40. Statistics 1

8/2/2019 40. Statistics 1

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Mathematics


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Statistics


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Session Objectives


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Session Objectives

1. Intervals

1. Basic properties of inequalities

1. Definition and solution of linear inequation

1. Solution of modulus inequations

1. Solution of two variable inequations

1. Inequalities related to AM, GM and HM


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Interval

(i) Open interval:

{ } <


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Interval

(iii) Open closed interval:

{ } < (a, b] = x R : a x b

(iv) Closed open interval:

{ }


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Basic Properties of Inequalities

(i) If a > b, b > c, then a > c.

(ii) If a > b, then a + m > b + m.

(iii)If a > b, then am > bm for m > 0 and

am < bm for m < 0.

(iv) If a > b > 0, then > >n n1 1 1 1a b , a b , ..., a b

+ + + + > + + +n n1 1 1 1 1a a a ... a b b ... b

(vi) If , then

for all positive

numbers ai and bi for i = 1, 2, ... n.

> > > >n n1 1 1 1 1 1a b , a b , a b , ...a b

>n n1 1 1 1a a ...a b b ... b

(vii) If a > b > 0 and n > 0, then > >1 1

n n n na b and a b .


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Definition of Inequation

A statement involving variable(s) and

the sign of inequality, i.e. >, < f x or f x or f x or f x1 1 1 1


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Illustrative Example

Solve for x where x is non-negative integer.

(i) 2x + 8 = 20

(ii) 2x + 8 < 20

(iii) + x1 1 11

Solution:

(i) 2x + 8 = 20

= =x1 11 1 11

=x1

(ii) 2x + 8 < 20


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+ (iii) x1 111

x1 11

x 1

Possible values of x are , , , , ,1 1 1 1 1 , .1 1

An inequation may be linear or quadratic or cubic,etc., containing one or more variables.


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Solutions of Linear Inequations inOne Variable

It is the process of obtaining all possiblesolutions of an inequation.

Solution set

The set of all possible solutions of aninequation is known as its solutionset.

For example:

The solution set of the inequation x2 + 2 > 0 isthe set of all real numbers whereas the solutionset of the inequation x2 + 2 < 0 is the null set.


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Rules for Solving Inequations

1. Adding or subtracting the same number

or expression from each side of aninequation does not change theinequality.

2. Multiplying or dividing each side of an

inequation by the same positive numberdoes not change the inequality.

3. Multiplying each side of an inequation by

the same negative number reverse theinequality.


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Rules for Solving Inequations

Example :

Solve the inequations

+ +x x11 1 1 1

x x1 1 1 11

x1 11 x .11

Answer is ] , . ]11

.1 1 1


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Rules for Solving Absolute ValueFunction Inequations

1. Factorize the expression.

2. Get the roots of the expression or say critical point (critical pointsare the points where the expression becomes zero or ).Expression only changes its sign at critical value.

3. Make various interval on number line.

4. Assign the sign of each bracket in these intervals and check thesign of expression.

5. List out the intervals where expression is positive or negativeseparately.

6. Left as it is, when expression is positive and multiply with 1when function is negative to make

it positive.


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Example

|x + 1| > 4

Step 1: Already in factorized form, i.e x + 1

Step 2: x + 1 = 0

x = 1

Step 3: 1 1

So intervals are .( ) ( ) , and ,1 1


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Step 4:

Sign of x + 1 in their intervalis negative. To get the sign of x + 1 inthe interval we generally a value of x

which is less than 1. Say it as 1.008.Then obviously x + 1 will be negativeand in the interval expression x + 1is positive. This can be realized bytaking the value of x = 0.98 which

lies in the interval

( ) , 1

( ) ,1

( ) , .1


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Step 5: So in , (x + 1) is negative.( ) , 1and in , (x + 1) is positive.( ) ,1

Step 6:( )

+ < + = + < < x x

11x 1

x x1 1

Minus sign is added here as x + 1 is negativein this interval and we are interested in positivevalue of the expression.


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Now solve the equation.

< For x1

|x + 1| > 4

But the value of |x + 1| in this interval is x + 1.

So x + 1 > 4 x > 3

< < So the solution is x1

< < For x 1the value of |x + 1| is (x + 1).( ) + >x 11 x 1 > 4


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>x 1

< x 1 ( ) x , 1

Possible solutions are

1 1

( ) ( ) , ,11 U


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Some Important Results on ModulusInequations or Absolute Value function

Result 1: If a is a positive real number,then

(i) ( )< < < x a a x a i.e. x a, a

(ii) [ ] x a a x a i.e. x a, a

(iii) > < >x a x a or x a

(iv) x a x a or x a


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Some Important Results on ModulusInequations or Absolute Value function

Result 2: Let r be a positive real numberand a be a fixed real number. Then

(i) ( ) < < < + +x a r a r x a r i.e. x a r, a r

(ii) [ ] + +x a r a r x a r i.e. x a r, a r

(iii) > < > +x a r x a r or x a r

(iv) +x a r x a r or x a r


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Graphical Solution of One or TwoVariable Equations

Let the equation is Ax + By + C < 0.

