If x¯ is the mean of n values of x , then ∑i=1(xi−x?). equal to ________. If a has any value other than x? , then ∑i=1(xi−a)2 is_______than (xi−a)2

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar I f x ¯ i s t h e m e a n o f n o b s e r v a t i o n s o f x , t h e n ∑ i = 1 n x i − x ¯ = 0 a n d i f ' a ' h a s t h e v a l u e o t h e r t h a n x ¯ , t h e n ∑ i = 1 n ( x i − x ¯ ) 2 i s l e s s t h a n t h a n ∑ ( x i − a ) 2 . H e n c e , t h e v a l u e o f t h e f i l l e r a r e 0 a n d l e s s .

Number of ways in which three distinct numbers can be selected between 1 and 20 both inclusive, whose sum is even is ____________

All even + 2 odd 1 even ¹? C? + ¹? C? × ¹? C? (10×9×8)/6 + (10×9)/2 × 10 120 + 450 = 570

A circle passes through the points (2, 2) and (9, 9) and touches the x-axis. The absolute value of the difference of possible x-coordinate of the point of contact is ______

Family of circle through (2, 2) and (9, 9), (x-2) (x-9) + (y-2) (y-9) + λ (y-x) = 0 Touches y=0 (x-axis) (x-2) (x-9) + 18 - λx = 0 x² - 11x - λx + 36 = 0 x² - (11+λ)x + 36 = 0 has repeated roots D = 0 ⇒ λ = 1, λ = -23 x = 6 or x = -6 Difference = 12

If a,b are any two perpendicular vectors of equal magnitude and |3a+4b| + |4a-3b| = 20, then |a| equals: ____________

|3a+4b|² = 9|a|² + 16|b|² + 24a·b But a·b = 0, |a|=|b|=k |3a+4b| = 5k |4a-3b| 10k = 20? k = 2 = |a| = |b|

The value of ∫₀^(π/2) |xsin²x - 1/2|dx is equal to aπ/b where a, b are co-prime numbers, then a.b is ____________

∫ (from 0 to π/2) (x/2)|cos (2x)|dx ∫ (from 0 to π/4) (x/2)cos (2x)dx - ∫ (from π/4 to π/2) (x/2)cos (2x)dx = [x sin (2x)/4 - ∫sin (2x)/4 dx] (from 0 to π/4) - [x sin (2x)/4 - ∫sin (2x)/4 dx] (from π/4 to π/2) = [x sin (2x)/4 + cos (2x)/8] (from 0 to π/4) - [x sin (2x)/4 + cos (2x)/8] (from π/4 to π/2) = (π/16 - 1/8) - (-1/8 - π/16) = π/8

lim (x→0) (ae2x - bcosx + c)/(xsinx) = 1, then a + b + c is equal to ____________

lim (x→0) (ae²? - bcosx + c) / (x sinx) = lim (x→0) (ae²? - bcosx + c) / x² * lim (x→0) x/sinx For limit to exist a - b + c = 0 2a + b = 0 (4a-b)/2 = 1 c = 2, b = 2, a = -1 a+b+c = 3

The mean and standard deviation of a set of n1 observations are x?1 and s1 , respectively, while the mean and standard deviation of another set of n2 observations are x?2 and s2 , respectively. Show that the standard deviation of the combined set of (n1+n2) observations is given by: S.D.=n1s12+n2s22+n1n2(x?1−x?2)2n1+n2

This is a short answer type question as classified in NCERT Exemplar L e t x i , i = 1 , 2 , 3 , 4 , … n 1 a n d y j , j = 1 , 2 , 3 , 4 , … n 2 ∴ x 1 ¯ = 1 n 1 ∑ i = 1 n x i a n d x 2 ¯ = 1 n 2 ∑ i = 1 n y j ⇒ σ 1 2 = 1 n 1 ∑ i = 1 n 1 ( x i − x 1 ¯ ) 2 a n d σ 2 2 = 1 n 2 ∑ j = 1 n 2 ( y j − x 2 ¯ ) 2 N o w m e a n o f t h e c o m b i n e d s e r i e s i s g i v e n b y x ¯ = 1 n 1 + n 2 [ ∑ i = 1 n x i + ∑ i = 1 n y j ] = n 1 x 1 ¯ + n 2 x 2 ¯ n 1 + n 2 T h e r e f o r e , σ 2 o f t h e c o m b i n e d s e r i e s i s σ 2 = 1 n 1 + n 2 [ ∑ i = 1 n 1 ( x i − x ¯ ) 2 + ∑ j = 1 n 2 ( y j − x ¯ ) 2 ] N o w , ∑ i = 1 n 1 ( x i − x ¯ ) 2 = ∑ i = 1 n 1 ( x i − x j ¯ + x j ¯ − x ¯ ) 2 = ∑ i = 1 n 1 ( x i − x j ¯ ) 2 + n 1 ( x j ¯ − x ¯ ) 2 + 2 ( x j ¯ − x ¯ ) ∑ i = 1 n 1 ( x i − x j ¯ ) 2 B u t ∑ i = 1 n ( x i − x i ¯ ) = 0 [ ? The algebraic sum of the deviation of values of first s e r i e s f r o m t h e i r m e a n i s z e r o . ] A l s o ∑ i = 1 n 1 ( x i − x ¯ ) 2 = n 1 s 1 2 + n 1 ( x 1 ¯ − x ¯ ) 2 = n 1 s 1 2 + n 1 d 1 2 w h e r e d 1 = ( x 1 ¯ − x ¯ ) S i m i l a r l y , w e h a v e ∑ j = 1 n 2 ( y j − x ¯ ) 2 = ∑ j = 1 n 2 ( y j − x i ¯ + x i ¯ − x ¯ ) 2 = n 2 s 2 2 + n 2 d 2 2 w h e r e d 2 = ( x 2 ¯ − x ¯ ) Now combined Standard Deviation(SD) σ = n 1 ( s 1 2 + d 1 2 ) + n 2 ( s 2 2 + d 2 2 ) n 1 + n 2 w h e r e d 1 = x 1 ¯ − x ¯ = x 1 ¯ − ( n 1 x 1 ¯ + n 2 x 2 ¯ n 1 + n 2 ) = n 2 ( x 1 ¯ − x 2 ¯ ) n 1 + n 2 d 2 = x 2 ¯ − x ¯ = x 2 ¯ − ( n 1 x 1 ¯ +

The following information relates to a sample of size 60: x2=18000 , ∑x=960 . The variance is: (a) 6.63 (b) 16 (c) 22 (d) 44

This is an Objective Type Questions as classified in NCERT Exemplar W e k n o w t h a t V a r i a n c e ( σ 2 ) = ∑ x i 2 N − ( ∑ x i N ) 2 = 1 8 0 0 0 6 0 − ( 9 6 0 6 0 ) 2 = 3 0 0 − 2 5 6 = 4 4 H e n c e , t h e c o r r e c t o p t i o n i s ( d ) .

Mean deviation for n observations x1,x2,…,xn from their mean x? is given by (a) 1n∑i=1?xi−x?? (b) 1n∑i=1(xi−x?) (c) 1n∑i=1(xi−x?)2 (d) ∑ n ( x i − x ? ) 2 n

This is an Objective Type Questions as classified in NCERT Exemplar M D = 1 n ∑ i = 1 n | x i − x ¯ | H e n c e , t h e c o r r e c t o p t i o n i s ( b ) .

