|is the largest value of |𝑓′′(𝑥)| in [𝑎, 𝑏].

Exercise-1

T Que.1 State Trapezoidal rule with n = 10 and evaluate ∫ e−x2

dx.1

0

[𝟎. 𝟕𝟒𝟔𝟐]

Nov-10

C Que.2 Write Trapezoidal rule for numerical integration. Jun-13

C Que.3 State Trapezoidal rule with n = 10 and evaluate ∫ ex1

0dx.

[𝟏. 𝟕𝟏𝟗𝟕]

Jun-11

H Que.4 State Trapezoidal rule with n = 10 and using it, evaluate ∫ 2ex1

0 dx.

[ 𝟑. 𝟒𝟑𝟗𝟒 ]

Dec-14

T Que.5 Evaluate ∫

dx

1+x2

1

0using trapezoidal rule with h = 0.2.

[𝟎. 𝟕𝟖𝟑𝟕]

Nov-11

Jun-11

Dec-15

H Que.6 Evaluate ∫

dx

1+x2

1

0using trapezoidal rule.

[𝟎. 𝟕𝟖𝟓𝟒]

May-15



C Que.7

Given the data below, find the isothermal work done on the gas as it

is compressed from v1 = 22L to v2 = 2L use w = − ∫ Pv2

v1 dv

V,L 2 7 12 17 22 P(atm.) 12.20 3.49 2.04 1.44 1.11

Use Trapezoidal rule. [𝟔𝟖. 𝟏𝟐𝟓]

Dec-12

H Que.8

Consider the following tabular values. x 25 25.1 25.2 25.3 25.4 25.5 25.6

y = f(x) 3.205 3.217 3.232 3.245 3.256 3.268 3.28

Determine the area bounded by the given curve and X-axis between x = 25 to x = 25.6 by Trapezoidal rule.

[𝟏. 𝟗𝟒𝟔𝟏]

Jun -12


dx

1+x2

6

0 by using Trapezoidal rule taking h = 1.

[𝟏. 𝟒𝟏𝟎𝟖]

Jun -14

C Que.10

Evaluate ∫ sin x dxπ

0 , taking n = 10.

[𝟏. 𝟗𝟖𝟑𝟓] Dec-15

T Que.11

Evaluate ∫ log10 x5

1 dx taking 8 subintervals by Trapezoidal rule.

[𝟏. 𝟕𝟓𝟎𝟓]

Dec-15

T Que.12

Use Trapezoidal rule to evaluate ∫ 𝑥31

0𝑑𝑥 considering five sub intervals.

[𝟎. 𝟐𝟔] Jun -16

Simpson’s 𝟏 𝟑⁄ - Rule ( If 𝐧 is a multiple of 𝟐)


𝐚

=𝐡

𝟑[(𝐲𝟎 + 𝐲𝐧) + 𝟒(𝐲𝟏 + 𝐲𝟑 + ⋯ ) + 𝟐(𝐲𝟐 + 𝐲𝟒 + ⋯ )]

=𝐡

𝟑[(𝐅𝐢𝐫𝐬𝐭 𝐭𝐞𝐫𝐦 + 𝐋𝐚𝐬𝐭 𝐭𝐞𝐫𝐦) + 𝟒(𝐎𝐝𝐝 𝐓𝐞𝐫𝐦𝐬) + 𝟐(𝐄𝐯𝐞𝐧 𝐓𝐞𝐫𝐦𝐬)]

𝐖𝐡𝐞𝐫𝐞 𝐧 𝐢𝐬 𝐧𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐬𝐮𝐛𝐢𝐧𝐭𝐞𝐫𝐯𝐚𝐥 & 𝐡 =𝐛 − 𝐚

𝐧

Error bounds :

If 𝑓4(𝑥) is continuous in [𝑎, 𝑏],then the error in the Simpson’s rule is no longer then


𝟏𝟖𝟎𝒉𝟒|𝒇𝟒(𝑴)|



Where |𝑓4(𝑀)|is the largest value of |𝑓4(𝑥)| in [𝑎, 𝑏].

Exercise-2

C Que.1 Evaluate ∫

dx

1+x

6

0 taking h = 1using Simpson’s

1

3 rule.Hence Obtain an

approximate value of loge 7. [𝟏. 𝟗𝟓𝟖𝟖]

Jun-11

T Que.2

Derive Trapezoidal rule and Evaluate ∫ ex21.3

0.5dx by using

Simpson’s 1

3

rd rule.

[𝟐. 𝟎𝟕𝟔𝟐]

Dec-15

H Que.3

Using Simpson’s 1

3 rule evaluate ∫ f(x)dx

2.5

1from the following data.

Take h = 0.3. X 1 1.3 1.6 1.9 2.2 2.5

f(x) 1 1.69 2.56 3.61 4.84 6.25

[𝟒. 𝟑𝟐𝟓]

Jun-13

H Que.4

Using Simpson’s13⁄ rule evaluate ∫ f(x)dx

6

0from following data.

h = 1.

x 0 1 2 3 4 5 6

f(x) 1 0.5 0.3333 0.25 0.2 0.1666 0.1428

[𝟏. 𝟗𝟓𝟖𝟔]

Dec-13

T Que.5

The speed , v meter per second , of a car , t seconds after it starts, is shown in the following table

t 0 12 24 36 48 60 72 84 96 108 120

v 0 3.6 10.08 18.90 21.

6 18.54 10.26 4.5 4.5 5.4 9

Using Simpson’s 1/3 rule, find the distance travelled by the car in

2minutes.

[𝟏𝟐𝟐𝟐. 𝟓𝟔]

Jun-10 Dec-10

C Que.6

A river is 80 meter wide. The depth ‘d’ in meters at a distance x meters from one bank is given by the following table, calculate the

area of cross-section of the river using Simpson’s 1

3 rule

x 0 10 20 30 40 50 60 70 80 d 0 4 7 9 12 15 14 8 3

[𝟕𝟏𝟎]

Dec-11 May-15

T Que.7

Consider the following tabular values. Find ∫ ydx 16

10by Simpson

1

3 rule.

x 10 11 12 13 14 15 16 y 1.02 0.94 0.89 0.79 0.71 0.62 0.55

[𝟒. 𝟕𝟐𝟑𝟑]

Jun-12



C Que.8 Evaluate ∫ (1 + x2)

3

26

−2 dx using Simpson’s

1

3 rule with taking 6

subintervals. Use four digits after decimal point for calculation. [𝟑𝟔𝟎. 𝟏𝟖𝟑𝟎]

Dec-12


1

1+x2

6

0dx by using Simpson’s

1

3 rule taking h = 1.

[𝟏. 𝟑𝟔𝟔𝟐] Jun-14

H Que.10

Evaluate ∫ sin x dxπ

0, Take n = 10

[𝟐. 𝟎𝟎𝟎𝟏] Dec-15

T Que.11

The velocity 𝜈 of a particle at distance 𝑠 from point on its path is

given by the Following table Find the time taken to travel 60 meter

Using Simpson’s 1/3 rule (Use =𝑑𝑠

𝑑𝑡 ).

[𝟑𝟒𝟓𝟎]

𝑠(𝑚𝑒𝑡𝑒𝑟) 0 10 20 30 40 50 60

𝜈(𝑚𝑒𝑡𝑒𝑟/sec) 47 58 64 65 61 52 38 May-15

T Que.12

Velocity of a car running on a straight road at intervals of 2 minutes

are given below:

Time 0 2 4 6 8 10 12

Velocity 0 22 30 27 18 7 0

Apply Simpson’s rule to find the distance covered by the car.

[𝟐𝟏𝟑. 𝟑𝟑]

Dec-16

Simpson’s 𝟑 𝟖⁄ - Rule ( If 𝐧 is a multiple of 𝟑.)


𝐚

=𝟑𝐡

𝟖[(𝐲𝟎 + 𝐲𝐧) + 𝟐(𝐲𝟑 + 𝐲𝟔 + ⋯ ) + 𝟑(𝐲𝟏 + 𝐲𝟐 + 𝐲𝟒 + 𝐲𝟓 + ⋯ )]

=𝟑𝐡

𝟖[(𝐅𝐢𝐫𝐬𝐭 𝐭𝐞𝐫𝐦 + 𝐋𝐚𝐬𝐭 𝐭𝐞𝐫𝐦) + 𝟐(𝐌𝐮𝐥𝐭𝐢𝐩𝐥𝐞 𝐨𝐟 𝟑) + 𝟑(𝐍𝐨𝐭 𝐌𝐮𝐥𝐭𝐢𝐩𝐥𝐞 𝐨𝐟 𝟑)]


𝐧

Error bounds:

If 𝑓4(𝑥) is continuous in [𝑎, 𝑏],then the error in the Simpson’s 3/8 rule is no longer then


𝟖𝟎𝒉𝟒|𝒇𝟒(𝑴)|



Where |𝑓4(𝑀)| is the largest value of |𝑓4(𝑥)| in [𝑎, 𝑏].

Exercise-3 C Que.1 Write the Simpson’s

3

8 rule for numerical integration. Dec-13

T Que.2

Evaluate ∫dx

1+x

3

0 with n = 6 by using Simpson’s

3

8 rule and hence

calculate loge2.Estimate the bound of error involved in the process.

[𝟏. 𝟑𝟖𝟖𝟖, 𝟎. 𝟎𝟓𝟔𝟑]

Jun-10 Jun-14

H Que.3

State Simpson’s 3

8 rule and evaluate ∫

1

1+x2

1

0dx taking h =

1

6.and also

by Simpson’s 1

3 rule taking h =

1

4.

[𝟎. 𝟕𝟖𝟓𝟒 ]

Dec-10 May-15 Dec-15

C Que.4

Evaluate the integral ∫ logex dx5.2

4 using Simpson’s

3

8 rule.

[𝟏. 𝟖𝟐𝟕𝟖]

Dec-11

C Que.5

Dividing the range into 10 equal part, evaluate ∫ sin xπ

0 dx by

simpson’s 3

8 rule. [𝟏. 𝟗𝟖𝟕𝟕]

May-15

T Que.6 Evaluate ∫

dx

1+x

1

0 using Simpson’s

3

8 rule. [𝟎. 𝟔𝟗𝟑𝟕] Dec-15

T Que.7

The speed, v m/s, of a car, t seconds after it starts, is shown in table:

t 0 12 24 36 48 60 72 84 96 108 120

v 0 3.6 10.08 18.90 21.60 18.54 10.26 5.40 4.50 5.40 9.00

Using Simpson’s 1/3rd rule find the distance travelled by the car in 2

minutes.

. [𝟕𝟎. 𝟏𝟔𝟓𝟐]

May-16



H Que.8

A train is moving at the speed of 30 m/sec. Suddenly brakes are

applied. The speed Of the train per second after t seconds is given by

The following table. Apply Simpson’s three-eight rule to determine

the distance moved by the train in 30

[𝟏𝟎𝟐𝟑. 𝟕𝟓]

Time(t) 0 5 10 15 20 25 30

Speed(v) 30 24 19 16 13 11 10 May-16

H Que.9

Find ∫𝑒𝑥

1+𝑥

6

0𝑑𝑥 approximately using Simpson’s 3/8th rule with ℎ = 1.