Algorithm

Step 1: Convert the inequation into equation, i.e. Ax + By+ C = 0

Step 2: Draw the graph of Ax + By + C = 0.

Step 3: Take any point [generally (0, 0)] or any point not on theline whose position is known to you with respect to line.

Step 4: Put this point in the given inequation and check the validityof the inequation. If inequation satisfied then the correspondingside of the line where lies the chosen point is the graph of theinequation.


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Draw the graph of the inequation

2x y < 4.

Solution:

Step 1: 2x y = 4

Step 2:

( , )1 1

( , )0 0

1

y

y

x x


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Step 3: Take the point as (0, 0).

Step 4: 2 0 0 < 4

0 < 4, so it is valid. Hence,the shaded side is the requiredsolution.

( , )1 1

( , )0 0

1

y

y

x x


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Graphical Solution of Two Variable

Algorithm

(i) Draw the graph of two inequations.

(ii) Get the required common regionbounded by the two lines.


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Get the required region.

x 2y < 32x + y >1

Solution:

1

1 ,1 ( , )1 1

1

1

( , )1 1

1x

+y

= 1

x

1y

=1

1

Dotted part is for x 2y < 3.

Crossed part is for 2x + y > 1.

So the required region is thatregion where both cross anddots are present, i.e. the circledregion.


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Inequality related to AM, GM and HM

Let a and b be two real positive and

unequal numbers and A, G, H arearithmetic, geometric and harmonicmeans respectively between them.

+ = = =

+

a b ab1A , G ab and H

a b1

+ = = = + 1a b ab1Now AH ab G

a b1

=A G

...(i)G H

+ =

a bAgain A G ab

1


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+

= = >

1

a b ab a b11

1 1

>A G 1

>A G ...(ii)

( )> > >A

Again from (ii), A G G1 1G

Q

( ) = >A G

From i , 1G H

( ) > >G H H ...(iii)1Q

From (ii) and (iii), A > G > H, i.e. AM > GM > HM


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Cor: If the two numbers are equal,

i.e. a = b, then

= =

1

a bA G 1

1 =A G

= =A GAgain from (i), 1G H

=G H

Hence, A = G = H

Therefore, we can write equally holdswhen a = b.

AM GM HM


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Note: This inequality holds for n numbers

also.

+ + += n1 1 1

a a a ... aAM of 'n' numbers

n

( )=

1

nn0 0 0GM of 'n' numbers a a a ...a

=+ + + +

n1 1 1

nand HM of 'n' numbers

1 1 1 1...

a a a a

( )+ + +

+ + +

1n0 0

nn1 1

n1 1

a a ... a nAM GM HM a a ...a .

1 1 1n...

a a a


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Class Test


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Class Exercise - 1

<

x x x1 11 1Solve .

0 0 0


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Solution

Q We have, ( ) ( )

x 1

x 11

>x 1

1x 11

( )( ) >

x x1 1 1x1 1

( )

+ >

x x1 11

x1 1

( )

+ >

x 1

...(i)1x 1


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Solution contd..

( )( )

+>

x 1

From i , 1x 1

Q

Required solution region ( ) ( )= , ,11U

Here coefficient of x is positive. Now equating

numerator and denominator to zero, we getx = 5 and 5 respectively.

+ 1 1 +


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Class Exercise - 3

++ > x x1 1 11

...(i)1 1 1

+ x x1 1 11

From (i),1 1 1

+

>

x x11 1 11

1 1

>x11 11

>x ...(iii)1


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Solution contd...

From (ii),

+

1 1 1x y z 1

x y zProved.


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Class Exercise - 9

If a, b, c and d be four distinctpositive quantities in HP, then

show that

(i) a + d > b + c (ii) ad > bc


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Solution

(i) a, b, c and d are in HP.Q

For first three terms AM > HM +

>

a, b and c are in HPa cb

1 b is HM of a and c

Q

+ >a c b ...(i)1 and for last three terms

AM > HM +

>

Again b, c and d are in HPb dc

1 c is HM of b and d

+ >b d c ...(ii)1


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Solution contd..

From (i) and (ii),

a + c + b + d > 2b + 2c

Proved + > +a d b c

(ii) For first three terms, GM > HM

>ac b

> 1ac b ...(iii)

and for the last three terms >bd c

> 1bd c ...(iv)

From (iii) and (iv)

( ) ( ) > 1 1ac bd b c

>ad bc Pr oved.


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Class Exercise - 10

If a > 0, b > 0, c > 0, prove that

where

s = a + b + c.

+ + > 1 1 1 1,

s a s b s c s1


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Solution

As we know that for three positive

quantities, x, y and z, we have+ +

= =+ +

x y z 1A , H

1 1 11

x y z

and A > H.

Here (s a), (s b), (s c) are positive quantities.

( ) ( ) ( ) + + >

+ +

s a s b s c 1

1 1 11

s a s b s c


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Solution contd..

( ) + +

> + +

s a b c1 1

1 1 11

s a s b s c

[ ] > = + ++ +

s1 1

s a b c1 1 11

s a s b s c

Q

+ + > 1 1 1 1

Proved.s a s b s c s1


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Thank you

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40. Statistics 1