The statement A -> (B -> A) is equivalent to:

A -> (  ( A ↔ B )

Suppose y is a function of x that satisfies dy/dx = √(1-y 2 )/x 2 and y=0 at x = 2/π, then y(3/π) is equal to:

1/2

The image of the point (3, 5) in the line x – y + 1 = 0, lies on:

( x − 4 ) 2 + ( y − 4 ) 2 = 8

( x − 4 ) 2 + ( y + 2 ) 2 = 1 6

( x − 2 ) 2 + ( y − 4 ) 2 = 4

( x − 2 ) 2 + ( y − 2 ) 2 = 1 2

Let b 1 , b 2 , ................ be a geometric sequence such that b 1 + b 2 = a and Σ b i = 2. Given that b? < 0, then the value of b 1 is :

2+√2

2-√2

1+√2

4+√2

The constant term in the expansion of (log 10 (x^(log 10 x)) - log x 2 100) 12 is:

924

1050

The equation of the common tangent touching the circle (x-3) 2 + y 2 = 9 and the parabola y 2 = 4x above the x-axis, is

√3y = x+3

√3y = 3x+1

√3y = -(x+3)

All possible values of ∈ [ 0 , 2 π ]  for which sin 2q + tan 2q > 0 lie in:

( 0 , π 4 ) ∪ ( π 2 , 3 π 4 ) ∪ ( π , 5 π 4 ) ∪ ( 3 π 2 , 7 π 4 )

( 0 , π 2 ) ∪ ( π 2 , 3 π 4 ) ∪ ( π , 7 π 6 )

( 0 , π 4 ) ∪ ( π 2 , 3 π 4 ) ∪ ( 3 π 2 , 1 1 π 6 )

( 0 , π 2 ) ∪ ( π , 3 π 2 )

The value of the integral ∫ s i n θ s i n θ ( s i n 6 θ + s i n 4 θ + s i n 2 θ ) 2 s i n 4 θ + 3 s i n 2 θ + 6 1 − c o s 2 θ d θ (where c is a constant of integration)

1 1 8 [ 9 − c o s 6 θ − 3 c o s 4 θ − 6 c o s 2 θ ] 3 2 + c

1 1 8 [ 9 − 2 s i n 6 θ − 3 s i n 4 θ − 6 s i n 2 θ ] 3 2 + c

1 1 8 [ 1 1 − 1 8 c o s 2 θ + 9 c o s 4 θ − 2 c o s 6 θ ] 3 2 + c

1 1 8 [ 1 1 − 1 8 s i n 2 θ + 9 s i n 4 θ − 2 s i n 6 θ ] 3 2 + c

For the curve defined parametrically as y = 3sinθ.cosθ, x = e^θ.sinθ where θ∈[0,π], the tangent is parallel to x-axis when θ is :

π/4

3π/4

π/2

Maths NCERT Exemplar Solutions Class 11th Chapter Fifteen: Overview, Questions, Preparation

Updated on Sep 10, 2025 11:03 IST

By Payal Gupta, Retainer

Table of contents

Statistics Short Answers Type Question
Statistics Long Answers Type Question
Statistics Objective Type Question
Statistics Fill in the blank Type Question
JEE Mains 2021
JEE Mains
01 Mock Test 2025

Statistics Short Answers Type Question

1. Calculate the mean deviation about the mean of the set of first $n$ natural numbers when $n$ is an odd number.

\begin{array}{l} F i r s t n n a t u r a l n u m b e r s a r e 1, 2, 3, \dots, n . H e r e, n i s o d d . \\ ∴ M e a n \bar{x} = \frac{1 + 2 + 3 + \dots + n}{n} = \frac{\frac{n (n + 1)}{2}}{n} = \frac{n + 1}{2} \\ T h e d e v i a t i o n s o f n u m b e r s f r o m m e a n (\frac{n + 1}{2}) a r e \\ 1 - \frac{n + 1}{2}, 2 - \frac{n + 1}{2}, 3 - \frac{n + 1}{2}, \dots, n - \frac{n + 1}{2} \\ i . e ., - \frac{n - 1}{2}, - \frac{n - 3}{2}, \dots, - 2, - 1, 0, 1, 2, \dots, \frac{n - 1}{2} . \\ T h e a b s o l u t e v a l u e s o f d e v i a t i o n f r o m t h e m e a n i . e ., | x_{i} - \bar{x} | a r e \\ \frac{n - 1}{2}, \frac{n - 3}{2}, \dots, 2, 1, 0, 1, 2, \dots, \frac{n - 1}{2} . \\ T h e s u m o f a b s o l u t e v a l u e s o f d e v i a t i o n f r o m t h e m e a n i . e ., | x_{i} - \bar{x} | a r e \\ = 2 (1 + 2 + 3 + \dots t o \frac{n - 1}{2} t e r m s) \\ = 2 . \frac{\frac{n - 1}{2} (\frac{n - 1}{2} + 1)}{2} = \frac{n - 1}{2} . \frac{n + 1}{2} = \frac{n^{2} - 1}{4} . \\ ∴ M e a n d e v i a t i o n a b o u t t h e m e a n = \frac{\sum_{}^{} | x_{i} - \bar{x} |}{n} = \frac{\frac{n^{2} - 1}{4}}{n} = \frac{n^{2} - 1}{4 n} . \end{array}

2. Calculate the mean deviation about the mean of the set of first $n$ natural numbers when $n$ is an even number.

\begin{array}{l} F i r s t n n a t u r a l n u m b e r s a r e 1, 2, 3, 4, 5, 6, \dots, n . H e r e, n i s e v e n . \\ ∴ M e a n \bar{x} = \frac{1 + 2 + 3 + 4 + \dots + n}{n} = \frac{n (n + 1)}{2 n} = \frac{n + 1}{2} \\ ∴ M D = \frac{1}{n} [\begin{array}{l} | 1 - \frac{n + 1}{2} | + | 2 - \frac{n + 1}{2} | + | 3 - \frac{n + 1}{2} | + \dots + | \frac{n - 2}{2} - \frac{n + 1}{2} | + | \frac{n}{2} - \frac{n + 1}{2} | + \\ | \frac{n + 2}{2} - \frac{n + 1}{2} | + \dots + | n - \frac{n + 1}{2} | \end{array}] \\ = \frac{1}{n} [| \frac{1 - n}{2} | + | \frac{3 - n}{2} | + | \frac{5 - n}{2} | + \dots + | \frac{- 3}{2} | + | - \frac{1}{2} | + | \frac{1}{2} | + \dots + | \frac{n - 1}{2} |] \\ = \frac{1}{n} [\frac{1}{2} + \frac{3}{2} + \dots + \frac{n - 1}{2}] (\frac{n}{2}) t e r m s \\ = \frac{1}{n} {(\frac{n}{2})}^{2} = \frac{1}{n} . \frac{n^{2}}{4} = \frac{n}{4} [∵ S u m o f f i r s t o d d n n a t u r a l n u m b e r s = n^{2}] \\ H e n c e, t h e r e q u i r e d M D = \frac{n}{4} . \end{array}

