NCERT Exemplar for Class 11 Maths Chapter 15 - Statistics (Book Solutions)

Solved Examples

Example 1: Find the mean deviation about the mean of the following data:

Ans: Given: A discrete frequency distribution

For a discrete frequency distribution, mean deviation about mean is given by

\[M.D.(\bar x) = \dfrac{{\sum\limits_{i = 1}^n {{f_i}|{x_i} - \bar x|} }}{{\sum\limits_{i = 1}^n {{f_i}} }}\] where \[\bar x\] is the mean of distribution.

\[ \Rightarrow \bar x = \dfrac{{\sum\limits_{i = 1}^n {{f_i}} {x_i}}}{{\sum\limits_{i = 1}^n {{f_i}} }} = \dfrac{{336}}{{42}} = 8\]

\[ \Rightarrow M.D.(\bar x) = \dfrac{{\sum\limits_{i = 1}^n {{f_i}|{x_i} - \bar x|} }}{{\sum\limits_{i = 1}^n {{f_i}} }} = \dfrac{{124}}{{42}} = 2.95\]

So, Mean deviation about mean is  \[2.95\]

Example 2: Find the variance and standard deviation for the following data: \[57, 64, 43, 67, 49, 59, 44, 47, 61, 59\].

Ans: Given: Set of ungrouped data

We know that,variance of the data is given by \[{\sigma ^2} = \dfrac{{{{\sum\limits_{i = 1}^n {\left( {{x_i} - \bar x} \right)} }^2}}}{n}\] where n is number of observation and \[\bar x\] is the mean.  Standard deviation = \[\sigma \]

Then \[\bar x = \dfrac{{\sum\limits_{i = 1}^n {{x_i}} }}{n} = \dfrac{{57 + 64 + 43 + 67 + 49 + 59 + 44 + 47 + 61 + 59}}{{10}} = \dfrac{{550}}{{10}} = 55\]

\[{\sigma ^2} = \dfrac{{{{\sum\limits_{i = 1}^n {\left( {{x_i} - \bar x} \right)} }^2}}}{n} = \dfrac{{{2^2} + {9^2} + {{12}^2} + {{12}^2} + {6^2} + {4^2} + {6^2} + {4^2} + {{11}^2} + {8^2} + {{10}^2}}}{{10}}\]

\[ \Rightarrow {\sigma ^2} = \dfrac{{4 + 81 + 144 + 144 + 36 + 16 + 36 + 16 + 121 + 64 + 100}}{{10}} = \dfrac{{662}}{{10}} = 66.2\]

Standard deviation (\[\sigma \])  =  \[\sqrt {66.2}  = 8.13\]

So, Variance =  \[66.2\] and standard deviation = \[8.13\]

Example 3: Show that the two formulae for the standard deviation of ungr ouped data \[\sigma  = \sqrt {\dfrac{{{{\left( {{x_i} - \bar x} \right)}^2}}}{n}} \text{ and }\sigma ' = \sqrt {\dfrac{{{x_i}^2}}{n} - {\bar x}^{2}} \] are equivalent.

Ans: \[\sigma  = \sqrt {\dfrac{{{{\left( {{x_i} - \bar x} \right)}^2}}}{n}}  = \sqrt {\dfrac{{{x_i}^2 - 2{x_i}\bar x + {{\bar x}^2}}}{n}} \]

\[ \Rightarrow \sigma  = \sqrt {\dfrac{{{x_i}^2}}{n} - \dfrac{{2\bar x}}{n}\left( {n\bar x} \right) + \dfrac{{{{\bar x}^2}}}{n}}  = \sqrt {\dfrac{{{x_i}^2}}{n} - 2{{\bar x}^2} + {{\bar x}^2}}  = \sqrt {\dfrac{{{x_i}^2}}{n} - {\bar x}^{2}}  = \sigma '\]

Hence, proved.

Example 4:  Calculate variance of the following data :

Ans:  Given: A continuous frequency distribution

Variance of the continuous frequency distribution is given by  

\[{\sigma ^2} = \dfrac{{{{\sum\limits_{i = 1}^n {{f_i}\left( {{x_i} - \bar x} \right)} }^2}}}{{\sum\limits_{i = 1}^n {{f_i}} }}\]

where \[\bar x\] is the mean of the data

\[\bar x = \dfrac{{\sum\limits_{i = 1}^n {{f_i}{x_i}} }}{{_{\sum\limits_{i = 1}^n {{f_i}} }}} = \dfrac{{260}}{{20}} = 13\]                        

\[{\sigma ^2} = \dfrac{{{{\sum\limits_{i = 1}^n {{f_i}\left( {{x_i} - \bar x} \right)} }^2}}}{{\sum\limits_{i = 1}^n {{f_i}} }} = \dfrac{{380}}{{20}} = 19\]

So, variance = \[19\]

Example 5:  Calculate mean, variation and standard deviation of the following frequency distribution:

Ans: Given: A continuous frequency distribution

Variance of the continuous frequency distribution is given by   

\[{\sigma ^2} = \dfrac{{{{\sum\limits_{i = 1}^n {{f_i}\left( {{x_i} - \bar x} \right)} }^2}}}{{\sum\limits_{i = 1}^n {{f_i}} }}\]

where \[\bar x\] is the mean of the data and  

\[\bar x = \dfrac{{\sum\limits_{i = 1}^n {{f_i}{x_i}} }}{{_{\sum\limits_{i = 1}^n {{f_i}} }}}\]

\[\bar x = \dfrac{{\sum\limits_{i = 1}^n {{f_i}{x_i}} }}{{_{\sum\limits_{i = 1}^n {{f_i}} }}} = \dfrac{{1505}}{{70}} = 21.5\]

\[{\sigma ^2} = \dfrac{{{{\sum\limits_{i = 1}^n {{f_i}\left( {{x_i} - \bar x} \right)} }^2}}}{{\sum\limits_{i = 1}^n {{f_i}} }} = \dfrac{{11280}}{{70}} = 161\]

Standard deviation (\[\sigma \]) = \[\sqrt {161}  = 12.7\]

So, mean =\[21.5\], variance = \[161\] and standard deviation = \[12.7\]

Example 6:  Life of bulbs produced by two factories A and B are given below:

The bulbs of which factory are more consistent from the point of view of length of life?

Ans: Given: Continuous frequency distribution of two data .

Data having less coefficient of variation is more consistent.

coefficient of variation (C.V.) = \[\dfrac{\sigma }{{\bar x}} \times 100\]

For factory A:

\[_{\sum\limits_{i = 1}^n {{f_i}}  = 120}\] , \[\sum\limits_{i = 1}^n {{f_i}{x_i} = } 98000\]  ,  \[\sum\limits_{i = 1}^n {{f_i}{{\left( {{x_i} - \bar x} \right)}^2}}  = 1451472.048\]

