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Essential Mathematics (For business and economic analysis)

Course: Business Mathematics II. Mat 211 Dr. Firoz

Chapter 8 Discrete and Continuous Distribution

Section 8.1 Random variables of the discrete type

In probability theory, a probability distribution is called discrete if it is characterized by a

probability mass function. Thus, the distribution of a random variable X is discrete, and X

is then called a discrete random variable, if ( ) 1p X

If a random variable is discrete, then the set of all values that it can assume with non-zero

probability is finite or countably infinite, because the sum of uncountably many positive

real numbers (which is the least upper bound of the set of all finite partial sums) always

diverges to infinity.

Given a random experiment with an outcome space S, a function X that assigns to each

element s in S one and only one real number X(s) = x is called a random variable, like a

function of s. The space of X is the set of real numbers { : ( ) , }x X s x s S , where s S

means the element s belongs to S. The probability mass function (pmf) ( )f x of a

discrete random variable X is a function that satisfies the following properties:

1. ( ) 0, ;f x x S

2. ( ) 1;x S

f x

3. ( ) ( ),x A

P X A f x A S

Example 1. Suppose that X has a discrete uniform distribution on {1,2,3,4,5,6}S and

its pmf is 1

( ) , 1,2,3,4,5,66

f x x .

As a general case we may write pmf as 1

( ) , 1,2,3,4, ,f x x mm

Example 2. Roll a 4 –sided die twice and let X equal the larger of the two outcomes if

they are different and common value if they are the same. The outcome space for this

experiment is 0 1 2 1 2{( , ); 1,2,3,4; 1,2,3,4}S d d d d , where we assume that each of

these 16 points has probability 1

16. Then

1 3 5( 1) [(1,1)] , ( 2) [{(1,2),(2,2), (2,1)] , ( 3)]

16 16 16P X P P X P P X and

7( 4)

16P X . Looking at the pattern one can easily find the pmf

2

2 1( ) , 1,2,3,4

16

xf x x

Exercise

1. Let the pmf of X be defined by ( ) , 2,3,49

xf x x , a) draw a bar graph and a

b) probability histogram

2. For each of the following, determine the constant c, so that ( )f x satisfies the

conditions of being a pmf for a random variable X,

a) ( ) , 1,2,3,4x

f x xc

b) ( ) , 1,2,3,4 10f x cx x

c) 1

( ) , 1,2,3,44

x

f x c x

d) 2( ) (1 ) , 0,1,2,3f x c x x

e) ( ) , 1,2,3,4, ,x

f x x nc

Section 8.3 Mathematical expectation

In probability theory and statistics, the expected value (or expectation value, or

mathematical expectation, or mean, or first moment) of a random variable is the

integral of the random variable with respect to its probability measure.

For discrete random variables this is equivalent to the probability-weighted sum of the

possible values.

The term "expected value" can be misleading. It must not be confused with the "most

probable value." The expected value is in general not a typical value that the random

variable can take on. It is often helpful to interpret the expected value of a random

variable as the long-run average value of the variable over many independent repetitions

of an experiment.

When it exists, mathematical expectation E satisfies the following properties:

a) If c is a constant, ( )E c c

b) If c is a constant and u is a function, [ ( )] [ ( )]E cu x cE u x

c) If 1c and 2c are constants and 1u and 2u are functions, than

1 1 2 2 1 1 2 2[ ( ) ( )] [ ( )] [ ( )]E c u X c u X c E u X c E u X

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Example 1. Let X have the pmf ( ) , 1,2,3,410

xf x x . Find ( )E X

Solution: 1

( ) ( ) 310

x

i i

x

xE X x f x x , verify.

Example 2. Let X have the pmf ( ) , 1,2,36

xf x x . Find mean = ( )E X and also

2( )E X and variance 2 2 2( ) ( ( ))E X E X and also standard deviation .

Solution: Mean = 1

7( ) ( )

6 3

x

i i

x

xE X x f x x

2 2 2

1

36( ) ( ) 6

6 6

x

i i

x

xE X x f x x

Variance

2

2 2 2 7 5( ) ( ( )) 6

3 9E X E X and

5

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Example 3. A politician can emphasize jobs or the environment in her election

campaign. The voters can be concerned about jobs or the environment. A payoff matrix

showing the utility of each possible outcome is shown.

Jobs Voters Environment

Jobs 25 10

Environment 15 30

The political analysts feel there is a 0.39 chance that the voters will emphasize

jobs. Which strategy should the candidate adopt to gain the highest utility

a) Environment b) Jobs

Explain mathematically.