Commonly asked questions

The mean and standard deviation of a set of $n_{1}$ observations are ${\overset{?}{x}}_{1}$ and $s_{1}$ , respectively, while the mean and standard deviation of another set of $n_{2}$ observations are ${\overset{?}{x}}_{2}$ and $s_{2}$ , respectively. Show that the standard deviation of the combined set of $(n_{1} + n_{2})$ observations is given by:

$S . D . = \sqrt[]{\frac{n_{1} s_{1}^{2} + n_{2} s_{2}^{2} + n_{1} n_{2} {({\overset{?}{x}}_{1} - {\overset{?}{x}}_{2})}^{2}}{n_{1} + n_{2}}}$

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} L e t x_{i}, i = 1, 2, 3, 4, \dots n_{1} \\ a n d y_{j}, j = 1, 2, 3, 4, \dots n_{2} \\ ∴ \bar{x_{1}} = \frac{1}{n_{1}} \sum_{i = 1}^{n} x_{i} a n d \bar{x_{2}} = \frac{1}{n_{2}} \sum_{i = 1}^{n} y_{j} \\ \Rightarrow σ_{1}^{2} = \frac{1}{n_{1}} \sum_{i = 1}^{n_{1}} {(x_{i} - \bar{x_{1}})}^{2} a n d σ_{2}^{2} = \frac{1}{n_{2}} \sum_{j = 1}^{n_{2}} {(y_{j} - \bar{x_{2}})}^{2} \\ N o w m e a n o f t h e c o m b i n e d s e r i e s i s g i v e n b y \\ \bar{x} = \frac{1}{n_{1} + n_{2}} [\sum_{i = 1}^{n} x_{i} + \sum_{i = 1}^{n} y_{j}] = \frac{n_{1} \bar{x_{1}} + n_{2} \bar{x_{2}}}{n_{1} + n_{2}} \\ T h e r e f o r e, σ^{2} o f t h e c o m b i n e d s e r i e s i s \\ σ^{2} = \frac{1}{n_{1} + n_{2}} [\sum_{i = 1}^{n_{1}} {(x_{i} - \bar{x})}^{2} + \sum_{j = 1}^{n_{2}} {(y_{j} - \bar{x})}^{2}] \\ N o w, \sum_{i = 1}^{n_{1}} {(x_{i} - \bar{x})}^{2} = \sum_{i = 1}^{n_{1}} {(x_{i} - \bar{x_{j}} + \bar{x_{j}} - \bar{x})}^{2} \\ = \sum_{i = 1}^{n_{1}} {(x_{i} - \bar{x_{j}})}^{2} + n_{1} {(\bar{x_{j}} - \bar{x})}^{2} + 2 (\bar{x_{j}} - \bar{x}) \sum_{i = 1}^{n_{1}} {(x_{i} - \bar{x_{j}})}^{2} \\ B u t \sum_{i = 1}^{n} (x_{i} - \bar{x_{i}}) = 0 [\begin{array}{l} ? T h e algebraic s u m o f t h e d e v i a t i o n o f v a l u e s o f f i r s t \\ s e r i e s f r o m t h e i r m e a n i s z e r o . \end{array}] \\ A l s o \sum_{i = 1}^{n_{1}} {(x_{i} - \bar{x})}^{2} = n_{1} s_{1}^{2} + n_{1} {(\bar{x_{1}} - \bar{x})}^{2} \\ = n_{1} s_{1}^{2} + n_{1} d_{1}^{2} \\ w h e r e d_{1} = (\bar{x_{1}} - \bar{x}) \\ S i m i l a r l y, w e h a v e \\ \sum_{j = 1}^{n_{2}} {(y_{j} - \bar{x})}^{2} = \sum_{j = 1}^{n_{2}} {(y_{j} - \bar{x_{i}} + \bar{x_{i}} - \bar{x})}^{2} = n_{2} s_{2}^{2} + n_{2} d_{2}^{2} w h e r e d_{2} = (\bar{x_{2}} - \bar{x}) \\ N o w c o m b i n e d Standard D e v i a t i o n (S D) \\ σ = \sqrt[]{\frac{n_{1} (s_{1}^{2} + d_{1}^{2}) + n_{2} (s_{2}^{2} + d_{2}^{2})}{n_{1} + n_{2}}} \\ w h e r e d_{1} = \bar{x_{1}} - \bar{x} = \bar{x_{1}} - (\frac{n_{1} \bar{x_{1}} + n_{2} \bar{x_{2}}}{n_{1} + n_{2}}) = \frac{n_{2} (\bar{x_{1}} - \bar{x_{2}})}{n_{1} + n_{2}} \\ d_{2} = \bar{x_{2}} - \bar{x} = \bar{x_{2}} - (n_{1} \bar{x_{1}} + \end{array}$

Calculate the mean deviation about the mean of the set of first $n$ natural numbers when $n$ is an odd number.

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} F i r s t n n a t u r a l n u m b e r s a r e 1, 2, 3, \dots, n . H e r e, n i s o d d . \\ ∴ M e a n \bar{x} = \frac{1 + 2 + 3 + \dots + n}{n} = \frac{\frac{n (n + 1)}{2}}{n} = \frac{n + 1}{2} \\ T h e d e v i a t i o n s o f n u m b e r s f r o m m e a n (\frac{n + 1}{2}) a r e \\ 1 - \frac{n + 1}{2}, 2 - \frac{n + 1}{2}, 3 - \frac{n + 1}{2}, \dots, n - \frac{n + 1}{2} \\ i . e ., - \frac{n - 1}{2}, - \frac{n - 3}{2}, \dots, - 2, - 1, 0, 1, 2, \dots, \frac{n - 1}{2} . \\ T h e a b s o l u t e v a l u e s o f d e v i a t i o n f r o m t h e m e a n i . e ., | x_{i} - \bar{x} | a r e \\ \frac{n - 1}{2}, \frac{n - 3}{2}, \dots, 2, 1, 0, 1, 2, \dots, \frac{n - 1}{2} . \\ T h e s u m o f a b s o l u t e v a l u e s o f d e v i a t i o n f r o m t h e m e a n i . e ., | x_{i} - \bar{x} | a r e \\ = 2 (1 + 2 + 3 + \dots t o \frac{n - 1}{2} t e r m s) \\ = 2 . \frac{\frac{n - 1}{2} (\frac{n - 1}{2} + 1)}{2} = \frac{n - 1}{2} . \frac{n + 1}{2} = \frac{n^{2} - 1}{4} . \\ ∴ M e a n d e v i a t i o n a b o u t t h e m e a n = \frac{\sum_{}^{} | x_{i} - \bar{x} |}{n} = \frac{\frac{n^{2} - 1}{4}}{n} = \frac{n^{2} - 1}{4 n} . \end{array}$

Calculate the mean deviation about the mean of the set of first $n$ natural numbers when $n$ is an even number.