For factory B:

\[_{\sum\limits_{i = 1}^n {{f_i}}  = 120}\],  \[\sum\limits_{i = 1}^n {{f_i}{x_i} = } 92400\]  ,  \[\sum\limits_{i = 1}^n {{f_i}{{\left( {{x_i} - \bar x} \right)}^2}}  = 1452000\]

\[{\bar x_A} = \dfrac{{\sum\limits_{i = 1}^n {{f_i}{x_i}} }}{{_{\sum\limits_{i = 1}^n {{f_i}} }}} = \dfrac{{98000}}{{120}} = 816.67\]

\[{\bar x_{\mathbf{B}}} = \dfrac{{\sum\limits_{i = 1}^n {{f_i}{x_i}} }}{{_{\sum\limits_{i = 1}^n {{f_i}} }}} = \dfrac{{92400}}{{120}} = 770\]

\[{\sigma _A} = \sqrt {\dfrac{{{{\sum\limits_{i = 1}^n {{f_i}\left( {{x_i} - \bar x} \right)} }^2}}}{{\sum\limits_{i = 1}^n {{f_i}} }}}  = \sqrt {\dfrac{{1451472.048}}{{120}}}  = \sqrt {12095.6}  = 109.98\]

\[{\sigma _B} = \sqrt {\dfrac{{{{\sum\limits_{i = 1}^n {{f_i}\left( {{x_i} - \bar x} \right)} }^2}}}{{\sum\limits_{i = 1}^n {{f_i}} }}}  = \sqrt {\dfrac{{1452000}}{{120}}}  = \sqrt {12100}  = 110\]

coefficient of variation (C.V.) = \[\dfrac{\sigma }{{\bar x}} \times 100\]

\[ \Rightarrow C{V_A} = \dfrac{{{\sigma _A}}}{{{{\bar x}_A}}} \times 100 = \dfrac{{109.98}}{{816.67}} \times 100 = 13.47\]

\[ \Rightarrow C{V_B} = \dfrac{{{\sigma _A}}}{{{{\bar x}_A}}} \times 100 = \dfrac{{110}}{{770}} \times 100 = 14.29\]

Since C.V. of factory B > C.V. of factory A

So, Factory B has more variability which means bulbs of factory A are more consistent.

Example 7:  The mean deviation of the data \[2, 9, 9, 3, 6, 9, 4\] from the mean is:

(A).\[2.23\]                

(B).\[2.57\]               

(C).\[3.23\]                    

(D).\[3.57\]

Ans: The correct option is (B).

Given: A set of data

Mean deviation about mean is given by 

\[M.D.(\bar x) = \dfrac{{\sum\limits_{i = 1}^n {|{x_i} - \bar x|} }}{n}\], where \[\bar x\] is the mean of distribution.

\[\bar x = \dfrac{{\sum\limits_{i = 1}^n {{x_i}} }}{n} = \dfrac{{2 + 9 + 9 + 3 + 6 + 9 + 4}}{7} = \dfrac{{42}}{7} = 6\]

\[M.D.(\bar x) = \dfrac{{\sum\limits_{i = 1}^n {|{x_i} - \bar x|} }}{n} = \dfrac{{4 + 3 + 3 + 3 + 0 + 3 + 2}}{7} = \dfrac{{18}}{7} = 2.57\]

Correct Option: B

Example 8:  Variance of the data \[2, 4, 5, 6, 8, 17\] is \[23.33\] . Then variance of \[4, 8, 10, 12, 16, 34\] will be:

(A).\[23.33\]            

(B).\[25.33\]           

(C).\[46.66\]

(D).\[48.66\]

Ans: The correct option is (C).

When values of data are multiplied by k , the variance also gets multiplied by k.

Here, k = 2, so, variance = \[23.33 \times 2 = 46.66\].

Correct Option: C

Example 9:  A set of \[n\] values \[{x_1},{x_2},.....,{x_n}\] has standard deviation \[\sigma \]. The standard deviation of n values \[{x_1} + k,{x_2} + k,.....,{x_n} + k\] will be:

(A) \[\sigma \]                

(B) \[\sigma  + k\]                

 (C) \[\sigma  - k\]                  

(D) \[k\sigma \]

Ans: The correct option is (C).

When each value is added with some constant, the variance remains unchanged.

Correct Option: C

Exercise:

Short Answer Type Questions

1.  Find the mean deviation about the mean of the distribution:

Ans: Given: A discrete frequency distribution

For a discrete frequency distribution, mean deviation about mean is given by

\[M.D.(\bar x) = \dfrac{{\sum\limits_{i = 1}^n {{f_i}|{x_i} - \bar x|} }}{{\sum\limits_{i = 1}^n {{f_i}} }}\] where \[\bar x\] is the mean of distribution.

\[ \Rightarrow \bar x = \dfrac{{\sum\limits_{i = 1}^n {{f_i}} {x_i}}}{{\sum\limits_{i = 1}^n {{f_i}} }} = \dfrac{{433}}{{20}} = 21.65\]

\[ \Rightarrow M.D.(\bar x) = \dfrac{{\sum\limits_{i = 1}^n {{f_i}|{x_i} - \bar x|} }}{{\sum\limits_{i = 1}^n {{f_i}} }} = \dfrac{{25}}{{20}} = 1.25\]

2. Find the mean deviation about the median of the following distribution:

Ans: Given: A discrete frequency distribution

For a discrete frequency distribution, mean deviation about mean is given by

\[M.D.(M) = \dfrac{{\sum\limits_{i = 1}^n {{f_i}|{x_i} - M|} }}{{\sum\limits_{i = 1}^n {{f_i}} }}\] ,where \[M\] is the mean of distribution

So, \[N = 20\] (even)

Median = \[\dfrac{1}{2} \times \left( {{{\left( {\dfrac{n}{2}} \right)}^{th}}{\text{observation}} + {{\left( {\dfrac{n}{2} + 1} \right)}^{th}}{\text{observation}}} \right)\]

\[ \Rightarrow M = \dfrac{1}{2} \times \left( {{{\left( {10} \right)}^{th}}{\text{observation}} + {{\left( {11} \right)}^{th}}{\text{observation}}} \right)\]

\[ \Rightarrow M = \dfrac{1}{2} \times \left( {12 + 12} \right) = 12\]

\[ \Rightarrow M.D.(M) = \dfrac{{\sum\limits_{i = 1}^n {{f_i}|{x_i} - M|} }}{{\sum\limits_{i = 1}^n {{f_i}} }} = \dfrac{{25}}{{20}} = 1.25\]

3. Calculate the mean deviation about the mean of the set of first \[n\] natural numbers when \[n\] is an odd number.

Ans:Mean deviation about mean is given by

\[M.D.(\bar x) = \dfrac{{\sum\limits_{i = 1}^n {|{x_i} - \bar x|} }}{n}\] where \[\bar x\] is the mean of distribution.

\[\bar x = \dfrac{{1 + 2 + 3 + ....... + n}}{n} = \dfrac{{\dfrac{{n\left( {n + 1} \right)}}{2}}}{n} = \dfrac{{n + 1}}{2}\]

\[M.D.(\bar x) = \dfrac{{\sum\limits_{i = 1}^n {|{x_i} - \bar x|} }}{n} \]

\[M.D.(\bar x)=\dfrac{1}{n}\left( \left( {1 - \dfrac{{n + 1}}{2}} \right) + \left( {2 - \dfrac{{n + 1}}{2}} \right) + \left( {3 - \dfrac{{n + 1}}{2}} \right) + ..... + \left( {n - \dfrac{{n + 1}}{2}}

\right) \right)\]

\[ \Rightarrow M.D.(\bar x)=\dfrac{1}{n}\left(\left({\dfrac{{1-n}}{2}}\right)+\left({\dfrac{{3 - n}}{2}} \right)+\left({\dfrac{{5-n}}{2}}\right)+ .... +\left({\dfrac{{1-n}}{2}}\right)+\left(