[𝟏𝟐𝟑𝟔. 𝟗𝟔]

May-16

Weddle’s Rule (If 𝐧 is multiple of 6.)

∫ 𝐟(𝐱)𝐛

𝐚

𝐝𝐱 =𝟑𝐡

𝟏𝟎[(𝐲𝟎+𝐲𝐧) + (𝟓𝐲𝟏 + 𝐲𝟐 + 𝟔𝐲𝟑 + 𝐲𝟒 + 𝟓𝐲𝟓) + (𝟐𝐲𝟔 + 𝟓𝐲𝟕 + 𝐲𝟖 + 𝟔𝐲𝟗 + 𝐲𝟏𝟎 + 𝟓𝐲𝟏𝟏)

+ ⋯ + (𝟐𝐲𝐧−𝟔 + 𝟓𝐲𝐧−𝟓 + 𝐲𝐧−𝟒 + 𝟔𝐲𝐧−𝟑 + 𝐲𝐧−𝟐 + 𝟓𝐲𝐧−𝟏)]


𝐧

Exercise-4

C Que.1

Consider the following tabular values.

x 25 25.1 25.2 25.3 25.4 25.5 25.6

f(x) 3.205 3.217 3.232 3.245 3.256 3.268 3.28

Determine the area bounded by the given curve and X-axis

between x = 25 to x = 25 .6 by Weddle’s rule.

[𝟏. 𝟗𝟒𝟔𝟎]

Jun -12

H Que.2

Consider the following tabular values. Find∫ ydx 16

10by Weddle’s

rule.

x 10 11 12 13 14 15 16

y 1.02 0.94 0.89 0.79 0.71 0.62 0.55

[𝟒. 𝟕𝟏𝟑]

Jun -12



Gaussian integration (Gaussian Quadrature)

One Point Gaussian Quadrature Formula (n = 1)

∫ 𝐟(𝐱)𝐝𝐱

𝟏

−𝟏

= 𝟐𝐟(𝟎)

Two Point Gaussian Quadrature Formula (n = 2)

∫ 𝐟(𝐱)𝐝𝐱 = 𝐟 (−𝟏

√𝟑)

𝟏

−𝟏

+ 𝐟 (𝟏

√𝟑)

Three Point Gaussian Quadrature Formula (n = 3)

∫ 𝐟(𝐱)𝐝𝐱

𝟏

−𝟏

=𝟖

𝟗𝐟(𝟎) +

𝟓

𝟗(𝐟 (−√

𝟑

𝟓) + 𝐟 (√

𝟑

𝟓))

You can also use following formula to find Gaussian Quadrature.


𝐚

=𝐛 − 𝐚

𝟐(𝐰𝟏𝐟(𝐲𝟏) + 𝐰𝟐 𝐟(𝐲𝟐)+ ⋯ + 𝐰𝐧 𝐟(𝐲𝐧)), 𝐖𝐡𝐞𝐫𝐞 𝐱 =

𝐛 − 𝐚

𝟐𝐲 +

𝐛 + 𝐚

𝟐

Table

𝑛 𝑤𝑖 𝑦𝑖 𝑛 𝑤𝑖 𝑦𝑖

1 2.0000 0.0000 0.65214 0.33998

2 1.0000 −0.57735 0.34785 0.86114

1.0000 0.57735 5 0.23693 −0.90618

3 0.55555 −0.77460 0.47863 −0.53847

0.88889 0.00000 0.56889 0.00000

0.55555 0.77460 0.47863 0.53847

4 0.34785 −0.86114 0.23693 0.90618

0.65214 −0.33998



Exercise-5

C Que.1

Evaluate ∫𝑑𝑥

1+𝑥2

1

−1 by using Gaussian quadrature formula with one

point, two points & three points.

[𝟐, 𝟏. 𝟓, 𝟏. 𝟓𝟖𝟑𝟑𝟑]

Dec-15

C Que.2

Evaluate 𝐼 = ∫𝑑𝑥

1+𝑥

1

0 by Gaussian formula with one point, two-point

and three- points.

[𝟎. 𝟔𝟔𝟔𝟔𝟕, 𝟎. 𝟔𝟗𝟐𝟑𝟏, 𝟎. 𝟔𝟗𝟑𝟏𝟐]

Nov-16

Dec-11

Dec-15

T Que.3

Evaluate ∫ 𝑒−𝑥21

0𝑑𝑥 by Gauss integration formula with 𝑛 = 3.

[𝟎. 𝟕𝟒𝟔𝟖𝟏]

Jun-10

Dec-14

May-15

T Que.4

Evaluate ∫ 𝑠𝑖𝑛𝑥3

1 𝑑𝑥 using Gauss Quadrature of five points. Compare

the result with analytic value.

[𝟏. 𝟓𝟑𝟎𝟑𝟏, 𝟏. 𝟓𝟑𝟎𝟐𝟗]

Nov-10

H Que.5

Evaluate integral ∫ (1 + 𝑥2)3

2⁄6

−2 𝑑𝑥 by the Gaussian formula for

𝑛 = 3.

[𝟑𝟓𝟖. 𝟔𝟗𝟐𝟑𝟔]

Dec-12

May-16

8. Linear Algebric Equation PAGE | 91


Matrix Equation

The matrix notation for following linear system of equation is as follow:

𝐚𝟏𝟏𝐱𝟏 + 𝐚𝟏𝟐𝐱𝟐 + 𝐚𝟏𝟑𝐱𝟑 + ⋯+ 𝐚𝟏𝐧𝐱𝐧 = 𝐛𝟏 𝐚𝟐𝟏𝐱𝟏 + 𝐚𝟐𝟐𝐱𝟐 + 𝐚𝟐𝟑𝐱𝟑 + ⋯+ 𝐚𝟐𝐧𝐱𝐧 = 𝐛𝟐𝐚𝟑𝟏𝐱𝟏 + 𝐚𝟑𝟐𝐱𝟐 + 𝐚𝟑𝟑𝐱𝟑 + ⋯+ 𝐚𝟑𝐧𝐱𝐧 = 𝐛𝟑…………………………………………………

𝐚𝐦𝟏𝐱𝟏 + 𝐚𝐦𝟐𝐱𝟐 + 𝐚𝐦𝟑𝐱𝟑 + ⋯+ 𝐚𝐦𝐧𝐱𝐧 = 𝐛𝐦}

Here 𝐀 =

[ 𝐚𝟏𝟏 𝐚𝟏𝟐 𝐚𝟏𝟑𝐚𝟐𝟏 𝐚𝟐𝟐 𝐚𝟐𝟑𝐚𝟑𝟏 𝐚𝟑𝟐 𝐚𝟑𝟑

⋯

𝐚𝟏𝐧𝐚𝟐𝐧𝐚𝟑𝐧

⋮ ⋱ ⋮𝐚𝐦𝟏 𝐚𝐦𝟐 𝐚𝐦𝟑 ⋯ 𝐚𝐦𝐧]

𝐁 =

[ 𝐛𝟏𝐛𝟐𝐛𝟑⋮𝐛𝐦]

and 𝐗 =

[ 𝐱𝟏𝐱𝟐𝐱𝟑⋮𝐱𝐧]

The above linear system is expressed in the matrix form as A ∙ X = B.

Elementary Transformation or Operation on a Matrix

Operation Meaning

Rij or Ri ↔ Rj Interchange of ith and jth rows

k ∙ Ri Multiplication of all the elements of ith row by non zero scalar k.

Rij(k) or Rj + k ∙ Ri Multiplication of all the elements of ith row by nonzero scalar k and added into jth row.

Row Echelon Form of Matrix

To convert the matrix into row echelon form follow the following steps:

1. Every zero row of the matrix occurs below the non zero rows.

2. Arrange all the rows in strictly decreasing order.

3. Make all the entries zero below the leading (first non zero entry of the row)

element of 1st row.

4. Repeat step-3 for each row.

Reduced Row Echelon Form of Matrix:

To convert the matrix into reduced row echelon form follow the following steps:

1. Convert given matrix into row echelon form.

2. Make all leading elements 1(one).

3. Make all the entries zero above the leading element 1(one) of each row.



Numerical Methods For Solultion Of A Linear Equation

1. Direct Methods

2. Iterative Methods

Direct Methods

This method produce the exact solution after a finite number of steps but are subject to

errors due to round-off and other factors.

We will discuss two direct methods :

1. Gauss Elimination method

2. Gauss-Jordan method

Indirect Method(Iterative Method)

In this method, an approximation to the true solution is assumed initially to start method. By

applying the method repeteadly, better and better approximation are obtained. For large systems,

iterative methods are faster than direct methods and roud-off error are also smaller. Any

1. Gauss seidel method

2. Gauss jacobi method

Gauss Elimination Method

To solve the given linear system using Gauss elimination method, follow the following steps:

1. Start with augmented matrix [A ∶ B].

2. Convert matrix A into row echelon form with leading element of each row is

one(1).

3. Apply back substitution for getting equations.

4. Solve the equations and find the unknown variables (i.e. solution).

Exercise-1 Solve the following system of equations by Gauss-elimination method.

H Que. 1 x + y + z = 9,2x − 3y + 4z = 13,3x + 4y + 5z = 40.

[𝟏, 𝟑, 𝟓 ] Jun-11

C Que. 2 2x + y + z = 10 , 3x + 2y + 3z = 18 , x + 4y + 9z = 16.

[𝟕, −𝟗, 𝟓]

T Que. 3 x + 2y + z = 3,2x + 3y + 3z = 10,3x − y + 2z = 13.

[𝟐, −𝟏, 𝟑]

H Que. 4 2x + 3y − z = 5,4x + 4y − 3z = 3,2x − 3y + 2z = 2.

[𝟏, 𝟐, 𝟑]

T Que. 5 2x + y − z = 1,5x + 2y + 2z = −4,3x + y + z = 5.

[𝟏𝟒,−𝟑𝟐,−𝟓] Dec-12

C Que. 6 8y + 2z = −7,3x + 5y + 2z = 8,6x + 2y + 8z = 26.

[𝟒, −𝟏,𝟏

𝟐]

Jun-14

H Que. 7 x + 4y − z = −5, x + y − 6z = −12,3x − y − z = 4.