Solution: For the environment the expected value is ( ) 15(.39) 30(1 .39) 12.45E x

On the other hand for jobs the expected value is ( ) 25(.39) 10(1 .39) 3.65E x . So

the preference will go for a) Environment (because of higher expected value).

Exercise

1. Find mean and standard deviation of the following:

a) 1

( ) , 5,10,15,20,255

f x x

b) ( ) 1, 5f x x

c) 4

( ) , 1,2,36

xf x x

2. Given 2( 4) 10, [( 4) ] 116E X E X , determine 2 Var ( 4)X and

mean= ( )E X

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Section 8.2 Bernoulli trials and the Binomial distribution

A Bernoulli experiment is a random experiment, the outcome of which can be classified

in but one of two mutually exclusive and exhaustive ways, say, success or failure (life or

death, head or tail, 3 or not 3 etc. The pmf of a Bernoulli trail is 1( ) (1 ) , 0,1x xf x p p x and we say that the random variable x has Bernoulli

distribution. The mean of Bernoulli trial is given as 1

1

0

( ) (1 )x x

x

E X xp p p , verify.

The variance of Bernoulli trial is 1

2 2 1

0

( ) ( ) (1 ) (1 ) , 1x x

x

Var X x p p p p p pq q p

Example 1. In the instant lottery with 20% winning tickets, if X is equal to the number of

winning tickets among n = 8 that are purchased, the probability of purchasing two

winning tickets is 2 68

(2) ( 2) (0.20) (1 0.2) 0.2936 29.36%2

f P X

One may use calculator as follows (TI)

2nd

DISTR 0 (binompdf) (8, 0.20, 2) will display 0.29360128

Example 2. In the instant lottery with 20% winning tickets, if X is equal to the number of

winning tickets among n = 8 that are purchased, the probability of purchasing at best 6

winning tickets is

7 1 88 8

( 6) 1 (7) (8) 1 (0.20) (1 0.2) (0.2) 0.999915527 8

P X f f

One may use calculator as follows (TI)

2nd

DISTR A (binomcdf) (8, 0.20, 6) will display 0.99991552

Example 3. In the instant lottery with 20% winning tickets, if X is equal to the number of

winning tickets among n = 8 that are purchased. Find the probability of purchasing at

least 6 winning tickets.

Hint. Find ( 6) ( 6) ( 7) ( 8)P X P X P X P X or

( 6) 1 ( 5) 1 (8,.2,5)P X P X binomcdf

Example 4. A quiz consists of 24 multiple choice questions. Each question has 5 possible

answers, only one of which is correct. If you answer the questions completely based on

guessing, what is the probability that

a) You will answer exactly 4 wrong?

b) You will answer exactly 4 correctly?

c) You will answer at least 20 correctly?

d) You will answer at most 3 wrong?

e) You will answer at most 3 correctly?

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Solution: The probability that you will answer one question wrong is 4

0.85

.

a) The probability of answering exactly 4 wrong is a binomial probability of

B(24, 0.8, 4), which is 4 20 1124

( 4) (24,0.8,4) (0.8) (0.2) 4.56 104

P X B ,

which is almost zero.

If you use TI calculator use binompdf (24, 0.8, 4). Check your calculator using the

following code:

2nd

DISTR 0 binompdf (24, .8, 4)

b) The probability that you will answer exactly 4 correct is B(24, 0.2, 4) = 0.196

c) At least 20 correct ( 20)P X

= 20 correct + 21 correct + 22 correct + 23 correct + 24 correct = 114.79 10 .

It is easy to use calculator with binomcdf as follows: 11( 20) 1 (24,0.2,19) 4.79 10P X binomcdf

d) At most three wrong: 12( 3) (24,0.8,3) 2.25 10P X binomcdf

e) At most three correct: ( 3) (24,0.2,3) 0.264P X binomcdf

Example 5. A computer manufacturer tests a random sample of 28 computers. The

probability that a computer is non defective is 91.3%. What is the probability that:

a) Exactly 7 computers are defective? Answer: 0.006605

b) At least two computers are defective? Answer: 0.7131689

c) At most two computers are defective? Answer: 0.555224

Example 6. A quiz consists of 10 multiple choice questions, each with 4 possible

choices. For someone who makes random guesses for all of the questions, find the

probability of passing if the minimum passing grade is 90%.