{\dfrac{{3-n}}{2}}\right)+\left({\dfrac{{n-1}}{2}}\right)}\right)\]

\[ \Rightarrow M.D.(\bar x) = \dfrac{2}{n}\left( {1 + 2 + 3 + ...... + \left( {\dfrac{{1 - n}}{2}} \right) + \left( {\dfrac{{3 - n}}{2}} \right) + \left( {\dfrac{{n - 1}}{2}} \right)} \right)\]

\[ \Rightarrow M.D.(\bar x) = \dfrac{2}{n} \times \dfrac{{\left( {\dfrac{{n - 1}}{2}} \right)\left( {\dfrac{{n - 1}}{2} + 1} \right)}}{2}\]

\[ \Rightarrow M.D.(\bar x) = \dfrac{1}{n} \times \left( {\dfrac{{n - 1}}{2}} \right) \times \left( {\dfrac{{n + 1}}{2}} \right)\]

\[ \Rightarrow M.D.(\bar x) = \dfrac{1}{n} \times \left( {\dfrac{{{n^2} - 1}}{4}} \right)\]

\[ \Rightarrow M.D.(\bar x) = \dfrac{{{n^2} - 1}}{{4n}}\]

4. Calculate the mean deviation about the mean of the set of first \[n\]natural numbers when \[n\] is an even number.

Ans: Given: \[n\] is an even number.

First we will calculate the mean of the \[n\] natural numbers. Then, we shall subtract the mean value from every number and evaluate the mean value of their absolute values. Finally, we shall divide it by the sum of \[n\] natural numbers.

${\text{Mean of first n natural numbers, where n is even:}}$

$x = \dfrac{{1 + 2 + 3 + ...... + n}}{n} = \dfrac{{\dfrac{{n(n + 1)}}{2}}}{n} = \dfrac{{n(n + 1)}}{{2n}} = \dfrac{{n + 1}}{2}$

${\text{Hence, mean deviation about the mean will be:}}$

$\overline {M.D.}  = \dfrac{{\left| {1 - \dfrac{{n + 1}}{2}} \right| + \left| {2 - \dfrac{{n + 1}}{2}} \right| + \left| {3 - \dfrac{{n + 1}}{2}} \right| + ..... + \left| {\dfrac{{n - 2}}{2} - \dfrac{{n + 1}}{2}} \right| + \left| {\dfrac{n}{2}-\dfrac{{n + 1}}{2}} \right|+\left| {\dfrac{{n + 2}}{2} - \dfrac{{n + 1}}{2}} \right| + ..... + \left| {n - \dfrac{{n + 1}}{2}} \right|}}{{n}}$

$= \dfrac{{\left| {\dfrac{{1 - n}}{2}} \right| + \left| {\dfrac{{3 - n}}{2}} \right| + \left| {\dfrac{{5 - n}}{2}} \right| + ..... + \left| {\dfrac{{ - 3}}{2}} \right| + \left| {\dfrac{{ - 1}}{2}} \right| + \left| {\dfrac{1}{2}} \right|..... + \left| {\dfrac{{n - 1}}{2}} \right|}}{{n}}$

$= \dfrac{{\dfrac{{n - 1}}{2} + \dfrac{{n - 3}}{2} + \dfrac{{n - 5}}{2} + ..... + \dfrac{3}{2} + \dfrac{1}{2} + \dfrac{1}{2} + ..... + \dfrac{{n - 1}}{2}}}{{n}}$

$= \dfrac{{2 \times \dfrac{{\dfrac{n}{2}}}{2}\left( {\dfrac{{n - 1}}{2} + \dfrac{1}{2}} \right)}}{{n}} = \dfrac{n}{{4}}$ 

5. Find the standard deviation of first \[n\] natural numbers.

Ans: Sum of first n natural numbers:

$\sum {{x_i} = 1 + 2 + ... + n = \dfrac{n}{2}(n + 1)} $

${\text{Sum of the squares of natural numbers:}}$ 

$\sum {{x_i}^2 = {1^2} + {2^2} + ... + {n^2} = \dfrac{{n(n + 1)(2n + 1)}}{6}} $

${\text{Standard deviation:}}$ 

$\sigma  = \sqrt {\dfrac{{\sum {{x_i}^2} }}{n} - {{\left( {\dfrac{{\sum {{x_i}} }}{n}} \right)}^2}} $

$= \sqrt {\dfrac{{\dfrac{{n(n + 1)(2n + 1)}}{6}}}{n} - {{\left( {\dfrac{{\dfrac{{n(n + 1)}}{2}}}{n}} \right)}^2}}  = \sqrt {\dfrac{{(n + 1)(2n + 1)}}{6} - \dfrac{{{{(n + 1)}^2}}}{4}} $

$= \sqrt {\dfrac{{2{n^2} + 3n + 1}}{6} - \dfrac{{{n^2} + 2n + 1}}{4}}  = \sqrt {\dfrac{{4{n^2} + 6n + 2 - 3{n^2} - 6n - 3}}{{12}}}  = \sqrt {\dfrac{{{n^2} - 1}}{{12}}} $ 

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

Number of observations = \[25\], mean = \[18.2 seconds\], standard deviation = \[3.25 seconds\].

Further another set of \[15\] observations \[{x_1},{x_2},.....,{x_{15}}\], also in seconds, is now available and we have \[\sum\limits_{i = 1}^{15} {{x_i} = 279} \] and \[\sum\limits_{i = 1}^{15} {{x_i}^2}  = 5524\]. Calculate the standard deviation based on all 40 observations.

Ans: Given: Number of observations = \[25\], mean = \[18.2{\text{ seconds}}\], standard deviation = \[3.25{\text{ seconds}}\].

\[\sum\limits_{i = 1}^{15} {{x_i} = 279} \]

\[\sum\limits_{i = 1}^{15} {{x_i}^2}  = 5524\]

${\text{In the first set,}}$

Mean $= \overline {{x_i}}  = \dfrac{{\sum {{x_i}} }}{{25}} = 18.2$ 

$\sum {{x_i} = 25 \times 18.2 = 455} $

${\sigma _1}^2 = \dfrac{{\sum {{x_i}^2} }}{{25}} - {\left( {\dfrac{{\sum {{x_i}} }}{{25}}} \right)^2} = \dfrac{{\sum {{x_i}^2} }}{{25}} - {(18.2)^2}$

\[S.D. = \sqrt {\dfrac{{{n_1}{{({s_1})}^2} + {n_2}{{({s_2})}^2}}}{{{n_1} + {n_2}}} + \dfrac{{{n_1}{n_2}{{(\overline {{x_1}}  - \overline {{x_2}} )}^2}}}{{{{({n_1} + {n_2})}^2}}}} \]

${\text{For 40 observations,}}$

$\sum {{x_i}^2 = 8545.06 + 5524 = 14069.06} $

$\sum {{x_i} = 455 + 279 = 734} $

$\therefore \sigma  = \sqrt {\dfrac{{\sum {{x_i}^2} }}{n} - {{\left( {\dfrac{{\sum {{x_i}} }}{n}} \right)}^2}}  = \sqrt {\dfrac{{14069.06}}{{40}} - {{\left( {\dfrac{{734}}{{40}}} \right)}^2}}  = \sqrt {15.004}  = 3.87$ 

7. The mean and standard deviation of a set of \[{n_1}\] observations are \[\overline {{x_1}} \] and \[{s_1}\], respectively while the mean and standard deviation of another set of \[{n_2}\] observation are \[\overline {{x_2}} \] and \[{s_2}\] respectively. Show that the standard deviation of the combined set of \[({n_1} + {n_2})\] observations is given by

\[S.D. = \sqrt {\dfrac{{{n_1}{{({s_1})}^2} + {n_2}{{({s_2})}^2}}}{{{n_1} + {n_2}}} + \dfrac{{{n_1}{n_2}{{(\overline {{x_1}}  - \overline {{x_2}} )}^2}}}{{{{({n_1} + {n_2})}^2}}}} \]

Ans: Given: The mean and standard deviation of a set of\[{n_1}\]observations are \[\overline {{x_1}} \]and \[{s_1}\], respectively.The mean and standard deviation of another set of \[{n_2}\] observation are \[\overline {{x_2}} \] and \[{s_2}\] respectively.