[𝟏. 𝟔𝟒𝟕𝟕,−𝟏. 𝟏𝟒𝟎𝟖, 𝟐. 𝟎𝟖𝟒𝟓] May-15



H Que. 8 x + y + 2z = 4,3x + y − 3z = −4,2x − 3y − 5z = −5

[𝟏, −𝟏, 𝟐] May-15

C Que. 9

The following system of equations was generated by applying mess current law to the circuit. Use Gauss Elimination method to find the current in the circuit. 10𝑥1 + 𝑥2 + 𝑥3 = 12, 2𝑥1 + 2𝑥2 + 10𝑥3 =14, 2𝑥1 + 10𝑥2 + 𝑥3 = 15

[𝟎. 𝟗𝟖𝟑𝟏, 𝟏. 𝟐𝟎𝟕𝟐, 𝟎. 𝟗𝟔𝟏𝟗]

May-15

H Que. 10 𝑥1 + 2𝑥2 + 3𝑥3 = 10, 6𝑥1 + 5𝑥2 + 2𝑥3 = 30, 𝑥1 + 3𝑥2 + 𝑥3 = 10

[𝟑, 𝟐, 𝟏] May-16

T Que. 11

The following system of equations was generated by applying mess current law to the circuit. Use Gauss Elimination method to find the current in the circuit. 2𝐼1 − 𝐼2 + 3𝐼3 = 8, − 𝐼1 + 2𝐼2 + 𝐼3 = 4, 3𝐼1 +𝐼2 − 4𝐼3 = 0

[𝟐, 𝟐, 𝟐]

June-16

H Que. 12 Solve the following system of equations by Gauss elimination method: 3𝑥 + 𝑦 − 𝑧 = 3; 2𝑥 − 8𝑦 + 𝑧 = −5; 𝑥 − 2𝑦 + 9𝑧 = 8.

[𝟏, 𝟏, 𝟏] Nov-16

H Que. 13 Solve by Gauss Elimination method 𝑥 + 2𝑦 + 𝑧 = 3, 2𝑥 + 3𝑦 + 3𝑧 = 10, 3𝑥 − 𝑦 + 2𝑧 = 13.

[𝟐, −𝟏, 𝟑] Nov-16

Gauss Elimination Method With Partial Pivoting

To solve the given linear system using Gauss elimination method with partial pivoting, follow the following steps:

1. Find largest absolute value(pivot element) in first column.

2. Make the pivot element row to first row.

3. Eliminate x1 below the pivot element.

4. Again find pivot element in 2nd and 3rd row.

5. Make the pivot element row to second row.

6. Eliminate x2 below the pivot element.

7. Apply back substitution for getting equations.

8. Solve the equations and find the unknown variables (i.e. solution).

Exercise-2 Solve the following system of equations using partial pivoting by Gauss-elimination method. (With

Partial Pivoting )

C Que. 1 8x2 + 2x3 = −7 , 3x1 + 5x2 + 2x3 = 8 , 6x1 + 2x2 + 8x3 = 26.

[𝟒, −𝟏,𝟏

𝟐]

Dec-10 Jun-10

T Que. 2 x + y + z = 7 , 3x + 3y + 4z = 24 , 2x + y + 3z = 16.

[𝟑, 𝟏, 𝟑] Nov-11 Dec-15

H Que. 3 2x1 + 2x2 − 2x3 = 8 , −4x1 − 2x2 + 2x3 = −14 −2x1 + 3x2 + 9x3 = 9.

[𝟑, 𝟐, 𝟏]



T Que. 4 2x1 + 2x2 + x3 = 6,4x1 + 2x2 + 3x3 = 4, x1 + x2 + x3 = 0

[𝟓, 𝟏, −𝟔] Jun-15 Dec-15

Gauss-Jordan Method

This method is modification of the gauss elimination method. This method solves a given

system of equation by transforming the coefficient matrix into unit matrix.

Steps to solve Gauss-Jordan method:

1. Write the matrix form of the system of equations.

2. Write the augmented matrix.

3. Reduce the coefficient matrix to unit matrix by applying elementary row

transformations to the augmented matrix.

4. Write the corresponding linear system of equations to obtain the solution.

Exercise-3 Solve the following system of equations by Gauss-Jordan method

C Que. 1 10𝑥 + 𝑦 + 𝑧 = 12, 2𝑥 + 10𝑦 + 𝑧 = 13, 𝑥 + 𝑦 + 𝑧 = 7

[𝟏, 𝟏, 𝟏, ] May-16

H Que. 2 𝑥 + 2𝑦 + 𝑧 = 8 , 2𝑥 + 3𝑦 + 4𝑧 = 20 , 4𝑥 + 3𝑦 + 2𝑧 = 16.

[𝟏, 𝟐, 𝟑]

H Que. 3 10𝑥1 + 𝑥2 + 𝑥3 = 12 , 𝑥1 + 10𝑥2 − 𝑥3 = 10 , 𝑥1 − 2𝑥2 + 10𝑥3 = 9.

[𝟏, 𝟏, 𝟏]

T Que. 4 2𝑥 + 5𝑦 − 3𝑧 = 1 , 5𝑥 + 𝑦 + 4𝑧 = 2 , 7𝑥 + 3𝑦 + 𝑧 = 4.

[𝟓𝟑

𝟔𝟗,−𝟐𝟑

𝟔𝟗,−𝟐𝟔

𝟔𝟗]



Gauss Jacobi Method

This method is applicable to the system of equations in which leading diagonal elements of the coefficient matrix are dominant (large in magnitude) in their respective rows.

Consider the system of equations.

𝐚𝟏𝐱 + 𝐛𝟏𝐲 + 𝐜𝟏𝐳 = 𝐝𝟏

𝐚𝟐𝐱 + 𝐛𝟐𝐲 + 𝐜𝟐𝐳 = 𝐝𝟐

𝐚𝟑𝐱 + 𝐛𝟑𝐲 + 𝐜𝟑𝐳 = 𝐝𝟑

Where co-efficient matrix 𝐀 must be diagonally dominant,

|𝐚𝟏| ≥ |𝐛𝟏| + |𝐜𝟏|

|𝐛𝟐| ≥ |𝐚𝟐| + |𝐜𝟐|

|𝐜𝟑| ≥ |𝐚𝟑| + |𝐛𝟑| …… (𝟏)

And the inequality is strictly greater than for at least one row.

Solving the system (𝟏) for 𝐱, 𝐲, 𝐳 respectively,we obtain

𝐱 =𝟏

𝐚𝟏(𝐝𝟏 − 𝐛𝟏𝐲 − 𝐜𝟏𝐳)

𝐲 =𝟏

𝐛𝟐(𝐝𝟐 − 𝐚𝟐𝐱 − 𝐜𝟐𝐳)

𝐳 =𝟏

𝐜𝟑(𝐝𝟑 − 𝐚𝟑𝐱 − 𝐛𝟑𝐲)…… (𝟐)

We start with 𝐱𝟎 = 𝟎, 𝐲𝟎 = 𝟎 & 𝐳𝟎 = 𝟎 in equ.(𝟐)

∴ 𝐱𝟏 =𝟏

𝐚𝟏(𝐝𝟏 − 𝐛𝟏𝐲𝟎 − 𝐜𝟏𝐳𝟎)

∴ 𝐲𝟏 =𝟏

𝐛𝟐(𝐝𝟐 − 𝐚𝟐𝐱𝟎 − 𝐜𝟐𝐳𝟎)

∴ 𝐳𝟏 =𝟏

𝐜𝟑(𝐝𝟑 − 𝐚𝟑𝐱𝟎 − 𝐛𝟑𝐲𝟎)

Again substituting these value x1, y1, z1 in Eq. (2), the next approximation is obtained.

This process is continued till the values of 𝐱, 𝐲, 𝐳 are obtained to desired degree of accuracy.



Gauss Seidel Method

This is a modification of Gauss-Jacobi method. In this method we replace the approximation by the corresponding new ones as soon as they are calculated.

Consider the system of equations.

𝐚𝟏𝐱 + 𝐛𝟏𝐲 + 𝐜𝟏𝐳 = 𝐝𝟏

𝐚𝟐𝐱 + 𝐛𝟐𝐲 + 𝐜𝟐𝐳 = 𝐝𝟐

𝐚𝟑𝐱 + 𝐛𝟑𝐲 + 𝐜𝟑𝐳 = 𝐝𝟑

Where co-efficient matrix 𝐀 must be diagonally dominant,

|𝐚𝟏| ≥ |𝐛𝟏| + |𝐜𝟏|

|𝐛𝟐| ≥ |𝐚𝟐| + |𝐜2|

|𝐜𝟑| ≥ |𝐚𝟑| + |𝐛𝟑| …… (𝟏)

And the inequality is strictly greater than for at least one row.

Solving the system (𝟏) for 𝐱, 𝐲, 𝐳 respectively,we obtain

𝐱 =𝟏

𝐚𝟏(𝐝𝟏 − 𝐛𝟏𝐲 − 𝐜𝟏𝐳)

𝐲 =𝟏

𝐛𝟐(𝐝𝟐 − 𝐚𝟐𝐱 − 𝐜𝟐𝐳)

𝐳 =𝟏

𝐜𝟑(𝐝𝟑 − 𝐚𝟑𝐱 − 𝐛𝟑𝐲)…… (𝟐)

We start with 𝐱𝟎 = 𝟎, 𝐲𝟎 = 𝟎 & 𝐳𝟎 = 𝟎 in equ.(𝟐)

∴ 𝐱𝟏 =𝟏

𝐚𝟏(𝐝𝟏 − 𝐛𝟏𝐲𝟎 − 𝐜𝟏𝐳𝟎)

Now substituting 𝐱 = 𝐱𝟏 & 𝐳 = 𝐳𝟎 in the second equ. Of (𝟐)

∴ 𝐲𝟏 =𝟏

𝐛𝟐(𝐝𝟐 − 𝐚𝟐𝐱𝟏 − 𝐜𝟐𝐳𝟎)

Now substituting 𝐱 = 𝐱𝟏 & 𝐲 = 𝐲1 in the third equ. Of (𝟐)

∴ 𝐳𝟏 =𝟏

𝐜𝟑(𝐝𝟑 − 𝐚𝟑𝐱𝟏 − 𝐛𝟑𝐲𝟏)

This process is continued till the values of 𝐱, 𝐲, 𝐳 are obtained to desired degree of accuracy.



Exercise-4

C Que. 1 Solve the following system of equations by Gauss-Seidel method. 10𝑥1 + 𝑥2 + 𝑥3 = 6 , 𝑥1 + 10𝑥2 + 𝑥3 = 6 , 𝑥1 + 𝑥2 + 10𝑥3 = 6.

[𝟎. 𝟓, 𝟎. 𝟓, 𝟎. 𝟓]

Jun-10 Dec-15 Nov-16

T Que. 2 Use Gauss seidel method to determine roots of the following equations. 2𝑥 − 𝑦 = 3 , 𝑥 + 2𝑦 + 𝑧 = 3 , −𝑥 + 𝑧 = 3.

[𝟏, −𝟏, 𝟒] Jun-13

H Que. 3 Use Gauss seidel method to find roots of the following equations. 8𝑥 + 𝑦 + 𝑧 = 5 , 𝑥 + 8𝑦 + 𝑧 = 5 , 𝑥 + 𝑦 + 8𝑧 = 5.

[𝟎. 𝟓, 𝟎. 𝟓, 𝟎. 𝟓]

Dec-13 May-15

H Que. 4

State the Direct & iterative method to solve system of linear equations. Arrange following system of equations into diagonally dominant form and solve it using Gauss Seidel method. 10𝑥1 + 𝑥2 + 𝑥3 = 12 , 2𝑥1 + 10𝑥2 + 𝑥3 = 13, 2𝑥1 + 2𝑥2 + 10𝑥3 = 14.