Solution: 5( 9) 1 (10,0.25,8) 2.95639 10P X binomcdf

Example 8. A student claims that he has extrasensory perception (ESP). A coin is flipped

25 times, and a student is asked to predict the outcome in advance. He gets 20 out of 25

correct. What was the probability that he would have done at least this well if he had no

EPS?

Solution: ( 20) 1 (25,0.5,19) 0.002038658P X binomcdf

Exercise 1. Toss a fair coin 12 times. How many possible outcomes do you have? What

is the probability of getting a) exactly 7 heads, b) at least 7 heads, c) at most 7 heads?

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Exercise 2. A student claims that he has extrasensory perception (ESP). A coin is flipped

30 times, and a student is asked to predict the outcome in advance. He gets 25 out of 30

correct. What was the probability that he would have done at least this well if he had no

EPS?

Exercise 3. A quiz consists of 20 multiple choice questions, each with 5 possible choices.

For someone who makes random guesses for all of the questions, find the probability of

passing if the minimum passing grade is 80%.

Example 4. A computer manufacturer tests a random sample of 30 computers. The

probability that a computer is defective is 7

78

%. What is the probability that:

a) Exactly 7 computers are defective?

b) At least two computers are defective?

c) At most two computers are defective?

Exercise 5. In the instant lottery with 10% winning tickets, if X is equal to the number of

winning tickets among 20 tickets that are purchased, find the probability of purchasing

a) at best 7 winning tickets,

b) at least 7 winning tickets,

c) no more than 6 winning tickets,

d) no less than 6 winning tickets

Exercise 6. The rates of on-time flights for commercial jets are continuously tracked by

the U.S Department of transportation. Recently, Southwest Air had the best rate with

80% of its flights arriving on time. A test is conducted by randomly selected 16

Southwest flights and observing whether they arrive on time. Find

a) the probability that exactly 4 flights arrive on time

b) The probability that at least 4 flights arrive on time

c) At best 4 flights arrive on time

Appendix P.1 Random variable of the continuous type

A random variable is a function X that assigns to each element s in the outcome space S

one and only one corresponding real number X(s) = x. The space of X is the set of real

numbers { : ( ) , }S x X s x s S is an interval. In discrete case the S is the set of discrete

points. In the continuous case we call the integrable function ( )f x , a probability

density function (pdf) which satisfies the following:

a) ( ) 0,f x x S b) ( ) 1S

f x dx

c) The probability of the event X A is ( ) ( )A

P X A f x dx

Probability Distribution Function: A function F is a distribution function of the

random variable X iff the following conditions are satisfied:

a) F is non decreasing i.e., 0F , or ( ) ( )F x F y for all x y

b) F is continuous c) F is normalized i.e., lim ( ) 0; lim ( ) 1x x

F x F x

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Example 1. Evaluate the integral / 20

020

xedx

Solution: / 20

/ 20

00

lim 120

xb

x

b

edx e

Example 2. Show that ( ) , 0mxe

f x xm

is a probability density function.

Solution (Hint): Show that ( ) 0f x and /

0

1x me

dxm

Example 3. Let Y be a continuous random variable with pdf ( ) 2 , 0 1g y y y and the

distribution function is defined by

2

0

0 0

( ) 2 0 1

1 1

y

y

G y t dt y y

y

Find mean

1

0

2( ) ( )

3E Y yg y dy (check the integral) and

Variance

1

2 2 2 2 2

0

1( ) ( ) ( )

18Var Y E Y y g y dy (check the integral). Find

also the standard deviation .

Example 4. The probability density function of a continuous random variable x is given

as ( ) 1 1 , 0 2f x x x . Find its corresponding distribution function, mean,

variance and standard deviation, interval of one standard deviation of mean, two standard

deviation of mean and three standard deviation of mean.

Solution: The distribution function is the integral of the 1

pdf function over the real line.

Draw the graph of the pdf and notice that F(0) = 0,

0 1 2

F(1) = 1/2 and F(2) = 1. We also notice that distribution function is zero, i.e., (0) 0F

when x < 0.

The distribution function over the interval 0 1x is 2

( ) (1 1) (1 (1 ) , 0,as (0) 02

xF x x dx x dx x dx c c F

The distribution function over the interval 1 2x is 2 1

( ) (1 1) (1 1) (2 ) 2 , 1, (1)2 2

xF x x dx x dx x dx x c c as F

The distribution function over the interval 2 x is

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( ) ( ) 0 1, as (2) 1F x f x dx dx c F

Thus we have the probability distribution function defined as follows:

2

2

0 0

, 0 12

( )

2 1, 1 22

1 2

x

xx

F xx

x x

x

You can calculate mean, standard deviation and variance. Look at example 3.