${\text{For the first and second set,}}$

$\overline {{x_1}}  = \dfrac{1}{{{n_1}}}\sum\limits_{i = 1}^n {{x_i}} ,\overline {{x_2}}  = \dfrac{1}{{{n_2}}}\sum\limits_{j = 1}^n {{y_j}} $

${\text{Mean of the combines series,}}$ 

$\overline x  = \dfrac{1}{{{n_1} + {n_2}}}\left[ {\sum\limits_{i = 1}^n {{x_i}}  + \sum\limits_{j = 1}^n {{y_j}} } \right] = \dfrac{{{n_1}\overline {{x_1}}  + {n_2}\overline {{x_2}} }}{{{n_1} + {n_2}}} -------- (1)$

${\sigma _1}^2 = \dfrac{1}{{{n_1}}}\sum\limits_{i = 1}^n {{{({x_i} - \overline x )}^2}} ,{\sigma _2}^2 = \dfrac{1}{{{n_2}}}\sum\limits_{j = 1}^n {({y_j} - } \overline x {)^2}$ 

${\text{For the total set,}}$

${\sigma ^2} = {\sigma _1}^2 + {\sigma _2}^2 = \dfrac{1}{{{n_1} + {n_2}}}\left( {\sum\limits_{i = 1}^n {{{({x_i} - \overline x )}^2}}  + \sum\limits_{j = 1}^n {({y_j} - } \overline x {)^2}} \right) -------- (2)$

$\sum\limits_{i = 1}^n {{{({x_i} - \overline x )}^2}}  = \sum\limits_{i = 1}^n {{{({x_i} - \overline {{x_j}} )}^2} + {n_1}(} \overline {{x_j}}  - \overline x {)^2} + 2(\overline {{x_j}}  - \overline x )\sum\limits_{i = 1}^n {{{({x_i} - \overline {{x_j}} )}^2}} $ 

${\text{Since, }}\sum\limits_{i = 1}^n {{{({x_i} - \overline {{x_j}} )}^2}}  = 0$

$\sum\limits_{i = 1}^n {{{({x_i} - \overline x )}^2}}  = {n_1}{s_1}^2 + {n_1}{(\overline {{x_j}}  - \overline x )^2} -------- (3)$

$\because {d_1} = \overline {{x_1}}  - \overline x$

${\text{From (1),}}$

${d_1} = \overline {{x_1}}  - \dfrac{{{n_1}\overline {{x_1}}  + {n_2}\overline {{x_2}} }}{{{n_1} + {n_2}}} = \dfrac{{{n_1}\overline {{x_1}}  + {n_2}\overline {{x_1}}  - {n_1}\overline {{x_1}}  - {n_2}\overline {{x_2}} }}{{{n_1} + {n_2}}} = \dfrac{{{n_2}(\overline {{x_1}}  - \overline {{x_2}} )}}{{{n_1} + {n_2}}}$ 

${\text{Finally, }}\sum\limits_{i = 1}^n {{{({x_i} - \overline x )}^2}}  = {n_1}{s_1}^2 + \dfrac{{{n_1}{n_2}{}^2{{(\overline {{x_1}}  - \overline {{x_2}} )}^2}}}{{{{({n_1} + {n_2})}^2}}}$

${\text{Similarly,}}$

$\sum\limits_{j = 1}^n {({y_j} - } \overline x {)^2} = {n_2}{s_2}^2 + \dfrac{{{n_2}{n_1}^2{{(\overline {{x_2}}  - \overline {{x_1}} )}^2}}}{{{{({n_1} + {n_2})}^2}}}$

$\therefore {\sigma ^2} = {\sigma _1}^2 + {\sigma _2}^2 = \dfrac{1}{{{n_1} + {n_2}}}\left( {\sum\limits_{i = 1}^n {{{({x_i} - \overline x )}^2}}  + \sum\limits_{j = 1}^n {({y_j} - } \overline x {)^2}} \right)$ 

$= \dfrac{1}{{{n_1} + {n_2}}}\left( {{n_1}{s_1}^2 + \dfrac{{{n_1}{n_2}{}^2{{(\overline {{x_1}}  - \overline {{x_2}} )}^2}}}{{{{({n_1} + {n_2})}^2}}} + {n_2}{s_2}^2 + \dfrac{{{n_2}{n_1}^2{{(\overline {{x_2}}  - \overline {{x_1}} )}^2}}}{{{{({n_1} + {n_2})}^2}}}} \right)$

$= \dfrac{1}{{{n_1} + {n_2}}}\left( {{n_1}{s_1}^2 + {n_2}{s_2}^2 + \dfrac{{{n_1}{n_2}{}^2{{(\overline {{x_1}}  - \overline {{x_2}} )}^2}}}{{{{({n_1} + {n_2})}^2}}} + \dfrac{{{n_2}{n_1}^2{{(\overline {{x_1}}  - \overline {{x_2}} )}^2}}}{{{{({n_1} + {n_2})}^2}}}} \right)$ 

\[ = \dfrac{1}{{{n_1} + {n_2}}}\left( {{n_1}{s_1}^2 + {n_2}{s_2}^2 + \dfrac{{{n_1}{n_2}{{(\overline {{x_1}}  - \overline {{x_2}} )}^2}}}{{{{({n_1} + {n_2})}^2}}}({n_2} + {n_1})} \right)\]

\[{\text{Hence, }}{\sigma}^{2}  = \dfrac{{{n_1}{s_1}^2 + {n_2}{s_2}^2}}{{{n_1} + {n_2}}} + \dfrac{{{n_1}{n_2}{{(\overline {{x_1}}  - \overline {{x_2}} )}^2}}}{{{{({n_1} + {n_2})}^2}}}\]

8. Two sets of \[20\] observations have the same standard deviation \[5\]. The first set has a mean \[17\] and the second a mean \[22\]. Determine the standard deviation of the set obtained by combining the given \[2\] sets.

Ans: Given: \[{n_1} = {n_2} = 20\], \[\overline {{x_1}}  = 17\], \[\overline {{x_2}}  = 22\] ,\[{\sigma _1} = {\sigma _2} = 5\]

$\sigma  = \sqrt{\dfrac{{{n_1}{s_1}^2 + {n_2}{s_2}^2}}{{{n_1} + {n_2}}} + \dfrac{{{n_1}{n_2}{{(\overline {{x_1}}  - \overline {{x_2}} )}^2}}}{{{{({n_1} + {n_2})}^2}}}}$

$= \sqrt {\dfrac{{20 \times 25 + 20 \times 25}}{{40}} + \dfrac{{400{{( - 5)}^2}}}{{{{40}^2}}}}  = \sqrt {25 + \dfrac{{25}}{4}}  = \sqrt {31.25}  = 5.59$ 

9. The frequency distribution:

where A is a positive integer, has a variance of 160. Determine the value of A.