[𝟏, 𝟏, 𝟏]

Dec-10 Dec-15 June-16

C Que. 5

Solve by Gauss-Seidel & Gauss-Jacobi method correct up to two decimal places. 20𝑥 + 2𝑦 + 𝑧 = 30, 𝑥 − 40𝑦 + 3𝑧 = −75,2𝑥 − 𝑦 + 10𝑧 = 30.

[𝟏. 𝟏𝟒, 𝟐. 𝟏𝟑, 𝟐. 𝟗𝟗]

Jun-11

H Que. 6

Solve this system of linear equations using Jacobi’s method in three iterations first check the co-efficient matrix of the following systems is diagonally dominant or not? 20𝑥 + 𝑦 − 2𝑧 = 17, 2𝑥 − 3𝑦 + 20𝑧 = 25, 3𝑥 + 20𝑦 − 𝑧 = −18

[𝟏, −𝟏, 𝟏]

Dec-15

H Que. 7 Solve the following system of equations by Gauss-Seidel method. 20𝑥 + 𝑦 − 2𝑧 = 17, 2𝑥 − 3𝑦 + 20𝑧 = 25, 3𝑥 + 20𝑦 − 𝑧 = −18

[𝟏, −𝟏, 𝟏] Jun-14

C Que. 8 Solve by Gauss-Seidel method correct up to three decimal places. 2𝑥 + 𝑦 + 54𝑧 = 110,27 𝑥 + 6𝑦 − 𝑧 = 85,6𝑥 + 15𝑦 + 2𝑧 = 72.

[𝟐. 𝟒𝟐𝟐, 𝟑. 𝟓𝟖𝟎, 𝟏. 𝟖𝟖𝟏] Nov-11

H Que. 9 Solve by Gauss-Seidel Method. 9𝑥 + 2𝑦 + 4𝑧 = 20, 𝑥 + 10𝑦 + 4𝑧 = 6,2𝑥 − 4𝑦 + 10𝑧 = −15.

[𝟐. 𝟕𝟒, 𝟎. 𝟗𝟗,−𝟏. 𝟔𝟓]

H Que. 10 Solve by Gauss-Seidel method correct up to three decimal places. 10𝑥 − 5𝑦 − 2𝑧 = 3,4𝑥 − 10𝑦 + 3𝑧 = −3, 𝑥 + 6𝑦 + 10𝑧 = −3.

[𝟎. 𝟑𝟒𝟐, 𝟎. 𝟐𝟖𝟓,−𝟎. 𝟓𝟎𝟓]

T Que. 11

Check whether the following system is diagonally dominant or not. If not, re-arrange the equations so that it becomes diagonally dominant and hence solve the system of simultaneous linear equation by Gauss sidle Method. −100𝑦 + 130𝑧 = 230,−40𝑥 + 150𝑦 − 100𝑧 = 0,60𝑥 − 40𝑦 = 200.

[𝟕. 𝟕𝟖, 𝟔. 𝟔𝟕, 𝟔. 𝟗𝟎]

Dec-12

H Que. 12

Solve the following system of equations using Gauss-Seidel method correct up to three decimal places. 60𝑥 − 4𝑦 + 6𝑧 = 150 ; 2𝑥 + 2𝑦 + 18𝑧 = 30 ; 𝑥 + 17𝑦 − 2𝑧 = 48

[𝟐. 𝟓𝟖𝟎, 𝟐. 𝟕𝟗𝟖, 𝟏. 𝟎𝟔𝟗]

Dec-13



T Que. 13 State diagonal dominant property .Using Gauss-seidel method solve 6𝑥 + 𝑦 + 𝑧 = 105 ; 4𝑥 + 8𝑦 + 3𝑧 = 155 ; 5𝑥 + 4𝑦 − 10𝑧 = 65

[𝟏𝟓, 𝟏𝟎, 𝟓] May-15

T Que. 14

By gauss Seidel method solve the following system upto six iteration 12𝑥1 + 3𝑥2 − 5𝑥3 = 1 ; 𝑥1 + 5𝑥2 + 3𝑥3 = 28 ; 3𝑥1 + 7𝑥2 + 13𝑥3 = 76 Use initial condition (𝑥1 𝑥2 x3) = (1 0 1).

[𝟏, 𝟑, 𝟒]

May-15

H Que. 15 By gauss Seidel method solve the following system 2x + y + 6z = 9 ; 8x + 3y + 2z = 13 ; x + 5y + z = 7

[𝟏, 𝟏, 𝟏] May-15

H Que. 16

State the Direct and iterative methods to solve system of linear equations. Using Gauss-Seidel method, Solve up to 3 iteration 2x1 − x2 = 7 ; −x1 + 2x2 − x3 = 1 ; −x2 + 2x3 = 1

[𝟓. 𝟑𝟏𝟐𝟓, 𝟒. 𝟑𝟏𝟐𝟓, 𝟐. 𝟔𝟓𝟔𝟑]

Dec-15

H Que. 17 Solve the following system of equations by Gauss – Seidel method 10𝑥 − 𝑦 − 𝑧 = 13, 𝑥 + 10𝑦 + 𝑧 = 36, 𝑥 + 𝑦 − 10𝑧 = −35

[𝟐, 𝟑, 𝟒] May-16

C Que. 18 Use Gauss Seidel method to solve 83𝑥 + 11𝑦 − 4𝑧 = 95, 7𝑥 + 52𝑦 + 13𝑧 = 104, 3𝑥 + 8𝑦 + 29𝑧 = 71

[𝟏. 𝟎𝟔, 𝟏. 𝟑𝟕, 𝟏. 𝟗𝟔] May-16

T Que. 19

Using Gauss-Seidel iteration method solve the system of equations: 10𝑥 − 2𝑦 − 𝑧 − 𝑤 = 3, −2𝑥 + 10𝑦 − 𝑧 − 𝑤 = 15, −𝑥 − 𝑦 + 10𝑧 − 2𝑤 = 27, −x − y − 2z + 10w = −9.

[𝟏, 𝟐, 𝟑, 𝟎]

Nov-16

9. Roots Of Non-Linear Equation PAGE | 99


Introduction

An expression of the form 𝒇(𝒙) = 𝒂𝟎𝒙𝒏 + 𝒂𝟏𝒙𝒏−𝟏 + 𝒂𝟐𝒙𝒏−𝟐 + ⋯ + 𝒂𝒏−𝟏𝒙 + 𝒂𝒏,

where 𝒂𝟎, 𝒂𝟏, 𝒂𝟐 … 𝒂𝒏 are constants and 𝒏 is a positive integer , is called an algebraic

polynomial of degree n if 𝒂𝟎 ≠ 𝟎. The equation 𝒇(𝒙) = 𝟎 is called an algebraic

equation if 𝒇(𝒙) is an algebraic polynomial, e.g. 𝒙𝟑 − 𝟒𝒙 − 𝟗 = 𝟎. If 𝒇(𝒙) contains

functions such as trigonometric, logarithmic, exponential, etc. , then 𝒇(𝒙) is called a

transcendental equation, e.g. , 𝟐𝒙𝟑 − 𝐥𝐨𝐠(𝒙 + 𝟑) 𝒕𝒂𝒏𝒙 + 𝒆𝒙 = 𝟎.

In general, an equation is solved by factorization. But in many cases, the method of

factorization fails. In such cases, numerical methods are used. There are some

methods to solve the equation 𝒇(𝒙) = 𝟎, which are given as follows.

SR.NO. TOPIC NAME

1. Bisection Method

2. Secant Method

3. Regula-Falsi Method (False Position Method)

4. Newton-Raphson Method

Methods to find Roots of equations

Bisection MethodRegula-Falsi Method

Secant Method Newton-Raphson Method



Bisection Method

Let , an equation which is f(x) = 0

If f(a) > 0 and f(b) < 0 ,Where a and b are consecutive integer, then

x1 =a + b

2

Check f(x1) > 0 OR f(x1) < 0.

If f(x1) > 0,then x2 =x1+b

2

OR

If f(x1) < 0, then we find x2 =a+x1

2.

Check f(x2) > 0 OR f(x2) < 0.

If f(x1) > 0, f(x2) > 0 then x3 =x2+b

2 OR f(x1) < 0, f(x2) > 0 then x3 =

x2+x1

2

OR

If f(x1) > 0, f(x2) < 0 then x3 =x1+x2

2 OR f(x1) < 0, f(x2) < 0 then x3 =

a+x2

2

Processing like this when latest two consecutive values of x are not same.

Exercise-1

C Que.1 Discuss the method Bisection to find the root of an equation 𝑓(𝑥) = 0. Nov-16

C Que.2

Find the positive root of x = cos x correct up to three decimal places

by bisection method.

[𝟎. 𝟕𝟑𝟗]

Jun-10

Jun-14

C Que.3

Explain bisection method for solution of equation. Using this method

find the approximate solution x3 + x − 1 = 0 of correct up to three

decimal points.

[𝟎. 𝟔𝟖𝟑]

Dec-13

H Que.4

Perform three iterations of Bisection method to obtain a root of the

equation 𝑓(𝑥) = cos 𝑥 − 𝑥𝑒𝑥 = 0 in the interval (0.5, 1).

[𝟎. 𝟓𝟑𝟏𝟐]

Nov-16

T Que.5

Perform the five iterations of the bisection method to obtain a root of

the equation f(x) = cos x − xex = 0.

[𝟎. 𝟓𝟑𝟏𝟐]

Nov-10



T Que.6

Find root of equation x3 − 4x − 9 = 0, using the bisection method in

four stages.

[𝟐. 𝟔𝟖𝟕𝟓]

Jun-11

H Que.7

Perform the five iteration of the bisection method to obtain a root of

the equation x3 − x − 1 = 0.

[𝟏. 𝟑𝟒𝟑𝟕𝟓]

Nov-11

C Que.8

Find the negative root of x3 − 7x + 3 = 0 bisection method up to

three decimal place.

[−𝟐. 𝟖𝟑𝟗]

Jun-12

T Que.9

Use bisection method to find a root of equation

x3 + 4x2 − 10 = 0 in the interval [1,2].Use four iteration.

[𝟏. 𝟑𝟏𝟐𝟓]

Dec-12

H Que.10

Perform three iterations of Bisection method to obtain root of the

equation 2 sin x − x = 0.

[𝟏. 𝟖𝟕𝟓]

May-15

T Que.11

Explain bisection method for solving an equation f(x) = 0.Find the real

root of equation x2 − 4x − 10 = 0 by using this method correct to

three decimal places.

[𝟓. 𝟕𝟒𝟐]

Dec-15

May-16

Method of False Position or Regula-Falsi Method

This is the oldest method to finding the real root of an equation

𝑓(𝑥) = 0 and closely resembles the bisection method.

Here we choose two points 𝑥0 and 𝑥1 such that 𝑓(𝑥0) and 𝑓(𝑥1) are of opposite signs i.e.

the graph of 𝑦 = 𝑓(𝑥) crosses the x-axis between these points. This indicates that a root

lies between 𝑥0 and 𝑥1 and consequently 𝑓(𝑥0)𝑓(𝑥1) < 0.