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} F i r s t n n a t u r a l n u m b e r s a r e 1, 2, 3, 4, 5, 6, \dots, n . H e r e, n i s e v e n . \\ ∴ M e a n \bar{x} = \frac{1 + 2 + 3 + 4 + \dots + n}{n} = \frac{n (n + 1)}{2 n} = \frac{n + 1}{2} \\ ∴ M D = \frac{1}{n} [\begin{array}{l} | 1 - \frac{n + 1}{2} | + | 2 - \frac{n + 1}{2} | + | 3 - \frac{n + 1}{2} | + \dots + | \frac{n - 2}{2} - \frac{n + 1}{2} | + | \frac{n}{2} - \frac{n + 1}{2} | + \\ | \frac{n + 2}{2} - \frac{n + 1}{2} | + \dots + | n - \frac{n + 1}{2} | \end{array}] \\ = \frac{1}{n} [| \frac{1 - n}{2} | + | \frac{3 - n}{2} | + | \frac{5 - n}{2} | + \dots + | \frac{- 3}{2} | + | - \frac{1}{2} | + | \frac{1}{2} | + \dots + | \frac{n - 1}{2} |] \\ = \frac{1}{n} [\frac{1}{2} + \frac{3}{2} + \dots + \frac{n - 1}{2}] (\frac{n}{2}) t e r m s \\ = \frac{1}{n} {(\frac{n}{2})}^{2} = \frac{1}{n} . \frac{n^{2}}{4} = \frac{n}{4} [? S u m o f f i r s t o d d n n a t u r a l n u m b e r s = n^{2}] \\ H e n c e, t h e r e q u i r e d M D = \frac{n}{4} . \end{array}$

Find the standard deviation of the first $n$ natural numbers.

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} \sum_{}^{} x_{i} = 1 + 2 + 3 + 4 + \dots + n = \frac{n (n + 1)}{2} \\ \sum_{}^{} x_{i}^{2} = 1^{2} + 2^{2} + 3^{2} + 4^{2} + \dots + n^{2} = \frac{n (n + 1) (2 n + 1)}{6} \\ ∴ S . D . (σ) = \sqrt[]{\frac{\sum_{}^{} x_{i}^{2}}{n} - {(\frac{\sum_{}^{} x_{i}}{n})}^{2}} \\ = \sqrt[]{\frac{n (n + 1) (2 n + 1)}{6 n} - \frac{n^{2} {(n + 1)}^{2}}{4 n^{2}}} \\ = \sqrt[]{\frac{(n + 1) (2 n + 1)}{6} - \frac{{(n + 1)}^{2}}{4}} \\ = \sqrt[]{\frac{2 n^{2} + 3 n + 1}{6} - \frac{n^{2} + 2 n + 1}{4}} \\ = \sqrt[]{\frac{4 n^{2} + 6 n + 2 - 3 n^{2} - 6 n - 3}{12}} = \sqrt[]{\frac{n^{2} - 1}{12}} \\ H e n c e, t h e r e q u i r e d S D = \sqrt[]{\frac{n^{2} - 1}{12}} . \end{array}$

The mean and standard deviation of some data for the time taken to complete a test are calculated with the following results:

Number of observations = 25, mean = 18.2 seconds, standard deviation = 3.25 seconds. Further, another set of 15 observations $x_{1}, x_{2}, \dots, x_{15}$ , also in seconds, is now available, and we have: $\sum_{i = 1} x_{i} = 279$ $\sum_{i = 1} x_{i}^{2} = 5524$ Calculate the standard deviation based on all 40 observations.

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t n_{1} = 25, \bar{x_{1}} = 18.2 a n d σ_{1} = 3.25 \\ a n d n_{2} = 15, \sum_{i = 1} x_{i} = 279 a n d \sum_{i = 1} x_{i}^{2} = 5524 \\ F o r t h e f i r s t s e t, w e h a v e \\ \sum_{}^{} x_{1} = 25 \times 18.2 = 455 \\ ∴ σ_{1}^{2} = \frac{\sum_{}^{} x_{i}^{2}}{25} - {(18.2)}^{2} \\ \Rightarrow {(3.25)}^{2} = \frac{\sum_{}^{} x_{i}^{2}}{25} - 331.24 \Rightarrow 10.5625 + 331.24 = \frac{\sum_{}^{} x_{i}^{2}}{25} \\ \Rightarrow \sum_{}^{} x_{i}^{2} = 25 \times (10.5625 + 331.24) \\ = 25 \times 341.8025 = 8545.06 \\ F o r t h e c o m b i n e d s t a n d a r d d e v i a t i o n o f t h e 40 o b s e r v a t i o n, n = 40 \\ a n d \sum_{}^{} x_{i}^{2} = 5524 + 8545.06 = 14069.06 \\ \Rightarrow \sum_{}^{} x_{i} = 455 + 279 = 734 \\ ∴ S D = \sqrt[]{\frac{14069.06}{40} - {(\frac{734}{40})}^{2}} = \sqrt[]{351.7265 - {(18.35)}^{2}} \\ = \sqrt[]{351.7265 - 336.7225} = \sqrt[]{15.004} = 3.87 \\ H e n c e, t h e r e q u i r e d S D = 3.87 \end{array}$

Two sets, each of 20 observations, have the same standard deviation 5. The first set has a mean of 17, and the second set has a mean of 22. Determine the standard deviation of the set obtained by combining the two given sets.

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t n_{1} = 20, \bar{x_{1}} = 17 a n d σ_{1} = 5 \\ a n d n_{2} = 20, \bar{x_{2}} = 22 a n d σ_{2} = 5 \\ W e k n o w t h a t f o r t h e c o m b i n e d t w o s e r i e s t h a t \\ ∴ σ = \sqrt[]{\frac{n_{1} s_{1}^{2} + n_{2} s_{2}^{2}}{n_{1} + n_{2}} + \frac{n_{1} n_{2} {(\bar{x_{1}} - \bar{x_{2}})}^{2}}{{(n_{1} + n_{2})}^{2}}} = \sqrt[]{\frac{20 \times {(5)}^{2} + 20 \times {(5)}^{2}}{20 + 20} + \frac{20 \times 20 {(17 - 22)}^{2}}{{(20 + 20)}^{2}}} \\ = \sqrt[]{\frac{1000}{40} + \frac{400 \times 25}{1600}} = \sqrt[]{25 + \frac{25}{4}} = \sqrt[]{\frac{125}{4}} = \sqrt[]{31.25} = 5.59 \\ H e n c e, t h e r e q u i r e d S D = 5.59 \end{array}$

The mean life of a sample of 60 bulbs was 650 hours, and the standard deviation was 8 hours. A second sample of 80 bulbs has a mean life of 660 hours and a standard deviation of 7 hours. Find the overall standard deviation.