Example 5. Show the following function is a probability distribution function

2

2

0 0

, 0 12

( )

2 1 1 22

1 2

x

xx

F xx

x x

x

Solution: We need to check the following properties:

a) Check that F is non-decreasing.

0 00, 0

, 0 1( ) 1 | 1|, 0 2

2 1 20, 2

0 2

xx

x xF x x x

x xx

x

shows that F is not decreasing.

b) For continuity check that 0 0

lim ( ) lim ( ) (0) 0x x

F x F x F and

2 2lim ( ) lim ( ) (2) 1x x

F x F x F

c) F is normalized: lim ( ) 0; lim ( ) 1x x

F x F x

The function F(x) defined above is probability distribution function.

Exercise:

1. For each of the following functions, i) find the constant c so that f (x) is a pdf of

the random variable X, ii) find the distribution function F(x) ( )P X x and iii)

sketch f (x) and F(x), iv) find also 2, , .

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a) 3

( ) , 04

xf x x c

b) 23

( ) ,16

xf x c x c

c) ( ) 4 , 0 1cf x x x

d) ( ) , 0 4f x c x x

2. Sketch the graph of the following pdf f (x), then find and sketch the probability

distribution function F(x) on the real line. Review example 4.

a) 23

( ) , 1 12

xf x x

b) 1

( ) , 1 12

f x x

c) 1 , 1 0

( )1 , 0 1

x xf x

x x

Section 8.5 The Normal Distribution

A normal distribution of a random variable X with mean and variance 2 is a statistic

distribution with probability density function (pdf)

(1)

on the domain . While statisticians and mathematicians uniformly use the term

"normal distribution" for this distribution, physicists sometimes call it a Gaussian

distribution and, because of its curved flaring shape, social scientists refer to it as the

"bell curve."

De Moivre developed the normal distribution as an approximation to the binomial

distribution, and it was subsequently used by Laplace in 1783 to study measurement

errors and by Gauss in 1809 in the analysis of astronomical data (Havil 2003, p. 157).

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The normal distribution is an extremely important probability distribution in many

fields. It is a family of distributions of the same general form, differing in their location

and scale parameters: the mean ("average") and standard deviation ("variability"),

respectively. The standard normal distribution is the normal distribution with a mean

of zero and a standard deviation of one (the green curve in the plots below). It is often

called the bell curve because the graph of its probability density resembles a bell.

If a random variable X has this distribution, we write ~ . If and , the

distribution is called the standard normal distribution and the probability density

function reduces to

Area under a normal curve:

For a standard normal variate z, the normal distribution has mean zero and standard

deviation one with pdf 22 / 21 1

( ) exp / 22 2

zf x z e

The area under the standard normal distribution curve for 21

( ) exp22

zu

f Z z du . We have now the difficulty to evaluate the integral

without having the knowledge of multivariable calculus and polar coordinate form. But

this difficulty we can manage using standard values from the table 5 of Normal

distribution at page # 423 or using our calculator. Look at example 3.

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Important Information: All normal density curves satisfy the following property which

is often referred to as empirical rule:

1. 68.26% of the observations fall within 1 standard deviation of mean.

2. 95.44% of the observations fall within 2 standard deviation of mean

3. 99.74% of the observations fall within 3 standard deviation of mean

Note: Within 5 standard deviation of mean we assume 100% data points.

Example 1. Find the mean and standard deviation of the normal distribution whose pdf is

given as 21 ( 7)

( ) exp128128

xf x

Solution: Compare with the standard formula of pdf for the normal distribution and find

that 8, 7 .

Example 2. Write the pdf of a normal distribution with mean 3 and variance 16.

Solution: We have 4, 3 , the pdf of the normal distribution is given as

21 ( 3)( ) exp

324 2

xf x

Example 3. Find the area under the normal curve with mean zero and standard deviation

one for the standard variate 1.24z .

Solution: From table 5a:

( 1.24) 0.8925 89.25%P z

For this value choose row with 1.2 and column 0.04.

0 1.24

Using calculator: ( 1.24) 0.8925 89.25%P z

The calculator code:

2nd

DISTR 2 normalcdf (-5, 1.24) =0.8925120

Example 4. Find the area under the normal curve with mean zero and standard deviation

one for the standard variate 1.24z .