Ans: Given: Variance = \[160\].

${\text{Variance = }}{\sigma ^2} = \dfrac{{\sum {{f_i}{x_i}^2} }}{n} - {\left( {\dfrac{{\sum {{f_i}{x_i}} }}{n}} \right)^2}$

$160=\dfrac{{2{A^2}+1.(4{A^2})+1.(9{A^2})+1.(16{A^2})+1.(25{A^2})+1.(36{A^2})}}{7}-{\left({\dfrac{{2A+2A+3A+4A+5A+6A}}{7}}\right)^2}$

 $\Rightarrow 160 = \dfrac{{92{A^2}}}{7} - \dfrac{{484{A^2}}}{{49}}$

On Solving, we get

A = 7

10. For the frequency distribution:

Find the standard distribution.

Ans: ${\text{For standard deviation:}}$

$\sigma  = \sqrt {\dfrac{{\sum {{f_i}{x_i}^2} }}{n} - {{\left( {\dfrac{{\sum {{f_i}{x_i}} }}{n}} \right)}^2}}$

$= \sqrt {\dfrac{{({{4\times 2}^2} + {{9\times 3}^2} + {{16\times 4}^2} + {{14\times 5}^2} + {{11\times 6}^2} + {{6\times 7}^2})}}{{4 + 9 + 16 + 14 + 11 + 6}} - {{\left( {\dfrac{{8 + 27 + 64 + 70 + 66 + 42}}{{60}}} \right)}^2}} $

$= \sqrt {\dfrac{{1393}}{{60}} - \dfrac{{{{277}^2}}}{{3600}}}  = 1.379 \approx 1.38$ 

11. There are 60 students in a class. The following is the frequency distribution of marks obtained by the students in a test:

Where x, is positive integer. Determine the mean and standard deviation of the marks.

Ans: $\sum {{f_i} = 60} $

$\Rightarrow (x - 2) + x + {x^2} + {(x + 1)^2} + 2x + x + 1 = 60$

$\Rightarrow 2{x^2} + 7x - 60 = 0$

Solving the above equation, we get $x =  - \dfrac{{15}}{2},4$

$\therefore x = 4.$

Let the mean is 3. 

Mean: 

$\bar{x}=A+\left(\dfrac{\sum f_{i} d_{i}}{n}\right)$ $=3+\left(\dfrac{-12}{60}\right)=3-0.2=2.8$

So the mean is 2.8.

Standard deviation: 

$\sigma=\sqrt{\dfrac{\sum f_{i} d_{i}^{2}}{n}-\left(\dfrac{\sum f_{i} d_{i}}{n}\right)^{2}}$ 

$=\sqrt{\dfrac{78}{60}-\left(\dfrac{-12}{60}\right)^{2}}=\sqrt{1.3-\dfrac{144}{3600}}=\sqrt{1.26}=1.122$ 

So the Standard deviation $=1.122$ 

12. 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 \[650\] hours and a standard deviation of \[7\] hours. Find the overall standard deviation.

Ans: Given: Mean life of a sample of \[60\] bulbs was \[650\] hours and the standard deviation was \[8\] hours is $ 8$ hours.

With the help of the below formula,

\[\sigma    = \sqrt {\dfrac{{{n_1}{s_1}^2 + {n_2}{s_2}^2}}{{{n_1} + {n_2}}} + \dfrac{{{n_1}{n_2}{{\left( {\overline {{x_1}}  - \overline {{x_2}} } \right)}^2}}}{{{{\left( {{n_1} + {n_2}} \right)}^2}}}} \]

Then,

${n_1} = 60 ,  {{\bar x}_1} = 650 ,  {s_1} = 8; {n_2} = 80,{{\bar x}_2} = 660,  {s_2} = 7$

$\sigma  = \sqrt{\dfrac{{{n_1}{s_1}^2 + {n_2}{s_2}^2}}{{{n_1} + {n_2}}} + \dfrac{{{n_1}{n_2}{{(\overline {{x_1}}  - \overline {{x_2}} )}^2}}}{{{{({n_1} + {n_2})}^2}}}}$ 

$\Rightarrow \sigma= \sqrt {\dfrac{{60 \times {8^2} + 80 \times {7^2}}}{{60 + 80}} + \dfrac{{60 \times 80{{\left( {650 - 660} \right)}^2}}}{{{{\left( {60 + 80} \right)}^2}}}}$        

Since, $=\dfrac{{60 \times 64 + 80 \times 49}}{{140}} = 20\dfrac{{\left[ {3(64) + 4(49)} \right]}}{{140}}$

$=  \dfrac{{\left[ {3(64) + 4(49)} \right]}}{7} =  \dfrac{{192 + 196}}{7}$ 

$\sigma= \sqrt {\dfrac{{60 \times 64 + 80 \times 49}}{{140}} + \dfrac{{60 \times 80{{\left( {650 - 660} \right)}^2}}}{{{{140}^2}}}}$

$\Rightarrow \sigma= \sqrt {\dfrac{{60 \times 64 + 80 \times 49}}{{140}} + \dfrac{{60 \times 80{{\left( {650 - 660} \right)}^2}}}{{140 \times 140}}}$

$\dfrac{{60 \times 80{{\left( {650 - 660} \right)}^2}}}{{140 \times 140}} = \dfrac{{60 \times 80 \times 100}}{{140 \times 140}}$

$= \dfrac{{1200}}{{49}}$

$\Rightarrow \sigma   =  \sqrt {\dfrac{{192 + 196}}{7} + \dfrac{{1200}}{{49}}}  = \sqrt {\dfrac{{2716 + 1200}}{{49}}}$

$= \sqrt {\dfrac{{3916}}{{49}}}  = \dfrac{{62.58}}{7} = 8.9$

13. Mean and standard deviation of \[100\] items are \[50\] and \[{\text{4}}\] respectively. Find the sum of the entire item and the sum of the squares of the items. 

Ans: Given:

(a)  Mean 

(b)  The standard deviation of respectively.

With the help of the below formula,

\[{\sigma ^2} = \dfrac{{\sum {f_i}{x_i}^2}}{{\sum {f_i}}} - {\left( {\dfrac{{\sum {f_i}{x_i}}}{{\sum {f_i}}}} \right)^2}\]

$\bar x = 50,n = 100 {\text{and}} \sigma  = 4$

$\dfrac{{\sum {x_i}}}{{100}} = 50$

$\sum\limits_{}^{} {{x_i} = 5000} $

${\sigma ^2} = \dfrac{{\sum {f_i}{x_i}^2}}{{\sum {f_i}}} - {\left( {\dfrac{{\sum {f_i}{x_i}}}{{\sum {f_i}}}} \right)^2}$ 

\[\Rightarrow {4^2} = \dfrac{{\sum {f_i}{x_i}^2}}{{\sum {f_i}}} - {\left( {50} \right)^2}\]

$\Rightarrow {4^2} + {\left( {50} \right)^2} = \dfrac{{\sum {f_i}{x_i}^2}}{{\sum {f_i}}}$

$\Rightarrow \dfrac{{\sum {f_i}{x_i}^2}}{{\sum {f_i}}} = 16 + 2500$

$\Rightarrow \dfrac{{\sum {f_i}{x_i}^2}}{{100}} = 2516$

$\Rightarrow \sum {f_i}{x_i}^2 = 251600$ 

14. If for distribution\[\sum \left( {x - 5} \right) = 3 \text{ and } {\sum\left( {x - 5} \right)^2} = 43\] and the total number of items is \[18\]. Find the mean and standard deviation.