Equation of the chord joining the points 𝐴[𝑥0, 𝑓(𝑥0)] and 𝐵[𝑥1, 𝑓(𝑥1)] is

𝑦 − 𝑓(𝑥0) =𝑓(𝑥1)−𝑓(𝑥0)

𝑥1−𝑥0(𝑥 − 𝑥0)



Fig. 2.6

The method consists in replacing the curve AB by means of the chord AB and taking the

point of intersection of the chord with the x-axis as an approximation to the root. So the

abscissa of the point where the chord cuts the x-axis (𝑦 = 0) is given by

𝒙𝟐 = 𝒙𝟎 −𝒙𝟏−𝒙𝟎

𝒇(𝒙𝟏)−𝒇(𝒙𝟎)𝒇(𝒙𝟎) … (1)

Which is an approximation to the root.

If now 𝑓(𝑥0) and 𝑓(𝑥2) are of opposite signs, then the root lies between 𝑥0 and 𝑥2. So

replacing 𝑥1 by 𝑥2 in (1) , we obtain the next approximation 𝑥3. (The root could as well

lie between 𝑥1 and 𝑥2 and we would obtain 𝑥3 accordingly ). This procedure is repeated

till the root is found to the desired accuracy. The iteration process based on (1) is known

as the method of false position and its rate of convergence is faster than that of the

bisection method.

Secant Method

This method is an improvement over the method of false position as it does not require

the condition 𝑓(𝑥0)𝑓(𝑥1) < 0 of that method. Here also the graph of the function 𝑦 = 𝑓(𝑥)

is approximated by a secant line but at each iteration, two most recent approximations to

the root are used to find the next approximation. Also it is not necessary that the interval

must contain the root.

Taking 𝑥0 , 𝑥1 as the initial limits of the interval, we write the equation of the chord joining

these as

𝒚 − 𝒇(𝒙𝟏) =𝒇(𝒙𝟏) − 𝒇(𝒙𝟎)

𝒙𝟏 − 𝒙𝟎(𝒙 − 𝒙𝟏)



Then the abscissa of the point where it crosses the 𝑋 −axis (𝑦 = 0) is given by

𝒙𝟐 = 𝒙𝟏 −𝒙𝟏−𝒙𝟎

𝒇(𝒙𝟏)−𝒇(𝒙𝟎)𝒇(𝒙𝟏)

Which is an approximation to the root. The general formula for successive

approximations is, therefore, given by

𝒙𝒏+𝟏 = 𝒙𝒏 −𝒙𝒏−𝒙𝒏−𝟏

𝒇(𝒙𝒏)−𝒇(𝒙𝒏−𝟏)𝒇(𝒙𝒏) , 𝑛 ≥ 1.

Note :- If at any iteration 𝑓(𝑥𝑛) = 𝑓(𝑥𝑛−1), this method fails and shows that it does not

converge necessarily. This is drawback of secant method over the method of false

position which always converges

Exercise-2

C Que.1

Find a positive root of 𝑥𝑒𝑥 − 2 = 0 by the method of False

position.

[𝟎. 𝟖𝟓𝟐𝟔]

Nov-16

C Que.2

Apply False Position method to find the negative root of the equation x3 − 2x + 5 = 0 correct to four decimal places.

[−𝟐. 𝟎𝟗𝟒𝟓] May-15

H Que.3

Find a root of the equation x3 − 4x − 9 = 0 using False-position method correct up to three decimal.

[𝟐. 𝟕𝟎𝟔𝟓] Dec-15



T Que.4

Find an approximate value of the root the equation 𝑥3 + 𝑥 − 1 = 0

using the method of false position.

[𝟎. 𝟔𝟕𝟐]

Nov-16

H Que.5

Explain False position method for finding the root of the equation f(x) = 0.Use this method to find the root of an equation x = e−x correct up to three decimal places.

[𝟎. 𝟓𝟔𝟕]

Dec-15

Nov-16

T Que.6 Using method of False-position, compute the real root of the equation x𝑙𝑜𝑔10x − 1.2 = 0 correct to four decimals. where(𝑥 > 0)

[𝟐. 𝟕𝟒𝟎𝟐𝟏]

Dec-15

C Que.7 Find the positive solution of f(x) = x − 2sinx = 0 by the secant method, starting form x0 = 2, x1 = 1.9.

[𝟏. 𝟖𝟗𝟓𝟓]

Nov-10

Jun-14

T Que.8 Derive Secant method and solve xex − 1 = 0 correct up to three decimal places between 0 and 1.

[𝟎. 𝟓𝟔𝟕]

Jun-12

H Que.9

Find the real root of the following by secant method. a) x2 − 4x − 10 = 0 (using x0 = 4, x1 = 2,upto six iteration) b) x3 − 2x − 5 = 0 (using x0 = 2, x1 = 3, upto four iteration)

[𝟓. 𝟕𝟒𝟏𝟏, 𝟐. 𝟎𝟗𝟐𝟖]

Nov-16

C Que.10 Use Secant method to find the roots of cos x − xex = 0 correct upto 3 decimal places of decimal. [𝟎. 𝟓𝟏𝟖]

May-15

May-16

Nov-16

Newton-Raphson Method (Newton’s Method)

Let , an equation which is f(x) = 0

If f(a) ∙ f(b) < 0

x0 = a when |f(a)| < |f(b)| OR x0 = b when |f(b)| < |f(a)|. then

𝐱𝐧+𝟏 = 𝐱𝐧 −𝐟(𝐱𝐧)

𝐟′(𝐱𝐧); 𝐧 = 𝟎, 𝟏, 𝟐, 𝟑 …

Where f ′(xn) ≠ 0

Processing like this when latest two consecutive values of x are not same.

Explanation:

Le x1be the root of f(x) = 0 and x0 be an approximation to x1. If h = x1 − x0, then by

Taylor’s Series,

𝐟(𝐱𝟎 + 𝐡) = 𝐟(𝐱𝟎) + 𝐡 𝐟′(𝐱𝟎) +𝐡𝟐

𝟐! 𝐟′′(𝐱𝟎) + ⋯



Since, x1 = x0 + h is root. f(x1) = f(x0 + h) = 0

If h is chosen too small enough, then we can neglect 2nd , 3rd and higher powers of h.

We have,

𝟎 = 𝐟(𝐱𝟎) + 𝐡 𝐟′(𝐱𝟎) ⟹ 𝐡 = −𝐟(𝐱𝟎)

𝐟′(𝐱𝟎) ; 𝐟′(𝐱𝟎) ≠ 𝟎.

Suppose that, x1 = x0 + h be the better approximation.

⟹ 𝐱𝟏 = 𝐱𝟎 −𝐟(𝐱𝟎)

𝐟′(𝐱𝟎)

By repeating the process,

⟹ 𝐱𝟐 = 𝐱𝟏 −𝐟(𝐱𝟏)

𝐟′(𝐱𝟏)

⟹ 𝐱𝟑 = 𝐱𝟐 −𝐟(𝐱𝟐)

𝐟′(𝐱𝟐)

In general,


𝐟′(𝐱𝐧); 𝐧 = 𝟎, 𝟏, 𝟐, 𝟑 …

Where f ′(xn) ≠ 0

This is called Newton-Raphson Formula.

Note

If the function is linear ,( f ′(xn) = 0)then N-R method has to be failed.

Prove that NR procedure is second order convergent.

Or

Rate of convergence of the NR methods.

Proof : Let 𝛼 be exact root of 𝑓(𝑥) = 0 and let 𝑥𝑛,𝑥𝑛+1 be two successive approximations

to the actual root. If ∈𝑛 and ∈𝑛+1 are the corresponding errors then

𝒙𝒏= 𝜶 +∈𝒏

𝒙𝒏+𝟏= 𝜶 +∈𝒏+𝟏

Substituting in equation of ‘NR METHOD FORMULA’

𝜶 +∈𝒏+𝟏 = 𝜶 +∈𝒏 −𝒇(𝜶 +∈𝒏 )

𝒇′(𝜶 +∈𝒏 )

∈𝒏+𝟏=∈𝒏 −𝒇(𝜶 +∈𝒏 )

𝒇′(𝜶 +∈𝒏 )



=∈𝒏 −𝒇(𝜶) +∈𝒏 𝒇′(𝜶) +

∈𝒏𝟐

𝟐! 𝒇′′(𝜶) + ⋯

𝒇′(𝜶) +∈𝒏 𝒇′′(𝜶) + ⋯

=∈𝒏 −∈𝒏 𝒇′(𝜶) +

∈𝒏𝟐

𝟐! 𝒇′′(𝜶) + ⋯

𝒇′(𝜶) +∈𝒏 𝒇′′(𝜶)

(∵ 𝒇(𝜶) = 𝟎)

Neglecting the derivative of order higher then two

∈𝒏+𝟏=∈𝒏 −∈𝒏 𝒇′(𝜶) +

∈𝒏𝟐

𝟐! 𝒇′′(𝜶) + ⋯

𝒇′(𝜶) +∈𝒏 𝒇′′(𝜶) + ⋯

=𝟏

𝟐[

∈𝒏𝟐 𝒇′′(𝜶)

𝒇′(𝜶) +∈𝒏 𝒇′′(𝜶)]

=∈𝒏

𝟐

𝟐[

𝒇"(𝜶)𝒇′(𝜶)

𝟏 +∈𝒏 𝒇"(𝜶)𝒇′(𝜶)

]

=∈𝒏

𝟐

𝟐

𝒇"(𝜶)

𝒇′(𝜶)

The last equation shows that the error at stage is proportional to the squares of the error

in the previous stage. Hence, the NR method has a quadratic convergence and the

convergence is of the order 2

Formula for 𝒒𝒕𝒉 root

Let, = 𝑵𝟏

𝒒 , where 𝑞 = 1, 2, 3, … … 𝑛 and N is an natural number.

⟹ xq − N = 0 ⟹ f(x) = xq − N

⟹ f ′(x) = qxq−1

By the general formula we get,



𝐱𝐧+𝟏 =𝟏

𝒒[(𝒒 − 𝟏)𝒙𝒏 +

𝑵

𝒙𝒏𝒒−𝟏

]

Where 𝑛 = 0, 1, 2 … …

Find the iterative formula for √𝐍 and 𝟏

𝐍 by N-R method.

Formula for √𝐍

Let, 𝐱 = √𝐍 ⟹ 𝐱𝟐 − 𝐍 = 𝟎 ⟹ 𝐟(𝐱) = 𝐱𝟐 − 𝐍

⟹ 𝐟′(𝐱) = 𝟐𝐱

By N-R formula,


𝐟′(𝐱𝐧)-

= 𝐱𝐧 −𝐱𝐧

𝟐 − 𝐍

𝟐𝐱𝐧=

𝟐𝐱𝐧𝟐 − 𝐱𝐧

𝟐 + 𝐍

𝟐𝐱𝐧

=𝐱𝐧

𝟐 + 𝐍

𝟐𝐱𝐧=

𝟏

𝟐(𝐱𝐧 +

𝐍

𝐱𝐧)

Formula for 𝟏

𝐍

Let, 𝐱 =𝟏

𝐍⟹

𝟏

𝐱− 𝐍 = −𝟎 ⟹ 𝐟(𝐱) =

𝟏

𝐱− 𝐍

⟹ 𝐟′(𝐱) = −𝟏

𝐱𝟐

By N-R formula,


𝐟′(𝐱𝐧)= 𝐱𝐧 −

𝟏

𝐱𝐧−𝐍

−𝟏

𝐱𝐧𝟐

= 𝐱𝐧 + (𝟏

𝐱𝐧− 𝐍) 𝐱𝐧

𝟐 = 𝟐𝐱𝐧 − 𝐍𝐱𝐧𝟐

𝐱𝐧+𝟏 = 𝐱𝐧(𝟐 − 𝐍𝐱𝐧) , where 𝑛 = 0, 1, 2, … …



Exercise-3

C Que.1 Derive the Newton Raphson iterative scheme by drawing appropriate figure.