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t n_{1} = 60, \bar{x_{1}} = 650 a n d s_{1} = 8 \\ a n d n_{2} = 80, \bar{x_{2}} = 660 a n d s_{2} = 7 \\ W e k n o w t h a t f o r t h e c o m b i n e d t w o s e r i e s t h a t \\ ∴ σ = \sqrt[]{\frac{n_{1} s_{1}^{2} + n_{2} s_{2}^{2}}{n_{1} + n_{2}} + \frac{n_{1} n_{2} {(\bar{x_{1}} - \bar{x_{2}})}^{2}}{{(n_{1} + n_{2})}^{2}}} = \sqrt[]{\frac{60 \times {(8)}^{2} + 80 \times {(7)}^{2}}{60 + 80} + \frac{60 \times 80 {(650 - 660)}^{2}}{{(60 + 80)}^{2}}} \\ = \sqrt[]{\frac{6 \times 64 + 8 \times 49}{14} + \frac{60 \times 80 \times 100}{140 \times 140}} = \sqrt[]{\frac{192 + 196}{7} + \frac{1200}{49}} \\ = \sqrt[]{\frac{388}{7} + \frac{1200}{49}} = \sqrt[]{\frac{2716 + 1200}{49}} = \sqrt[]{\frac{3916}{49}} = \frac{62.58}{7} = 8.9 \\ H e n c e, t h e r e q u i r e d S D = 8.9 \end{array}$

Mean and standard deviation of 100 items are 50 and 4, respectively. Find the sum of all the items and the sum of the squares of the items.

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t \bar{x} = 50, n = 100 a n d S D (σ) = 4 \\ \bar{x} = \frac{\sum_{}^{} x_{i}}{N} \Rightarrow 50 = \frac{\sum_{}^{} x_{i}}{100} \Rightarrow \sum_{}^{} x_{i} = 5000 \\ a n d variance σ^{2} = \frac{\sum_{}^{} f_{i} x_{i}^{2}}{N} - {(\frac{\sum_{}^{} f_{i} x_{i}}{N})}^{2} \\ {(4)}^{2} = \frac{\sum_{}^{} f_{i} x_{i}^{2}}{100} - {(50)}^{2} \Rightarrow 16 = \frac{\sum_{}^{} f_{i} x_{i}^{2}}{100} - 2500 \\ ∴ \sum_{}^{} f_{i} x_{i}^{2} = (2500 + 16) \times 100 \\ \Rightarrow \sum_{}^{} f_{i} x_{i}^{2} = 2516 \times 100 = 251600 \\ H e n c e, t h e r e q u i r e d s u m a r e 5000 a n d 251600 . \end{array}$

If for a distribution: ∑ $(x - 5) = 3$ $\sum {(x - 5)}^{2} = 43$ and the total number of items is 18, find the mean and standard deviation.

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t n = 18, \sum_{}^{} (x - 5) = 3, \sum_{}^{} {(x - 5)}^{2} = 43 \\ ∴ M e a n = A + \frac{\sum_{}^{} (x - 5)}{n} = 5 + \frac{3}{18} = \frac{93}{18} = 5.166 = 5.17 \\ a n d S . D . (σ) = \sqrt[]{\frac{\sum_{}^{} {(x - 5)}^{2}}{n} - {(\frac{\sum_{}^{} (x - 5)}{n})}^{2}} = \sqrt[]{\frac{43}{18} - {(\frac{3}{18})}^{2}} \\ = \sqrt[]{2.39 - {(0.166)}^{2}} = \sqrt[]{2.39 - 0.027} = 1.54 \\ H e n c e, t h e r e q u i r e d m e a n i s 5.17 a n d S D = 1.54 \end{array}$

Statistics Long Answers Type Question

1. Mean and standard deviation of 100 observations were found to be 40 and 10, respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the correct standard deviation.

\begin{array}{l} G i v e n t h a t n = 100, \bar{x} = 40, σ = 10 \\ ∴ \bar{x} = \frac{\sum_{}^{} x_{i}}{n} \Rightarrow 40 = \frac{\sum_{}^{} x_{i}}{100} \Rightarrow \sum_{}^{} x_{i} = 4000 \\ C o r r e c t e d \sum_{}^{} x_{i} = 4000 - 30 - 70 + 3 + 27 = 3930 \\ a n d C o r r e c t e d m e a n = \frac{3930}{100} = 39.3 \\ N o w σ^{2} = \frac{\sum_{}^{} x_{i}^{2}}{n} - {(40)}^{2} \Rightarrow 100 = \frac{\sum_{}^{} x_{i}^{2}}{100} - 1600 \\ \Rightarrow \sum_{}^{} x_{i}^{2} = 1700 \times 100 \Rightarrow \sum_{}^{} x_{i}^{2} = 170000 \\ ∴ C o r r e c t e d \sum_{}^{} x_{i}^{2} = 170000 - {(30)}^{2} - {(70)}^{2} + {(3)}^{2} + {(27)}^{2} \\ = 170000 - 900 - 4900 + 9 + 729 = 164938 \\ ∴ C o r r e c t S D = \sqrt[]{\frac{164938}{100} - {(39.3)}^{2}} = \sqrt[]{1649.38 - 1544.49} = \sqrt[]{104.89} = 10.24 \\ H e n c e, t h e r e q u i r e d S D = 10.24 \end{array}

2. While calculating the mean and variance of 10 readings, a student wrongly used the reading 52 for the correct reading 25. He obtained the mean and variance as 45 and 16, respectively. Find the correct mean and variance.

\begin{array}{l} G i v e n t h a t n = 10, \bar{x} = 45, σ^{2} = 16 \\ ∴ \bar{x} = \frac{\sum_{}^{} x_{i}}{n} \Rightarrow 45 = \frac{\sum_{}^{} x_{i}}{10} \Rightarrow \sum_{}^{} x_{i} = 450 \\ C o r r e c t e d \sum_{}^{} x_{i} = 450 - 52 + 25 = 423 \\ a n d C o r r e c t e d m e a n = \frac{423}{10} = 42.3 \\ N o w σ^{2} = \frac{\sum_{}^{} x_{i}^{2}}{n} - {(\frac{\sum_{}^{} x_{i}}{n})}^{2} \Rightarrow 16 = \frac{\sum_{}^{} x_{i}^{2}}{10} - {(45)}^{2} \\ \Rightarrow 16 = \frac{\sum_{}^{} x_{i}^{2}}{10} - 2025 \Rightarrow \frac{\sum_{}^{} x_{i}^{2}}{10} = 2041 \\ \Rightarrow \sum_{}^{} x_{i}^{2} = 2041 \times 10 \Rightarrow \sum_{}^{} x_{i}^{2} = 20410 \\ ∴ C o r r e c t e d \sum_{}^{} x_{i}^{2} = 20410 - {(52)}^{2} + {(25)}^{2} \\ = 20410 - 2704 + 625 = 18331 \\ a n d C o r r e c t e d v a r i a n c e \\ σ^{2} = \frac{18331}{10} - {(42.3)}^{2} = 1833.1 - 1789.3 = 43.8 \\ H e n c e, t h e r e q u i r e d m e a n = 42.3 a n d v a r i a n c e = 43.8 \end{array}

Commonly asked questions

While calculating the mean and variance of 10 readings, a student wrongly used the reading 52 for the correct reading 25. He obtained the mean and variance as 45 and 16, respectively. Find the correct mean and variance.