Ans: Given: \[\sum\left( {x - 5} \right) = 3 {\text{ and }} {\sum\left( {x - 5} \right)^2} = 43\]

The total number of items is \[18\] .

Mean   \[ = {\text{A}} + \dfrac{{\sum \left( {x - 5} \right)}}{n}\]

Standard deviation   \[\sigma  = \sqrt {\dfrac{{\sum {{\left( {x - 5} \right)}^2}}}{n} - {{\left( {\dfrac{{\sum \left( {x - 5} \right)}}{n}} \right)}^2}} \].

Then,

\[\sum \left( {x - 5} \right) = 3 {\text{ and }} {\sum \left( {x - 5} \right)^2} = 43\]

Mean \[ = {\text{A}} + \dfrac{{\sum \left( {x - 5} \right)}}{n}\]

$ = 5 + \dfrac{3}{{18}}$

$= 5 + 0.1666$

$= 5.17$ 

Standard deviation   \[\sigma  = \sqrt {\dfrac{{\sum {{\left( {x - 5} \right)}^2}}}{n} - {{\left( {\dfrac{{\sum \left( {x - 5} \right)}}{n}} \right)}^2}} \]

$= \sqrt {\dfrac{{43}}{{18}} - {{\left( {\dfrac{3}{{18}}} \right)}^2}} $

$= \sqrt {2.3889 - 0.0277}  = 1.53$ 

15. Find the mean and variance of the frequency distribution given below:

Ans: By using the below basic formulas, we can find the mean and variance.

Mean \[ = \dfrac{{\sum {f_i}{y_i}}}{{{f_i}}}\]

Standard deviation   \[\sigma  = \sqrt {\dfrac{{\sum {f_i}{y_i}^2}}{n} - {{\left( {\dfrac{{\sum {f_i}{y_i}}}{n}} \right)}^2}} \]

Mean  \[ = \dfrac{{\sum {f_i}{y_i}}}{{{f_i}}}\]

$= \dfrac{{66.5}}{{16}}$

$= 4.16$ 

Standard deviation   \[\sigma  = \sqrt {\dfrac{{\sum {f_i}{y_i}^2}}{n} - {{\left( {\dfrac{{\sum {f_i}{y_i}}}{n}} \right)}^2}} \]

$= \sqrt {\dfrac{{340.25}}{{16}} - {{\left( {\dfrac{{66.5}}{{16}}} \right)}^2}} $

$= \sqrt {21.26 - 17.22}  = 4.04$ 

Long Answer Type :

16. Calculate the mean deviation about the mean for the following frequency distribution:

Ans: We know that,\[X = A + h\left( {\dfrac{1}{N}\sum {f_i}{d_i}} \right)\]

Mean deviation about mean = \[ = \dfrac{1}{N}\left( {\sum {f_i}\left| {{x_i} - X} \right|} \right)\]

Let us assume \[{\text{A  =  10}}\] and \[{\text{h  =  4}}\]

$X = A + h\left( {\dfrac{1}{N}\sum {f_i}{d_i}} \right)$

$= 10 + 4\left( {\dfrac{1}{{25}}\left( { - 5} \right)} \right)$

$= 10 - \dfrac{4}{5} = 10 - 0.8$

$= 9.2$ 

Mean deviation about mean = \[ = \dfrac{1}{N}\left( {\sum {f_i}\left| {{x_i} - X} \right|} \right) = \dfrac{{96}}{{25}} = 3.84\]

17. Calculate the mean deviation from the median of the following data:

Ans: With the help of the basic formula,

Median \[ = l + \dfrac{{\dfrac{N}{2} - C}}{f} \times h\]

$= 12 + \dfrac{{(10 - 9)}}{3} \times 6 = 14$

Since, $\dfrac{N}{2} = 10$ ${\text{median class is }}12 - 18$

${\text{Median}} = l + \dfrac{{\dfrac{N}{2} - C}}{f} \times h$ 

${\text{Median}} = 12 + \dfrac{{10 - 9}}{3} \times 6=14$ 

${\text{MD}} = \dfrac{{\sum {f_i}{d_i}}}{{\sum {f_i}}} = \dfrac{{140}}{{20}} = 7$. 

So the required mean deviation is 7.

18: Determine the mean and standard deviation of the following distribution: 

Ans: ${\text{Mean}} = \dfrac{{\sum {f_i}{x_i}}}{{\sum {f_i}}}$

  $\sigma  = \sqrt {\dfrac{{\sum {f_i}{x_i}^2}}{{\sum {f_i}}} - \left( {\dfrac{{\sum {f_i}{x_i}}}{{\sum {f_i}}}} \right)}$ 

${\text{Mean}} = \dfrac{{\sum {f_i}{x_i}}}{{\sum {f_i}}} = \dfrac{{239}}{{40}} = 5.975 = 6$

$\sigma  = \sqrt {\dfrac{{\sum {f_i}{d_i}^2}}{{\sum {f_i}}} - \left( {\dfrac{{\sum {f_i}{d_i}}}{{\sum {f_i}}}} \right)} $

$= \sqrt {\dfrac{{325}}{{40}} - {{\left( {\dfrac{{ - 1}}{{40}}} \right)}^2}} $

$= \sqrt {8.124}  = 2.85$ 

19: The weight of coffee in \[70\] jars is shown in the following table:

Determine variance and standard deviation of the above distribution. 

Ans: We know that,

\[{\sigma ^2} = \dfrac{{\sum {f_i}{d_i}^2}}{{\sum {f_i}}} - {\left( {\dfrac{{\sum {f_i}{d_i}}}{{\sum {f_i}}}} \right)^2}\]

The mean $\bar{x}=\dfrac{200.5\times 13+201.5\times 27+202.5\times 18+203.5\times 10+204.5\times 1+205.5\times 1}{70}=\dfrac{14137}{70}=202.5$

${\sigma ^2} = \dfrac{{\sum {f_i}{d_i}^2}}{{\sum {f_i}}} - {\left( {\dfrac{{\sum {f_i}{d_i}}}{{\sum {f_i}}}} \right)^2}$

$= \dfrac{{102}}{{70}} - {\left( {\dfrac{{ - 38}}{{70}}} \right)^2}$

$= 1.4571 - 0.2946 = 1.1655$

$\sigma  = \sqrt {1.1655}  = 1.08gm $ 

20: Determine the mean and standard deviation of first n terms of an A.P. whose first term is \[a\] and the common difference is \[d.\]

Ans: Given: First-term of an A.P is \[a\].