Nov-10 May-15

C Que.2 Obtain Newton-Raphson formula from Taylor’s theorem. Jun-12

C Que.3

Explain Newton’s method for solving equation f(x) = 0. Apply this method to find the approximate solution of x3 + x − 1 = 0 correct up to three decimal.

[𝟎. 𝟔𝟖𝟐]

Jun-13

T Que.4

Find the real root of the equation 𝑥4 − 𝑥 − 9 = 0 by Newton-Raphson

method, correct to three decimal places.

[𝟏. 𝟖𝟏𝟑]

Nov-16

H Que.5 Use Newton-Raphson method to find smallest positive root of f(x) = x3 − 5x + 1 = 0 correct to four decimals.

[𝟎. 𝟐𝟎𝟏𝟔𝟒]

Dec-15

C Que.6 Find the positive root of x = cos x correct up to three decimal places by N-R method.

[𝟎. 𝟕𝟑𝟗]

Jun-11

T Que.7 Find to four decimal places, the smallest root of the equation sin x = e−x Using the N-R starting with x0 = 0.6.

[𝟎. 𝟓𝟖𝟖𝟓]

Dec-11

H Que.8 Using Newton-Raphson method find a root of the equation xex = 2 Correct to three decimal places.

[𝟎. 𝟓𝟏𝟖]

Dec-15

C Que.9 Find a root of x4 − x3 + 10x + 7 = 0 correct up to three decimal places between a = −2 & b = −1 by N-R method.

[−𝟏. 𝟒𝟓𝟒]

Jun-12

T Que.10 Find a zero of function f(x) = x3 − cos x with starting point x0 = 1 by NR Method could x0 = 0 be used for this problem ?

[𝟎. 𝟖𝟔𝟓𝟓]

Dec-12

T Que.11 Discuss the rate of convergence of NR Method. Jun-12

H Que.12

Explain Newton’s method for solving equation f(x) = 0.

Apply this method to Find an iterative formula to find √N and hence

find √7 Correct up to three decimal points. [𝟐. 𝟔𝟒𝟔]

Dec-13

T Que.13 Set up a Newton iteration for computing the square root x of a given positive number c and apply it to c = 2.

[𝟏. 𝟒𝟏𝟒𝟐]

Nov -10

C Que.14 Derive iterative formula to find √𝑁. Use this formula to find √28.

[𝟐. 𝟐𝟖𝟗𝟒] Nov-16

H Que.15 Derive an iterative formula to find √N hence find approximate value

of √65 and √3, correct up to three decimal places. [𝟖. 𝟎𝟔𝟐, 𝟏. 𝟕𝟑𝟐]

Dec-14



T Que.16

Derive Newton Raphson’s formula for finding the cube root of

positive no. N, hence find √123

.

[𝟐. 𝟐𝟖𝟗𝟒

May-16

C Que.17

Derive an iterative formula for finding cube root of any positive number using Newton Raphson method and hence find approximate

value of √583

. [𝟑. 𝟖𝟕𝟎𝟖]

May-15

H Que.18 Find the √10 correct to three decimal places by using Newton-Raphson iterative method.

[𝟑. 𝟏𝟔𝟐𝟑]

Dec-15

C Que.19

Find an iterative formula to find 1

N(Where N is positive number) and

hence evaluate 1

3 ,

1

19 ,

1

23.

[𝟎. 𝟑𝟑𝟑, 𝟎. 𝟎𝟓𝟐𝟔, 𝟎. 𝟎𝟒𝟑𝟓]

10. Numerical methods for O.D.E. PAGE | 110

N.S.M. 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

Methods for O.D.E.

Picard's Method

Taylor's Method

Euler's Method(R-K 1st Order Method)

Improved Euler's Method

(R-K 2nd order method)(Heun's method)

Runge-Kutta 4th Order Method

Predictor -Corrector Method

SR. NO. TOPIC NAME

1 Picard's Method

2 Taylor 's Method

3 Euler's Method (R-K 1st Order Method)

4 Improved Euler's Method/Modified Euler's Method

5 Runge-Kutta 4th Order Method

6 Predictor -Corrector Method



Picard’s Method

If 𝐝𝐲

𝐝𝐱= 𝐟(𝐱, 𝐲) 𝐚𝐧𝐝 𝐲(𝐱𝟎) = 𝐲𝟎 ,then

Picard’s formula

𝐲𝐧 = 𝐲𝟎 + ∫ 𝐟(𝐱,𝐱

𝐱𝟎

𝐲𝐧−𝟏)𝐝𝐱 ; 𝐧 = 𝟏, 𝟐, 𝟑, …

Note

We stop the process, When yn = yn−1, up to the desired decimal places. This method is applicable only to a limited class of equations in which successive

integrations can be performed easily.

Exercise-1

C Que 1. Using Picard’s method solve

dy

dx− 1 = xy with initial condition

y(0) = 1, compute y(0.1) correct to three decimal places. [𝐲(𝟎. 𝟏) = 𝟏. 𝟏𝟎𝟓]

T Que 2. Solve

dy

dx= 3 + 2xy , Where y(0) = 1, for x = 0.1 by Picard’s

method. [𝐲(𝟎. 𝟏) = 𝟏. 𝟑𝟏𝟐𝟏]

Jun-12

T Que 3.

Obtain Picard’s second approximation solution of the initial

value problem dy

dx= x2 + y2 for x = 0.4 correct places, given

that y(0) = 0. [𝐲(𝟎. 𝟒) = 𝟎. 𝟎𝟐𝟏𝟒]

C Que 4.

Obtain Picard’s second approximate solution of the initial

value problem 𝑑𝑦

𝑑𝑥=

𝑥2

𝑦2+1, 𝑦(0) = 0

[𝐲𝟐 =𝟏

𝟑𝐱𝟑 −

𝐱𝟗

𝟖𝟏+

𝐱𝟏𝟓

𝟏𝟐𝟏𝟓−]

Nov-16

H Que 5.

Using Picard’s method solve dy

dx= x + y2, y(0) = 1.

[𝐲𝟐 = 𝟏 + 𝐱 +𝟑

𝟐𝐱𝟐 +

𝟐

𝟑𝐱𝟑 +

𝐱𝟒

𝟒+

𝐱𝟓

𝟐𝟎]



Taylor Series

Taylor’s series expansion

If 𝐝𝐲


𝐲(𝐱) = 𝐲(𝐱𝟎) +(𝐱 − 𝐱𝟎)

𝟏! 𝐲′(𝐱𝟎) +

(𝐱 − 𝐱𝟎)𝟐

𝟐! 𝐲′′(𝐱𝟎) + ⋯

Putting x = x1 = x0 + h ⇒ x − x0 = h

∴ 𝐲(𝐱𝟏) = 𝐲(𝐱𝟎 + 𝐡) = 𝐲(𝐱𝟎) +𝐡

𝟏! 𝐲′(𝐱𝟎) +

𝐡𝟐

𝟐! 𝐲′′(𝐱𝟎) + ⋯

So, 𝐲(𝐱𝟏) = 𝐲𝟏 = 𝐲𝟎 +𝐡

𝟏! 𝐲𝟎

′ +𝐡𝟐

𝟐!𝐲𝟎

′′ + ⋯

In general,

𝐲(𝐱𝐧) = 𝐲𝐧 = 𝐲𝐧−𝟏 +𝐡

𝟏! 𝐲𝐧−𝟏

′ +𝐡𝟐

𝟐! 𝐲𝐧−𝟏

′′ +𝐡𝟑

𝟑! 𝐲𝐧−𝟏

′′′ + ⋯ ; 𝐧 = 𝟏, 𝟐, 𝟑 …

𝐖𝐡𝐞𝐫𝐞, 𝐱𝐧 = 𝐱𝟎 + 𝐧𝐡 ; 𝐧 = 𝟏, 𝟐, 𝟑, …

Here h is step size and n is the number of steps.

Exercise-2

C Que 1.

Find by Taylor’s series method the value of y at 𝑥 = 0.1 and 𝑥 =

0.2 to four places of decimal, for 𝑑𝑦

𝑑𝑥= 𝑥2 − 1; 𝑦(0) = 1.

[𝐲(𝟎. 𝟏) = 𝟏. 𝟏𝟎𝟎𝟑, 𝐲(𝟎. 𝟐) = 𝟏. 𝟐𝟎𝟐𝟕]

May-16

T Que 2.

Using Taylor series method, find correct four decimal place, the

value of y(0.1), given dy

dx= x2 + y2 and y(0) = 1.

[𝐲(𝟎. 𝟏) = 𝟏. 𝟏𝟏𝟏𝟏]

Jun-11

C Que 3. Solve y′ = x2 + y2using the Taylor’s series method for the initial condition y(0) = 0.Where 0 ≤ x ≤ 0.4 and h = 0.2.

[𝐲(𝟎. 𝟐) = 𝟎. 𝟎𝟎𝟐𝟕, 𝐲(𝟎. 𝟒) = 𝟎. 𝟎𝟐𝟏𝟒]

H Que 4.

Using Taylor series method, find y(0.1) and y(0.2) correct to four

decimal places, if y(x) satisfies dy

dx= x − y2 , y(0) = 1.

[𝐲(𝟎. 𝟏) = 𝟎. 𝟗𝟏𝟑𝟖, 𝐲(𝟎. 𝟐) = 𝟎. 𝟖𝟓𝟏𝟓]

May-16

H Que 5.

Evaluate y(0.1) correct to four decimal places using Taylor’s

series method if dy

dx= y2 + x , y(0) = 1.

[𝐲(𝟎. 𝟏) = 𝟏. 𝟏𝟏𝟔]

May-15



T Que 6.

Using Taylor’s series method ,find y(1.1) correct to four decimal

places, given that dy

dx= xy

1

3, y(1) = 1.

[𝐲(𝟏. 𝟏) = 𝟏. 𝟏𝟎𝟔𝟖]

Dec-15

C Que 7.