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t n = 10, \bar{x} = 45, σ^{2} = 16 \\ ∴ \bar{x} = \frac{\sum_{}^{} x_{i}}{n} \Rightarrow 45 = \frac{\sum_{}^{} x_{i}}{10} \Rightarrow \sum_{}^{} x_{i} = 450 \\ C o r r e c t e d \sum_{}^{} x_{i} = 450 - 52 + 25 = 423 \\ a n d C o r r e c t e d m e a n = \frac{423}{10} = 42.3 \\ N o w σ^{2} = \frac{\sum_{}^{} x_{i}^{2}}{n} - {(\frac{\sum_{}^{} x_{i}}{n})}^{2} \Rightarrow 16 = \frac{\sum_{}^{} x_{i}^{2}}{10} - {(45)}^{2} \\ \Rightarrow 16 = \frac{\sum_{}^{} x_{i}^{2}}{10} - 2025 \Rightarrow \frac{\sum_{}^{} x_{i}^{2}}{10} = 2041 \\ \Rightarrow \sum_{}^{} x_{i}^{2} = 2041 \times 10 \Rightarrow \sum_{}^{} x_{i}^{2} = 20410 \\ ∴ C o r r e c t e d \sum_{}^{} x_{i}^{2} = 20410 - {(52)}^{2} + {(25)}^{2} \\ = 20410 - 2704 + 625 = 18331 \\ a n d C o r r e c t e d v a r i a n c e \\ σ^{2} = \frac{18331}{10} - {(42.3)}^{2} = 1833.1 - 1789.3 = 43.8 \\ H e n c e, t h e r e q u i r e d m e a n = 42.3 a n d v a r i a n c e = 43.8 \end{array}$

Mean and standard deviation of 100 observations were found to be 40 and 10, respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the correct standard deviation.

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t n = 100, \bar{x} = 40, σ = 10 \\ ∴ \bar{x} = \frac{\sum_{}^{} x_{i}}{n} \Rightarrow 40 = \frac{\sum_{}^{} x_{i}}{100} \Rightarrow \sum_{}^{} x_{i} = 4000 \\ C o r r e c t e d \sum_{}^{} x_{i} = 4000 - 30 - 70 + 3 + 27 = 3930 \\ a n d C o r r e c t e d m e a n = \frac{3930}{100} = 39.3 \\ N o w σ^{2} = \frac{\sum_{}^{} x_{i}^{2}}{n} - {(40)}^{2} \Rightarrow 100 = \frac{\sum_{}^{} x_{i}^{2}}{100} - 1600 \\ \Rightarrow \sum_{}^{} x_{i}^{2} = 1700 \times 100 \Rightarrow \sum_{}^{} x_{i}^{2} = 170000 \\ ∴ C o r r e c t e d \sum_{}^{} x_{i}^{2} = 170000 - {(30)}^{2} - {(70)}^{2} + {(3)}^{2} + {(27)}^{2} \\ = 170000 - 900 - 4900 + 9 + 729 = 164938 \\ ∴ C o r r e c t S D = \sqrt[]{\frac{164938}{100} - {(39.3)}^{2}} = \sqrt[]{1649.38 - 1544.49} = \sqrt[]{104.89} = 10.24 \\ H e n c e, t h e r e q u i r e d S D = 10.24 \end{array}$

Statistics Objective Type Question

1. Mean deviation for $n$ observations $x_{1}, x_{2}, \dots, x_{n}$ from their mean $\overset{ˉ}{x}$ is given by

(a) $\frac{1}{n} \sum_{i = 1} ∣ x_{i} - \overset{ˉ}{x} ∣$

(b) $\frac{1}{n} \sum_{i = 1} (x_{i} - \overset{ˉ}{x})$

(c) $\sqrt[]{\frac{1}{n} \sum_{i = 1} {(x_{i} - \overset{ˉ}{x})}^{2}}$

(d) $\frac{\sum^{n} {(x_{i} - \overset{ˉ}{x})}^{2}}{n}$

\begin{array}{l} M D = \frac{1}{n} \sum_{i = 1}^{n} | x_{i} - \bar{x} | \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}

2. Let $x_{1}, x_{2}, \dots, x_{n}$ be $n$ observations and $\overset{ˉ}{x}$ be their arithmetic mean. The formula for the standard deviation is given by:

(a) $\sqrt[]{\frac{1}{n} \sum_{i = 1} {(x_{i} - \overset{ˉ}{x})}^{2}}$

(b) $\frac{1}{n} \sum_{i = 1} {(x_{i} - \overset{ˉ}{x})}^{2}$

(c) $\sqrt[]{\frac{\sum^{n} x_{i}^{2}}{n} - {\overset{ˉ}{x}}^{2}}$

(d) $\sum_{i = 1} x_{i}^{2} + n {\overset{ˉ}{x}}^{2}$

\begin{array}{l} T h e f o r m u l a f o r S . D . = σ = \sqrt[]{\frac{\sum_{}^{} {(x_{i} - \bar{x})}^{2}}{n}} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}

Commonly asked questions

The mean of 100 observations is 50 and their standard deviation is 5. The sum of all squares of all the observations is:

(a) 50000

(b) 250000

(c) 252500

(d) 255000

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} H e r e \bar{x} = \frac{\sum_{}^{} x_{i}}{n} \\ 50 = \frac{\sum_{}^{} x_{i}}{100} \Rightarrow \sum_{}^{} x_{i} = 5000 \\ ∴ S . D . = \sqrt[]{\frac{\sum_{}^{} x_{i}^{2}}{n} - {(\frac{\sum_{}^{} x_{i}}{n})}^{2}} \\ 5 = \sqrt[]{\frac{\sum_{}^{} x_{i}^{2}}{100} - {(\frac{5000}{100})}^{2}} \Rightarrow 25 = \frac{\sum_{}^{} x_{i}^{2}}{100} - 2500 \\ \Rightarrow \frac{\sum_{}^{} x_{i}^{2}}{100} = 2500 + 25 \Rightarrow \frac{\sum_{}^{} x_{i}^{2}}{100} = 2525 \\ ∴ \sum_{}^{} x_{i}^{2} = 2525 \times 100 = 252500 \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

The following information relates to a sample of size 60: $x^{2} = 18000$ , $\sum x = 960$ . The variance is:

(a) 6.63

(b) 16

(c) 22

(d) 44

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} W e k n o w t h a t V a r i a n c e (σ^{2}) = \frac{\sum_{}^{} x_{i}^{2}}{N} - {(\frac{\sum_{}^{} x_{i}}{N})}^{2} \\ = \frac{18000}{60} - {(\frac{960}{60})}^{2} = 300 - 256 = 44 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

Mean deviation for $n$ observations $x_{1}, x_{2}, \dots, x_{n}$ from their mean $\overset{?}{x}$ is given by

(a) $\frac{1}{n} \sum_{i = 1} ? x_{i} - \overset{?}{x} ?$

(b) $\frac{1}{n} \sum_{i = 1} (x_{i} - \overset{?}{x})$

(c) $\sqrt[]{\frac{1}{n} \sum_{i = 1} {(x_{i} - \overset{?}{x})}^{2}}$

(d) $\sqrt[]{\frac{\sum^{n} {(x_{i} - \overset{?}{x})}^{2}}{n}}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} M D = \frac{1}{n} \sum_{i = 1}^{n} | x_{i} - \bar{x} | \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

Let $x_{1}, x_{2}, \dots, x_{n}$ be $n$ observations and $\overset{?}{x}$ be their arithmetic mean. The formula for the standard deviation is given by:

(a) $\frac{1}{n} \sum_{i = 1} {(x_{i} - \overset{?}{x})}^{2}$

(b) $\frac{1}{n} \sum_{i = 1} {(x_{i} - \overset{?}{x})}^{2}$

(c) $\frac{\sum^{n} x_{i}^{2}}{n} - {\overset{?}{x}}^{2}$

(d) $\sum_{i = 1} x_{i}^{2} + n {\overset{?}{x}}^{2}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T h e f o r m u l a f o r S . D . = σ = \sqrt[]{\frac{\sum_{}^{} {(x_{i} - \bar{x})}^{2}}{n}} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

Let $a, b, c, d, e$ be the observations with mean $m$ and standard deviation $s$ . The standard deviation of the observations $a + k, b + k, c + k, d + k, e + k$ is:

(a) $s$

(b) $k \cdot s$

(c) $s + k$

(d) $\frac{S}{K}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n o b s e r v a t i o n a r e a, b, c, d a n d e \\ ∴ M e a n = m = \frac{a + b + c + d + e}{5} \\ ∴ \sum_{}^{} x_{i} = 5 m \\ N o w m e a n o f a + K, b + K, c + K, d + K a n d e + K i s \\ = \frac{a + K + b + K + c + K + d + K + e + K}{5} \\ = \frac{(a + b + c + d + e) + 5 K}{5} = \frac{5 m + 5 K}{5} = m + K \\ ∴ S . D . = \sqrt[]{\frac{\sum_{}^{} {(x_{i} + K)}^{2}}{N} - {(\frac{\sum_{}^{} x_{i} + K}{N})}^{2}} \\ = \sqrt[]{\frac{\sum_{}^{} (x_{i}^{2} + K^{2} + 2 x_{i} K)}{N} - {(m + K)}^{2}} \\ = \sqrt[]{\frac{\sum_{}^{} x_{i}^{2}}{N} + \frac{\sum_{}^{} K^{2}}{N} + \frac{2 K \sum_{}^{} x_{i}}{N} - m^{2} - K^{2} - 2 m K} \\ = \sqrt[]{\frac{\sum_{}^{} x_{i}^{2}}{N} + K^{2} + 2 K m - m^{2} - K^{2} - 2 m K} = \sqrt[]{\frac{\sum_{}^{} x_{i}^{2}}{N} - m^{2}} [? \frac{\sum_{}^{} x_{i}}{N} = m] \\ = S \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

Let $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ be the observations with mean $m$ and standard deviation $s$ . The standard deviation of the observations $k x_{1}, k x_{2}, k x_{3}, k x_{4}, k x_{5}$ is:

(a) $k + s$

(b) s/k

(c) $k \cdot s$

(d) $s$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} H e r e m = \frac{\sum_{}^{} x_{i}}{N}, S = \sqrt[]{\frac{\sum_{}^{} x_{i}^{2}}{5} - {(\frac{\sum_{}^{} x_{i}}{5})}^{2}} \\ ∴ S D = \sqrt[]{\frac{K^{2} \sum_{}^{} x_{i}^{2}}{5} - {(\frac{K \sum_{}^{} x_{i}}{5})}^{2}} = \sqrt[]{\frac{K^{2} \sum_{}^{} x_{i}^{2}}{5} - K^{2} {(\frac{\sum_{}^{} x_{i}}{5})}^{2}} \\ = K \sqrt[]{\frac{\sum_{}^{} x_{i}^{2}}{5} - {(\frac{\sum_{}^{} x_{i}}{5})}^{2}} = K . S \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

Let $x_{1}, x_{2}, \dots, x_{n}$ be $n$ observations. Let $w_{i} = l x_{i} + k$ for $i = 1, 2, \dots, n$ , where $l$ and $k$ are constants. If the mean of $x_{i}$ â€™s is 48 and their standard deviation is 12, the mean of $w_{i}$ â€™s is 55 and the standard deviation of $w_{i}$ â€™s is 15, the values of $l$ and $k$ should be:

(a) $l = 1.25, k = - 5$

(b) $l = 2.5, k = - 5$

(c) $l = - 1.25, k = 5$

(d) $l = 2.5, k = 5$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t w_{i} = x_{i} + k, \bar{x_{i}} = 48, S D (x_{i}) = 12, \\ w_{i} = 55 a n d S D (w_{i}) = 15 \\ t h e n \bar{w_{i}} = \bar{x_{i}} + k (\bar{w_{i}} = m e a n o f w_{i}' s a n d \bar{x_{i}} i s t h e m e a n o f x_{i}' s) \\ \Rightarrow 55 = 48 + k \dots (i) \\ S D o f w_{i} = S D o f x_{i} \\ 15 = l \times 12 \\ \Rightarrow l = \frac{15}{12} = 1.25 \dots (i i) \\ f r o m e q n . (i) a n d (i i) w e h a v e \\ k = \bar{w_{i}} - \bar{x_{i}} = 55 - 1.25 \times 48 = 55 - 60 = - 5 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

Standard deviations for the first 10 natural numbers is:

(a) 5.5

(b) 3.87

(c) 2.97

(d) 2.87

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} W e k n o w t h a t S D o f f i r s t n n a t u r a l n u m b e r s \sqrt[]{\frac{n^{2} - 1}{12}} \\ H e r e n = 10 \\ ∴ S D = \sqrt[]{\frac{{(10)}^{2} - 1}{12}} = \sqrt[]{\frac{99}{12}} = \sqrt[]{8.25} = 2.87 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

Consider the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$ . If 1 is added to each number, the variance of the numbers so obtained is:

(a) 6.5

(b) 2.87

(c) 3.87

(d) 8.25

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n n u m b e r s a r e 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 \\ N u m b e r s o b t a i n e d w h e n 1 i s a d d e d t o t h e a b o v e n u m b e r s i s 2, 3, 4, 5, 6, 7, 8, 9, 10 a n d 11 . \\ ∴ \sum_{}^{} x_{i} = 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 \\ = \frac{10}{2} [2 \times 2 + (10 - 1) . 1] = 5 [4 + 9] = 5 \times 13 = 65 \\ N o w \sum_{}^{} x_{i}^{2} = 2^{2} + 3^{2} + 4^{2} + \dots + 1 1^{2} \\ = (1^{2} + 2^{2} + 3^{2} + 4^{2} + \dots + 1 1^{2}) - {(1)}^{2} \\ = \frac{n (n + 1) (2 n + 1)}{6} - 1 = \frac{11 \times 12 \times 23}{6} - 1 \\ = 22 \times 23 - 1 = 506 - 1 = 505 \\ ∴ V a r i a n c e (σ^{2}) = \frac{\sum_{}^{} x_{i}^{2}}{n} - {(\frac{\sum_{}^{} x_{i}}{n})}^{2} = \frac{505}{10} - {(\frac{65}{10})}^{2} \\ = 50.5 - {(6.5)}^{2} = 50.5 - 42.25 = 8.25 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

Consider the first 10 positive integers. If we multiply each number by -1 and then add 1 to each number, the variance of the numbers so obtained is:

(a) 8.25

(b) 6.5

(c) 3.87

(d) 2.87

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} F i r s t 10 p o s i t i v e integers a r e 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 \\ o n m u l t i p l y i n g e a c h n u m b e r b y - 1, w e g e t \\ - 1, - 2, - 3, - 4, - 5, - 6, - 7, - 8, - 9, - 10 \\ o n a d d i n g 1 t o e a c h o f t h e n u m b e r, w e g e t \\ 0, - 1, - 2, - 3, - 4, - 5, - 6, - 7, - 8, - 9 \\ ∴ \sum_{}^{} x_{i} = 0 - 1 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 \\ = - 45 \\ a n d \sum_{}^{} x_{i}^{2} = 0^{2} + {(- 1)}^{2} + {(- 2)}^{2} + {(- 3)}^{2} + {(- 4)}^{2} + \dots + {(- 9)}^{2} \\ = \frac{9 \times 10 \times 19}{6} = 285 [? \sum_{}^{} n^{2} = \frac{n (n + 1) (2 n + 1)}{6}] \\ ∴ S . D = \sqrt[]{\frac{\sum_{}^{} x_{i}^{2}}{N} - {(\frac{\sum_{}^{} x_{i}}{N})}^{2}} = \sqrt[]{\frac{285}{10} - {(\frac{- 45}{10})}^{2}} \\ = \sqrt[]{\frac{285}{10} - \frac{2025}{100}} = \sqrt[]{\frac{2850 - 2025}{100}} = \sqrt[]{8.25} \\ ∴ V a r i a n c e = {(S D)}^{2} = {(\sqrt[]{8.25})}^{2} = 8.25 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

Coefficient of variation of two distributions are 50 and 60, and their arithmetic means are 30 and 25 respectively. The difference of their standard deviations is:

(a) 0

(b) 1

(c) 1.5

(d) 2.5

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} H e r e, w e h a v e C V_{1} = 50, C V_{2} = 60 \\ \bar{x_{1}} = 30 a n d \bar{x_{2}} = 25 \\ ∴ C V_{1} = \frac{σ_{1}}{\bar{x_{1}}} \times 100 \Rightarrow 50 = \frac{σ_{1}}{30} \times 100 \Rightarrow σ_{1} = \frac{50 \times 30}{100} = 15 \\ a n d C V_{1} = \frac{σ_{2}}{\bar{x_{2}}} \times 100 \Rightarrow 60 = \frac{σ_{2}}{25} \times 100 \Rightarrow σ_{2} = \frac{60 \times 25}{100} = 15 \\ ∴ D i f f e r e n c e σ_{1} - σ_{2} = 15 - 15 = 0 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

The standard deviation of some temperature data in $?^{?} C$ is 5. If the data were converted into $?^{?} F$ , the variance would be:

(a) 81

(b) 57

(c) 36

(d) 25

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t σ_{C} = 5 \\ W e k n o w t h a t C = \frac{5}{9} (F - 32) \Rightarrow F = \frac{9 C}{5} + 32 \\ ∴ σ_{F} = \frac{9}{5} σ_{C} = \frac{9}{5} \times 5 = 9 \\ ∴ σ_{F}^{2} = {(9)}^{2} = 81 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

Statistics Fill in the blank Type Question

1. Coefficient of variation = .........../mean × 100

\begin{array}{l} G i v e n t h a t σ_{C} = 5 \\ C V = \frac{S D}{M e a n} \times 100 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s S D . \end{array}

2. If $\bar{x}$ is the mean of $n$ values of $x$ , then $\sum_{i = 1} (x_{i} - \overset{ˉ}{x}) .$ equal to ________

If $a$ has any value other than $\overset{ˉ}{x}$ , then $\sum_{i = 1} {(x_{i} - a)}^{2}$ is_______than ${(x_{i} - a)}^{2}$

\begin{array}{l} I f \bar{x} i s t h e m e a n o f n o b s e r v a t i o n s o f x, t h e n \sum_{i = 1}^{n} x_{i} - \bar{x} = 0 a n d i f' a' h a s t h e v a l u e o t h e r \\ t h a n \bar{x}, t h e n \sum_{i = 1}^{n} {(x_{i} - \bar{x})}^{2} i s l e s s t h a n t h a n \sum_{}^{} {(x_{i} - a)}^{2} . \\ H e n c e, t h e v a l u e o f t h e f i l l e r a r e 0 a n d l e s s . \end{array}

Commonly asked questions

Coefficient of variation = ............/mean × 100.

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t σ_{C} = 5 \\ C V = \frac{S D}{M e a n} \times 100 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s S D . \end{array}$

If $\bar{x}$ is the mean of $n$ values of $x$ , then $\sum_{i = 1} (x_{i} - \overset{?}{x}) .$ equal to ________. If $a$ has any value other than $\overset{?}{x}$ , then $\sum_{i = 1} {(x_{i} - a)}^{2}$ is_______than ${(x_{i} - a)}^{2}$

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} I f \bar{x} i s t h e m e a n o f n o b s e r v a t i o n s o f x, t h e n \sum_{i = 1}^{n} x_{i} - \bar{x} = 0 a n d i f' a' h a s t h e v a l u e o t h e r \\ t h a n \bar{x}, t h e n \sum_{i = 1}^{n} {(x_{i} - \bar{x})}^{2} i s l e s s t h a n t h a n \sum_{}^{} {(x_{i} - a)}^{2} . \\ H e n c e, t h e v a l u e o f t h e f i l l e r a r e 0 a n d l e s s . \end{array}$

If the variance of a data is 121, then the standard deviation of the data is ________

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$W e k n o w t h a t S D =$ =√variance= $\begin{array}{l} \sqrt[]{121} = 11 \end{array}$

$\begin{array}{l} H e n c e, t h e v a l u e o f t h e f i l l e r i s 11 . \end{array}$

The standard deviation of a data is ________ of any change in origin but is ________ on the change of scale.

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} Since t h e standard d e v i a t i o n o f a n y d a t a i s i n d e p e n d e n t o f a n y c h a n g e i n o r i g i n b u t i s \\ d e p e n d e n t o f a n y c h a n g e o f s c a l e . \\ H e n c e, t h e v a l u e o f t h e f i l l e r a r e i n d e p e n d e n t a n d d e p e n d e n t . \end{array}$

The sum of the squares of the deviations of the values of the variable is ________when taken about their arithmetic mean.

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T h e s u m o f t h e s q u a r e s o f t h e d e v i a t i o n s o f t h e v a l u e o f i s minimum w h e n t a k e n \\ a b o u t t h e i r a r i t h m e t i c m e a n . \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s minimum \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} . \end{array}$

The mean deviation of the data is ________ when measured from the median.

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T h e m e a n d e v i a t i o n o f t h e d a t a i s l e a s t w h e n m e a s u r e d f r o m t h e m e d i a n . \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s l e a s t . \end{array}$

The standard deviation is ________to the mean deviation taken from the arithmetic mean.

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T h e s t a n d a r d d e v i a t i o n s i s g r e a t e r t h a n o r e q u a l t o t h e m e a n d e v i a t i o n t a k e n f r o m t h e \\ a r i t h m e t i c m e a n . \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s g r e a t e r t h a n o r e q u a l . \end{array}$

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Maths NCERT Exemplar Solutions Class 11th Chapter Fifteen Exam

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Maths NCERT Exemplar Solutions Class 11th Chapter Fifteen: Overview, Questions, Preparation

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