In addition, the common difference is \[d.\]

Mean \[ = \dfrac{{\sum {x_i}}}{n} = \dfrac{1}{n}\left[ {\dfrac{n}{2}[2a + \left( {n - 1} \right)d]} \right] = a + \dfrac{{n - 1}}{2}d\]

Standard deviation \[\sigma  = \sqrt {\dfrac{{\sum {{\left( {{x_i} - a} \right)}^2}}}{n} - {{\left( {\dfrac{{\sum \left( {{x_i} - a} \right)}}{n}} \right)}^2}} \]

Mean \[ = \dfrac{{\sum {x_i}}}{n} = \dfrac{1}{n}\left[ {\dfrac{n}{2}[2a + \left( {n - 1} \right)d]} \right] = a + \dfrac{{n - 1}}{2}d\]

$\sum{\left( {{x_i} - a} \right)}=d\left[1+2+3+........+(n-1)\right]=\dfrac{d(n-1)n}{2}$

$\sum{\left( {{x_i} - a} \right)^{2}}={d}^{2}\left[1^2+2^2+3^2+........+(n-1)^{2}\right]= {d^2}\dfrac{n(n-1)(2n-1)}{6}$

Standard deviation \[\sigma  = \sqrt {\dfrac{{\sum {{\left( {{x_i} - a} \right)}^2}}}{n} - {{\left( {\dfrac{{\sum \left( {{x_i} - a} \right)}}{n}} \right)}^2}} \]

$= \sqrt {\dfrac{{{d^2}\left( {n - 1} \right)\left( {2n - 1} \right)}}{6} - \dfrac{{{d^2}{{\left( {n - 1} \right)}^2}}}{4}} $

 $= d\sqrt {\dfrac{{n - 1}}{2}\left[ {\dfrac{{\left( {2n - 1} \right)}}{3} - \dfrac{{\left( {n - 1} \right)}}{2}} \right]} $

$= d\sqrt {\dfrac{{n - 1}}{2}\left[ {\dfrac{{2\left( {2n - 1} \right) - 3(n - 1)}}{6}} \right]} $

$= d\sqrt {\dfrac{{n - 1}}{2}\left[ {\dfrac{{\left( {n + 1} \right)}}{6}} \right]}  = d\sqrt {\dfrac{{{n^2} - 1}}{{12}}} $ 

21: Following are the marks obtained, out of 100, by two students Ravi and Hashina in \[10\] tests. 

Who is more intelligent and who is more consistent? 

Ans: We know that,

\[\sigma  = \sqrt {\dfrac{{\sum {d_i}^2}}{n} - {{\left( {\dfrac{{\sum {d_i}^2}}{n}} \right)}^2}} \]

For Ravi

Let assumed mean =45

$\sigma  = \sqrt {\dfrac{{\sum {d_i}^2}}{n} - {{\left( {\dfrac{{\sum {d_i}}}{n}} \right)}^2}} $

$= \sqrt {\dfrac{{1699}}{{10}} - {{\left( {\dfrac{{ - 9}}{{10}}} \right)}^2}} $

$= \sqrt {169.9 - 0.81} $

$= \sqrt {169.09}  = 13.003$

$\overline x  = {\text{A}} + \dfrac{{\sum {d_i}}}{{\sum {f_i}}} = 45 - \dfrac{{9}}{{10}} = 44.1$ 

${\text{for Hashina,}}$

$\sigma  = \sqrt {\dfrac{{\sum {d_i}^2}}{n} - {{\left( {\dfrac{{\sum {d_i}}}{n}} \right)}^2}} $

$= \sqrt {\dfrac{{5698}}{{10}} - {{\left( {\dfrac{{20}}{{10}}} \right)}^2}} $

$= \sqrt {596.8 - 2} $

$= \sqrt {594.8}  = 24.38$ 

And, $\overline x  = {\text{A}} + \dfrac{{\sum {d_i}}}{{\sum {f_i}}} = 55 - \dfrac{{20}}{{10}} = 53$ 

For Ravi, \[\dfrac{\sigma }{{\overline x }} \times 100 = \dfrac{{13.003}}{{44.1}} \times 100 = 29.48\]

For hashina, \[\dfrac{\sigma }{{\overline x }} \times 100 = \dfrac{{24.38}}{{53}} \times 100 = 46.01\]

Hence hashina is more intelligent and consistent.

22: Mean and standard deviation \[100\] 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.

Ans: Given: $n = 100,\overline x  = 40$, $\sigma  = 10$

Two Observations were wrongly recorded as \[30{\text{ and }}70\] in the place of \[3{\text{ and }}27\]

We know that,${\sigma ^2} = \dfrac{{\sum {x_i}^2}}{n} - {\left( {\dfrac{{\sum {x_i}}}{n}} \right)^2}$

$n = 100,\overline x  = 40$,$\sigma  = 10$

$\dfrac{{\sum {x_i}}}{n} = 40 \Rightarrow \sum {x_i} = 40 \times 100$

$\sum {x_i} = 4000$ 

Corrected $\sum {x_i} = 4000 - 30 - 70 + 3 + 27$$ = 3930$

Corrected mean = $\dfrac{{3930}}{{100}} = 39.3$

${\sigma ^2} = \dfrac{{\sum {x_i}^2}}{n} - {\left( {\dfrac{{\sum {x_i}}}{n}} \right)^2}$

$100 = \dfrac{{\sum {x_i}^2}}{{100}} - 1600$

$\sum {x_i}^2 = 170000$ 

Corrected $\sum {x_i}^2 = 170000 - {\left( {30} \right)^2} - {\left( {70} \right)^2} + {3^2} + {\left( {27} \right)^2}$

$= 164938$

${\text{corrected }}\sigma  = \sqrt {\dfrac{{164938}}{{100}} - {{\left( {39.3} \right)}^2}}  = \sqrt {\left( {1649.38 - 1544.49} \right)} $

$= \sqrt {104.9}  = 10.24$ 

23: 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 the variance.

Ans: Given: Mistaken mean of the $10$ observations $ = 45$

Mistaken variance of the $10$ observations $ = 16$

The correct reading is $25$ instead of $52$.

Use the formula $\overline x  = \dfrac{{\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)}}{n}$ to find the sum of all the observations. Subtract the incorrect value and add the correct value to find the correct total. Here $\overline x $ is the mean.

Use the formula ${\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^n {{{\left( {{x_i}} \right)}^2}} }}{n} - {\left( {\overline x } \right)^2}$, where ${\sigma ^2}$ is the variance, to calculate $\sum\limits_{i = 1}^n {{{\left( {{x_i}} \right)}^2}}$. Now, subtract ${\left( {52} \right)^2}$ from $\sum\limits_{i = 1}^n {{{\left( {{x_i}} \right)}^2}}$ and add ${\left( {25} \right)^2}$.

The incorrect mean of the given $10$ observations is $45$. Assuming this mean as $\overline x $ and using the formula $\overline x  = \dfrac{{\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)}}{n}$,

$\Rightarrow 45 = \dfrac{{\left( {\sum\limits_{i = 1}^{10} {{x_i}} } \right)}}{{10}}$

$\Rightarrow \left( {\sum\limits_{i = 1}^{10} {{x_i}} } \right) = 450$ 

That means the sum of all the observations (including the incorrect one) is $450$. Since one reading is to be $25$ instead of $52$, so the correct sum must be,

$\Rightarrow {\left( {\sum\limits_{i = 1}^{10} {{x_i}} } \right)_{co}} = 450 - 52 + 25$

$\Rightarrow {\left( {\sum\limits_{i = 1}^{10} {{x_i}} } \right)_{co}} = 423$ 

Therefore, the correct mean is ${\overline x _{co}} = \dfrac{{{{\left( {\sum\limits_{i = 1}^{10} {{x_i}} } \right)}_{co}}}}{{10}} = 42.3$.