Solution: From table 5a:

( 1.24) 1 0.8925 10.75%P z

For this value choose row with 1.2 and column 0.04.

0 1.24

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Using calculator: ( 1.24) 0.1075 10.75%P z

The calculator code: 2nd

DISTR 2 normalcdf ( 1.24, 5) =0.1074875

Example 5. Find the area under the normal curve with mean zero and standard deviation

one for 0.12 1.24z .

Solution: From table 5a:

( 0.12 1.24) ( .12) ( 1.24) 1 44.03%P z P z P z

-0.12 0 1.24

Using calculator: ( 0.12 1.24) 0.4402707 44.03%P z

The calculator code:

2nd

DISTR 2 normalcdf (-0.12, 1.24) =0.4402707

Example 6. Suppose x is a normally distributed random variable with mean 10.2 and

standard deviation 1.5. Find each of the following probabilities.

a) (6.1 13.3)P x .

b) (9.4 13)P x

c) (15.5 13.1)P x

d) ( 11.6)P x

e) ( 14.4)P x

Draw normal curve and show the region bounded by the normal curve and the x values.

Solution: from calculator a)

(6.1 13.3) 6.1,13.3,10.2,1.5 0.9774824 97.75%P x normalcdf

try similar way for b), and c),

d) 11.6 10.2

( 11.6) ,5 17.53%1.5

P x normalcdf

Try for e).

6.1 10.2 13.3

Exercise Set

1. The physical fitness of an athlete is often measured by how much oxygen the

athlete takes in (which is recorded in millimeters per kilogram, ml/kg). The

maximum oxygen uptake for elite athletes has been found to be 80 with a standard

deviation 9.2. Assume that distribution is approximately normal.

a) What is the probability that an elite athlete has a maximum oxygen uptake of

at least 75 ml/kg? Answer: 70.66%

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b) What is the probability that an elite athlete has a maximum oxygen uptake of

65 ml/kg or lower? Answer: 5.15%

c) Consider someone with a maximum oxygen uptake of 26 ml/kg. Is it likely

that this person is an elite athlete? Answer: No

2. The combined score of SAT – 1 test are normally distributed with mean of 998

and a standard deviation of 202. If a college includes a minimum score of 800

among its requirements, what percentage of students do not satisfy that

requirement? Answer: 16.35%

3. IQ score are normally distributed with mean of 100 and a standard deviation 15.

Mensa is an international society that has one – and only one qualification for

membership, a score in the top 2 on an IQ test.

a) What IQ score should one have in order to be eligible for Mensa?

Answer: hint: (x-100)/15 = invnorm(0.98), x = 130.81

b) In a typical region of 90,000 people, how many are eligible for Mensa?

Answer: 90,000 (0.02) = 1800

4. Using diaries for many weeks, a study on the lifestyle of visually impaired

students was conducted. The students kept track of many lifestyle variables

including how many hours of sleep obtained on a typical day. Researchers found

that visually impaired students averaged 9.6 hours of sleep, with a standard

deviation of 2.56 hours. Assume that the number of hours of sleep for these

visually impaired students is normally distributed.

a) What is the probability that a visually impaired student gets less than 6.1

hours of sleep? Answer: 8.58%

b) What is the probability that a visually impaired student gets between 6.3 and

10.35 hours of sleep? Answer: 51.65%

c) Forty percent of students get less than how many hours of sleep on a typical

day? Answer: 8.95 hours

5. Healthy people have body temperatures that are normally distributed with a mean

of 98.20 degree Fahrenheit and a standard deviation of 0.62 degree Fahrenheit.

a) If a healthy person is randomly selected, what is the probability that he or she

has a body temperature above 98.9 degree Fahrenheit? Answer: 12.94%

b) A hospital wants to select a minimum temperature for requiring further

medical tests. What should that temperature be, if we want only 1% of healthy

people to exceed it? Answer: hint: (x-98.2)/.62 = invnorm(0.99), 99.64

6. The heights of a large group of people are assumed to be normally distributed.

Their mean height is 68 inches, and the standard deviation is 4 inches. What

percent of these people are taller than 73 inches? Answer: 10.56%

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7. Suppose a population is normally distributed with a mean of 24.6 and a standard

deviation of 1.3. What percent of the data will lie between 25.3 and 26.8?

Answer: 24.91%