Using Taylor’s series method, solve 𝑑𝑦

𝑑𝑥= 𝑥2 − 𝑦, 𝑦(0) = 1 at 𝑥 =

0.1, 0.2 𝑎𝑛𝑑 0.3. Also compare the values with exact solution. 𝐲(𝟎. 𝟏) 𝐲(𝟎. 𝟐) 𝐲(𝟎. 𝟑) Taylor’s series

1.1055 0.8242 0.7595

Exact solution

0.9052 0.8213 0.7492

Nov-16

Euler’s Method (RK 1st order method)

If 𝐝𝐲


𝐲𝐧+𝟏 = 𝐲𝐧 + 𝐡 [𝐟(𝐱𝐧, 𝐲𝐧)] ; 𝐧 = 𝟎, 𝟏, 𝟐, …

𝐖𝐡𝐞𝐫𝐞, 𝐱𝐧 = 𝐱𝟎 + 𝐧𝐡 ; 𝐧 = 𝟏, 𝟐, 𝟑, …

here h is step size and n is the number of steps.

Explanation:

Let [a, b] be the interval over which we want to find the solution of

𝐝𝐲

𝐝𝐱= 𝐟(𝐱, 𝐲) ; 𝐚 < 𝐱 < 𝐛 ; 𝐲(𝐱𝟎) = 𝐲𝟎 … (𝟏)

A set of points {(xn, yn)} is generated which are used for an approximation [i. e. y(xn) = yn]

For convenience, we divide [a, b] into n equal subintervals.

⟹ 𝐱𝐧 = 𝐱𝟎 + 𝐧𝐡 ; 𝐧 = 𝟏, 𝟐, … ; 𝐰𝐡𝐞𝐫𝐞, 𝐡 =𝐛−𝐚

𝐧 𝒐𝒓 𝐡 =

𝐱𝐧−𝐱𝟎

𝐧 is called the step size.

Now, y(x) is expand by using Taylor’s series about x = x0 as following

𝐲(𝐱) = 𝐲(𝐱𝟎) +(𝐱 − 𝐱𝟎)

𝟏 !𝐲′(𝐱𝟎) +

(𝐱 − 𝐱𝟎)𝟐

𝟐 !𝐲′′(𝐱𝟎) +

(𝐱 − 𝐱𝟎)𝟑

𝟑 !𝐲′′′(𝐱𝟎) + ⋯ … (𝟐)

𝐖𝐞 𝐡𝐚𝐯𝐞, [𝐝𝐲

𝐝𝐱]

𝐱=𝐱𝟎

= 𝐲′(𝐱𝟎) = 𝐟(𝐱𝟎, 𝐲𝟎) = 𝐟(𝐱𝟎, 𝐲(𝐱𝟎))

By Eq. (2),

𝐲(𝐱𝟏) = 𝐲(𝐱𝟎) +(𝐱𝟏 − 𝐱𝟎)

𝟏 !𝐲′(𝐱𝟎) +

(𝐱𝟏 − 𝐱𝟎)𝟐

𝟐 !𝐲′′(𝐱𝟎) +

(𝐱𝟏 − 𝐱𝟎)𝟑

𝟑 !𝐲′′′(𝐱𝟎) + ⋯



Taking, h = x1 − x0

𝐲(𝐱𝟏) = 𝐲(𝐱𝟎) +𝐡

𝟏 !𝐟(𝐱𝟎, 𝐲(𝐱𝟎)) +

𝒉𝟐

𝟐 !𝐲′′(𝐱𝟎) +

𝒉𝟑

𝟑 !𝐲′′′(𝐱𝟎) + ⋯

If the step size is chosen too small enough, then we may neglect the second order term

involving h2 and get

𝐲𝟏 = 𝐲(𝐱𝟏) = 𝐲(𝐱𝟎) +𝐡

𝟏 !𝐟(𝐱𝟎, 𝐲(𝐱𝟎)) = 𝐲𝟎 + 𝐡𝐟(𝐱𝟎, 𝐲𝟎)

Which is called Euler’s Approximation.

The process is repeated and generates a sequence of points that approximate the solution of

curve y = y(x).

The general step for Euler’s method is 𝐲𝐧+𝟏 = 𝐲𝐧 + 𝐡 [𝐟(𝐱𝐧, 𝐲𝐧)] ; 𝐧 = 𝟎, 𝟏, 𝟐, …

𝐖𝐡𝐞𝐫𝐞, 𝐱𝐧 = 𝐱𝟎 + 𝐧𝐡 ; 𝐧 = 𝟏, 𝟐, 𝟑, …

Here h is step size and n is the number of steps.

Exercise-3

C Que 1. Describe Euler’s Method for first order ordinary differential equation.

Dec-10 Jun-12

C Que 2.

Apply Euler’s method to find the approximate solution of dy

dx= x + y with y(0) = 0 and h = 2. Show your calculation up

to five iteration. [𝐲𝟓 = 𝟐𝟑𝟐]

Jun-13

H Que 3. Using Euler’s method solve for y at 𝑥 = 0.1 from

𝑑𝑦

𝑑𝑥= 𝑥 + 𝑦 +

𝑥𝑦. 𝑦(0) = 1, In five steps. 𝒚(𝟎. 𝟏) = 𝟏. 𝟏𝟏𝟐𝟒

May-16

C Que 4.

Derive Euler’s formula for initial value problem dy

dx=

f(x, y); y(x0) = y0. Hence , use it find the value of y for

dy

dx= x + y; y(0) = 1 when x = 0.1, 0.2 with step size h = 0.05.

Also Compare with analytic solution. 𝐲(𝟎. 𝟏) 𝐲(𝟎. 𝟐) Euler’s 𝟏. 𝟏𝟎𝟓𝟎 𝟏. 𝟐𝟑𝟏𝟏 Exact solution 𝟏. 𝟏𝟏𝟎𝟑 𝟏. 𝟐𝟒𝟐𝟖

May-15



T Que 5.

Apply Euler’s method to solve the initial value problem dy

dx= x + y , with y(0) = 0 with choosing h = 0.2 and compute

y1, y2, y3 … … y5 Compare your result with the exact solution. 𝐲(𝟎. 𝟐) 𝐲(𝟎. 𝟒) 𝐲(𝟎. 𝟔) 𝐲(𝟎. 𝟖) 𝐲(𝟏) Euler’s 𝟎 𝟎. 𝟎𝟒 𝟎. 𝟏𝟐𝟖 𝟎. 𝟐𝟕𝟑𝟔 𝟎. 𝟒𝟖𝟖𝟑 Exact solution

𝟎. 𝟎𝟐𝟏𝟒 𝟎. 𝟎𝟗𝟏𝟖 𝟎. 𝟐𝟐𝟐𝟏 𝟎. 𝟒𝟐𝟓𝟓 𝟎. 𝟕𝟏𝟖𝟑

Nov-16

H Que 6.

Using Euler’s method ,find an approximate value of y

corresponding to x = 1 given that dy

dx= x + y and y = 1 when

x = 0. [𝐅𝐨𝐫 𝐡 = 𝟎. 𝟐𝟓, 𝐲𝟒 = 𝟐. 𝟖𝟖𝟐𝟗]

Jun -13

T Que 7.

Use Euler method to find y(0.2) given that dy

dx= y −

2x

y ,

y(0) = 1. (Take h = 0.1) [𝐲(𝟎. 𝟐) = 𝟏. 𝟏𝟗𝟏𝟖]

Jun-11

H Que 8.

Explain Euler’s method for solving first order ordinary differential equation. Hence use this method, find y(2) for dy

dx= x + 2y with y(1) = 1.

[𝐲(𝟐) = 𝟓. 𝟕𝟓]

Dec-15

T Que 9. Solve initial value problem

dy

dx= x√y , y(1) = 1 and hence find

y(1.5) by taking h = 0.1 using Euler’s method. [𝐲(𝟏. 𝟓) = 𝟏. 𝟔𝟕𝟎𝟎]

May-15

T Que 10.

Using Euler’s method, find an approximate value of

corresponding to 𝑥 = 0.3, given that 𝑑𝑦

𝑑𝑥=

𝑦−𝑥

𝑦+𝑥; 𝑦(0) = 1. take

ℎ = 0.1. [𝒚(𝟏) = 𝟏. 𝟐𝟓𝟒𝟒]

Nov-16

H Que 11.

Use Euler’s method to find an approximation value of y at

x = 0.1 for the initial value problem dy

dx= x − y2; y(0) = 1.

[𝐲(𝟎. 𝟏) = 𝟎. 𝟗𝟏𝟑𝟑]

Dec-15

T Que 12.

Using Euler’s method compute 𝑦(0.3) for the initial value problem 𝑦′ = 𝑦2 − 𝑥2, 𝑦(0) = 1 taking the step size ℎ = 0.1.

[𝐲(𝟎. 𝟑) = 𝟏. 𝟑𝟔𝟒𝟖]

June-16



Improved Euler’s Method

(Heun’s Method/ Modified Euler’s Method / R-K 2nd OrderMethod)

If 𝐝𝐲


Improved Euler’s Formula

𝐲𝐧+𝟏 = 𝐲𝐧 +𝐡

𝟐[𝐟(𝐱𝐧, 𝐲𝐧) + 𝐟(𝐱𝐧 + 𝐡, 𝐲𝐧 + 𝐡𝐟(𝐱𝐧, 𝐲𝐧))]; 𝐧 = 𝟎, 𝟏, 𝟐, …

𝐖𝐡𝐞𝐫𝐞, 𝐱𝐧 = 𝐱𝟎 + 𝐧𝐡 ; 𝐧 = 𝟏, 𝟐, 𝟑, …


Exercise-4

C Que 1.

Using improved Euler’s method solves dy

dx= 1 − y with the initial

condition y(0) = 0 and tabulates the solution at x = 0.1 , 0.2. Compare the answer with exact solution.

𝐲(𝟎. 𝟏) 𝐲(𝟎. 𝟐) Improved Euler’s 𝟎. 𝟎𝟗𝟓𝟎 𝟎. 𝟏𝟖𝟏𝟎 Exact solution 𝟎. 𝟎𝟗𝟓𝟐 𝟎. 𝟏𝟖𝟏𝟑

Nov-11

H Que 2.

Using improved Euler’s method solves dy

dx+ 2xy2 = 0 with the initial

condition y(0) = 1 and compute y(1) taking h = 0.2 compare the answer with exact solution.

𝐲(𝟎) 𝐲(𝟎. 𝟐) 𝐲(𝟎. 𝟒) 𝐲(𝟎. 𝟔) 𝐲(𝟎. 𝟖) 𝐲(𝟏)

𝐈𝐦𝐩𝐫𝐨𝐯𝐞𝐝 𝐄𝐮𝐥𝐞𝐫’𝐬

𝟏 𝟎. 𝟗𝟔𝟎𝟎 𝟎. 𝟖𝟔𝟎𝟑 𝟎. 𝟕𝟑𝟓𝟎 𝟎. 𝟔𝟏𝟏𝟓 𝟎. 𝟓𝟎𝟑𝟑

𝐄𝐱𝐚𝐜𝐭 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧

𝟏 𝟎. 𝟗𝟔𝟏𝟓 𝟎. 𝟖𝟔𝟐𝟏 𝟎. 𝟕𝟑𝟓𝟑 𝟎. 𝟔𝟎𝟗𝟖 𝟎. 𝟓𝟎𝟎𝟎

Jun-10

T Que 3.

Given the equation dy

dx=

2y

x ; y(1) = 2 Estimate y(2) using Heun’s

method h = 0.25 and compare the results with exact answers. 𝐲(𝟏) 𝐲(𝟏. 𝟐𝟓) 𝐲(𝟏. 𝟓𝟎) 𝐲(𝟏. 𝟕𝟓) 𝐲(𝟐) Improved Euler’s

𝟐 𝟑. 𝟏𝟎𝟎𝟎 𝟒. 𝟒𝟒𝟑𝟑 𝟔. 𝟎𝟑𝟎𝟐 𝟕. 𝟖𝟔𝟎𝟖

Exact solution

𝟐 𝟑. 𝟏𝟐𝟓𝟎 𝟒. 𝟓𝟎𝟎𝟎 𝟔. 𝟏𝟐𝟓𝟎 𝟖. 𝟎𝟎𝟎𝟎

C Que 4.

Use Runge-Kutta second order method to find the approximate value

of y(0.2) given that dy

dx= x − y2and y(0) = 1 and h = 0.1.

[𝐲(𝟎. 𝟐) = 𝟎. 𝟖𝟓𝟐𝟑]

Dec-10 Dec-15



H Que 5.

Given that y = 1.3 when x = 1 and dy

dx= 3x + y use second order RK

method (i.e. Heun’s method) to approximate y ,when x = 1.2 use step size 0.1.

[𝐲(𝟏. 𝟐) = 𝟐. 𝟑𝟏𝟑𝟓]

Dec-12

T Que 6. Given that

𝑑𝑦

𝑑𝑥= 𝑥 + 𝑦2 and 𝑦 = 1 at 𝑥 = 0. Find an approximate value

of 𝑦 at 𝑥 = 0.3 by Modified Euler method. [𝐲(𝟎. 𝟑) = 𝟏. 𝟒𝟖𝟎𝟏]

Nov-16

Runge Kutta 4th Order Method

If dy

dx= f(x, y) and y(x0) = y0 ,then

RK 4th Order Formula 𝐲𝐧+𝟏 = 𝐲𝐧 +𝐡

𝟔(𝐊𝟏 + 𝟐𝐊𝟐 + 𝟐𝐊𝟑 + 𝐊𝟒); 𝐧 = 𝟎, 𝟏, 𝟐, …

𝐖𝐡𝐞𝐫𝐞, 𝐱𝐧 = 𝐱𝟎 + 𝐧𝐡 ; 𝐧 = 𝟏, 𝟐, 𝟑, …


𝐊𝟏 = 𝐟(𝐱𝐧, 𝐲𝐧)

𝐊𝟐 = 𝐟 (𝐱𝐧 +𝐡

𝟐, 𝐲𝐧 +

𝐊𝟏

𝟐)

𝐊𝟑 = 𝐟 (𝐱𝐧 +𝐡

𝟐, 𝐲𝐧 +

𝐊𝟐

𝟐)

𝐊𝟒 = 𝐟(𝐱𝐧 + 𝐡, 𝐲𝐧 + 𝐊𝟑)

Exercise-5

C Que 1. Write formula for Runge-Kutta method for order four. Jun-13

C Que 2.

Apply Runge-Kutta method of fourth order to calculate y(0.2)

given dy

dx= x + y , y(0) = 1 taking h = 0.1

[𝐲(𝟎. 𝟐) = 𝟏. 𝟐𝟒𝟐𝟖]

Dec-12 Nov-16

T Que 3.

Describe y(0.1) and y(0.2) Correct to four decimal places

from dy

dx= 2x + y , y(0) = 1 use fourth order R-K method.

[𝐲(𝟎. 𝟏) = 𝟏. 𝟏𝟏𝟓𝟓, 𝐲(𝟎. 𝟐) = 𝟏. 𝟐𝟔𝟒𝟐]

Jun-12 Nov-16

H Que 4.

Using Runge-Kutta fourth order method solve for 𝑦(0.1) and

𝑦(0.2) given that 𝑑𝑦

𝑑𝑥= 𝑥𝑦 + 𝑦2, 𝑦(0) = 1.

[𝐲(𝟎. 𝟏) = 𝟏. 𝟏𝟏𝟔𝟕, 𝐲(𝟎. 𝟐) = 𝟏. 𝟐𝟕𝟕𝟒] Nov-16



T Que 5.

Using Runge-Kutta method to fourth order, solve 𝑑𝑦

𝑑𝑥=

𝑦2−𝑥2

𝑦2+𝑥2

with 𝑦(0) = 1 at 𝑥 = 0.2 with a Step-size of 0.1 [𝐲(𝟎. 𝟐) = 𝟏. 𝟏𝟗𝟔𝟓]

May-16

H Que 6.

Use fourth order Runge Kutta method to find the value of 𝑦 at

𝑥 = 1 given that 𝑦′ =𝑦−𝑥

𝑦+𝑥 such that 𝑦(0) = 1. (take ℎ = 0.5)

[𝒚(𝟏) = 𝟏. 𝟒𝟗𝟗𝟐] May-16

T Que 7.

Apply Runge-Kutta fourth order method, to find an approximate value of ywhen x = 0.2 in steps of 0.1, if dy

dx= x + y2, given that y = 1 when x = 0.

[𝐲(𝟎. 𝟏) = 𝟏. 𝟏𝟏𝟔𝟓, 𝐲(𝟎. 𝟐) = 𝟏. 𝟐𝟕𝟑𝟔]

Jun -14

C Que 8.

Using the Range-Kutta method of fourth order , find y at x =

0.1 given diff. equation dy

dx= 3ex + 2y, y(0) = 0 by taking h =

0.1 and also compare the solution with exact solution.

𝐲(𝟎. 𝟏) R-K method 𝟎. 𝟑𝟒𝟖𝟕 Exact solution 𝟎. 𝟑𝟒𝟖𝟎

May -15

T Que 9.

Apply Runge-Kutta fourth order method to calculate y(0.2)

and y(0.4) given dy

dx= y −

2x

y, y(0) = 1.

[𝐲(𝟎. 𝟐) = 𝟏. 𝟏𝟖𝟑𝟐, 𝐲(𝟎. 𝟒) = 𝟏. 𝟑𝟒𝟏𝟔] Dec-15

H Que 10.

Solve initial value problem dy

dx= −2xy2; y(0) = 1 with h = 0.2

for y(0.2) using Runge-Kutta fourth order method. [𝐲(𝟎. 𝟐) = 𝟎. 𝟗𝟔𝟏𝟓]

Dec-15

H Que 11.

Use the Runge-Kutta 4th order method with ℎ = 0.1 to find the approximate solution

for 𝑦(1.1) for the initial value problem 𝑑𝑦

𝑑𝑥= 2𝑥𝑦, 𝑦(1) = 1.

[𝒚(𝟏. 𝟏) = 𝟏. 𝟐𝟑𝟑𝟕]

June-16



Predictor-Corrector Methods

Modified Euler’s Method (Heun’s Method)

𝐲𝐢+𝟏𝐩

= 𝐲𝐢 + 𝐡𝐟(𝐱𝐢, 𝐲𝐢)

𝐲𝐢+𝟏𝐜 = 𝐲𝐢 +

𝐡

𝟐[𝐟(𝐱𝐢, 𝐲𝐢) + 𝐟(𝐱𝐢+𝟏, 𝐲𝐢+𝟏

𝐩)]

Adams-Bashforth Method

𝐲𝐤+𝟏𝐩

= 𝐲𝐤 +𝐡

𝟐𝟒[−𝟗𝐟𝐤−𝟑 + 𝟑𝟕𝐟𝐤−𝟐 − 𝟓𝟗𝐟𝐤−𝟏 + 𝟓𝟓𝐟𝐤]

𝐲𝐤+𝟏𝐜 = 𝐲𝐤 +

𝐡

𝟐𝟒[𝐟𝐤−𝟐 − 𝟓𝐟𝐤−𝟏 + 𝟏𝟗𝐟𝐤 + 𝟗𝐟𝐤+𝟏

𝐩]

Milne’s Method

𝐲𝐤+𝟏𝐩

= 𝐲𝐤−𝟑 +𝟒𝐡

𝟑[𝟐𝐟𝐤−𝟐 − 𝐟𝐤−𝟏 + 𝟐𝐟𝐤]

𝐲𝐤+𝟏𝐜 = 𝐲𝐤−𝟏 +

𝐡

𝟑[𝐟𝐤−𝟏 + 𝟒𝐟𝐤 + 𝐟𝐤+𝟏

𝐩]

Exercise-6

C Que 1.

Solve dy

dx= y −

2x

y, y(0) = 1.Compute y(0.4), y(0.5) by Adam’s

method using y(0.1) = 1.0954, y(0.2) = 1.1832 y(0.3) = 1.2649.

[𝐲(𝟎. 𝟒) = 𝟏. 𝟑𝟒𝟏𝟔, 𝐲(𝟎. 𝟓) = 𝟏. 𝟒𝟏𝟒𝟐]

T Que 2.

Using Taylor’s method compute the approximate value of y at

x = 0.2,0.4 & 0.6 for the differential equation dy

dx= x − y2 with

the initial condition y(0) = 0. Now apply milne’s predictor-corrector method to find y at x = 0.8

[𝐲(𝟎. 𝟖) = 𝟎. 𝟑𝟎𝟒𝟕]

Dec-12

C Que 3.

Using Milne’s method, solve 𝑑𝑦

𝑑𝑥= 1 + 𝑦2 with 𝑦(0) = 0, 𝑦(0.2) =

0.2027, 𝑦(0.4) = 0.4228, 𝑦(0.6) = 0.6841, obtain 𝑦(0.8) and 𝑦(1).

[𝐲(𝟎. 𝟖) = 𝟐. 𝟎𝟓𝟗𝟕, 𝐲(𝟏) = 𝟑. 𝟒𝟏𝟗𝟗]

May-16



T Que 4.

Use Milne’s predictor-corrector to find y(0.4) for

dy

dx= x + y2, y(0) = 1 with h = 0.1.

x Y f = x + y2

0 1 1.0

0.1 1.1 1.31

0.2 1.231 1.715361

0.3 1.4025361 2.2671075

0.4 ?

[𝐲(𝟎. 𝟒) = 𝟏. 𝟕𝟎𝟑𝟎𝟏𝟑]

H Que 5.

State different predictor-corrector method. For the initial value

problem dy

dx= y + x2; y(0) = 1, use Milne's predictor-corrector

method to find y(0.8) by taking h = 0.2 from following data X 0 0.2 0.4 0.6 Y 1 1.2242 1.5155 1.9063

[𝐲𝐩 (𝟎. 𝟖) = 𝟐. 𝟒𝟑𝟔𝟏, 𝐲𝐜(𝟎. 𝟖) = 𝟐. 𝟒𝟑𝟔𝟔]

Dec-15

Documents

INDEX [] TECHNOLOGICAL UNIVERSITY CIVIL ENGINEERING (06) NUMERICAL AND STATISTICAL METHODS FOR CIVIL ENGINEERING SUBJECT CODE: 2140606 B.E. 4th Semester Type of course: Engineering