Now, using the formula ${\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^n {{{\left( {{x_i}} \right)}^2}} }}{n} - {\left( {\overline x } \right)^2}$ for the variance of the incorrect mean, where ${\sigma ^2} = 16$ is provided,

$\Rightarrow \dfrac{{\left( {\sum\limits_{i = 1}^{10} {{{\left( {{x_i}} \right)}^2}} } \right)}}{{10}} - {\left( {45} \right)^2} = 16$

$\Rightarrow \sum\limits_{i = 1}^{10} {{{\left( {{x_i}} \right)}^2}}  = 20410$ 

That means the incorrect sum of squares of all the observations is $20410$. Since $52$ is to be replaced with $25$, so the correct value of $\sum\limits_{i = 1}^{10} {{{\left( {{x_i}} \right)}^2}}$ is,

$\Rightarrow {\left( {\sum\limits_{i = 1}^{10} {{{\left( {{x_i}} \right)}^2}} } \right)_{co}} = 20410 - {\left( {52} \right)^2} + {\left( {25} \right)^2}$

$\Rightarrow {\left( {\sum\limits_{i = 1}^{10} {{{\left( {{x_i}} \right)}^2}} } \right)_{co}} = 18331$ 

Hence, the variance for the correct mean is,

$\Rightarrow {\left( {{\sigma ^2}} \right)_{co}} = \dfrac{{{{\left( {\sum\limits_{i = 1}^{10} {{{\left( {{x_i}} \right)}^2}} } \right)}_{co}}}}{{10}} - {\left( {\overline x } \right)^2}_{co}$

$\Rightarrow {\left( {{\sigma ^2}} \right)_{co}} = \dfrac{{18331}}{{10}} - {\left( {42.3} \right)^2}_{co}$

$\Rightarrow {\left( {{\sigma ^2}} \right)_{co}} = 43.81$ 

24: The mean deviation of the data $3$, $10$, $10$, $4$, $7$, $10$, $5$ from the mean is

(A) $2$

(B) $2.57$

(C) $3$

(D) $3.75$

Ans: The correct option is (B).

Given: The data $3$, $10$, $10$, $4$, $7$, $10$, $5$.

Use the formula $\overline x  = \dfrac{{\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)}}{n}$ to calculate the mean of the given data. Here $\overline x $ is the mean.

Now, use the formula $M.D. = \dfrac{{\sum\limits_{i = 1}^n {\left| {{x_i} - \overline x } \right|} }}{n}$ to calculate the mean deviation.

The total number of observations is $n = 7$. Using the formula $\overline x  = \dfrac{{\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)}}{n}$, where $\overline x $ is the mean,

$\Rightarrow \overline x  = \dfrac{{\left( {\sum\limits_{i = 1}^7 {{x_i}} } \right)}}{7}$

$\Rightarrow \overline x  = \dfrac{{3 + 10 + 10 + 4 + 7 + 10 + 5}}{7}$

$\Rightarrow \overline x  = 7$ 

Now, the mean deviation is given as $M.D. = \dfrac{{\sum\limits_{i = 1}^n {\left| {{x_i} - \overline x } \right|} }}{n}$. So for the given data,

$\Rightarrow M.D. = \dfrac{{\sum\limits_{i = 1}^7 {\left| {{x_i} - 7} \right|} }}{7}$

$\Rightarrow M.D. = \dfrac{{\left| {3 - 7} \right| + \left| {10 - 7} \right| + \left| {10 - 7} \right| + \left| {4 - 7} \right| + \left| {7 - 7} \right| + \left| {10 - 7} \right| + \left| {5 - 7} \right|}}{7}$

$\Rightarrow M.D. = 2.57$ 

Correct Answer: B

25: Mean deviation for $n$ observations ${x_1}, {x_2}, ...., {x_n}$ from their mean $\overline x $ is given by

(A) $\sum\limits_{i = 1}^n {\left( {{x_i} - \overline x } \right)}$

(B) $\dfrac{1}{n}\sum\limits_{i = 1}^n {\left| {{x_i} - \overline x } \right|}$

(C) ${\sum\limits_{i = 1}^n {\left( {{x_i} - \overline x } \right)} ^2}$

(D) $\dfrac{1}{n}{\sum\limits_{i = 1}^n {\left( {{x_i} - \overline x } \right)} ^2}$

Ans: The correct option is (B).

Given: $n$ observations ${x_1},{x_2},....,{x_n}$.

The mean value of these observations is $\overline x $.

Use the definition of the mean deviation which is ‘the ratio of the summation of absolute deviation of each data from the mean and the total number of observations given’.

According to the definition of the mean deviation it is the ratio of the summation of absolute deviation of each data from the mean and the total number of observations given.

Now, the absolute deviation of $n$ observations ${x_1},{x_2},....,{x_n}$ from their mean $\overline x $ is given as $\left| {{x_1} - \overline x } \right|,\left| {{x_2} - \overline x } \right|,....,\left| {{x_n} - \overline x } \right|$ respectively. Taking the sum of these deviations and dividing by the number of observations gives,

$\Rightarrow M.D. = \dfrac{{\sum\limits_{i = 1}^n {\left| {{x_i} - \overline x } \right|} }}{n}$

Correct Answer: B

26: When tested the lives (in hours) of $5$ bulbs were noted as follows:

$1357$, $1090$, $1666$, $1494$, $1623$

The mean deviation (in hours) from their mean is

(A) $178$

(B) $179$

(C) $220$

(D) $356$

Ans: The correct option is (A).

Given: The lives in hours of $5$ bulbs are $1357$, $1090$, $1666$, $1494$, $1623$.

Use the formula $\overline x  = \dfrac{{\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)}}{n}$ to calculate the mean of the given data. Here $\overline x $ is the mean.

Now, use the formula $M.D. = \dfrac{{\sum\limits_{i = 1}^n {\left| {{x_i} - \overline x } \right|} }}{n}$ to calculate the mean deviation.

The total number of observations is $n = 5$. Using the formula $\overline x  = \dfrac{{\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)}}{n}$, where $\overline x $ is the mean,

$\Rightarrow \overline x  = \dfrac{{\left( {\sum\limits_{i = 1}^5 {{x_i}} } \right)}}{5}$

$\Rightarrow \overline x  = \dfrac{{1357 + 1090 + 1666 + 1494 + 1623}}{5}$

$\Rightarrow \overline x  = 1446{\text{ hours}}$ 

Now, the mean deviation is given as $M.D. = \dfrac{{\sum\limits_{i = 1}^n {\left| {{x_i} - \overline x } \right|} }}{n}$. So for the given data,

$\Rightarrow M.D. = \dfrac{{\sum\limits_{i = 1}^5 {\left| {{x_i} - 1446} \right|} }}{5}$

$\Rightarrow M.D. = \dfrac{{\left| {1357 - 1446} \right| + \left| {1090 - 1446} \right| + \left| {1666 - 1446} \right| + \left| {1494 - 1446} \right| + \left| {1623 - 1446} \right|}}{5}$

$\Rightarrow M.D. = 178{\text{ hours}}$ 

Correct Answer: A

27: Following are the marks obtained by $9$ students in a mathematics test:

$50$, $69$, $20$, $33$, $53$, $39$, $40$, $65$, $59$

The mean deviation from the median is:

(A) $9$

(B) $10.5$

(C) $12.67$

(D) $14.76$

Ans: The correct option is (C).

Given: The marks obtained by $9$ students in the test are $50$, $69$, $20$, $33$, $53$, $39$, $40$, $65$, $59$.

First arrange the marks in ascending order as $20, 33, 39, 40, 50, 53, 59, 65, 69$ and find the median $\left( M \right)$ which will be ${\left( {\dfrac{{n + 1}}{2}} \right)^{th}}$ term after the arrangement.

Now, use the formula $M.D. = \dfrac{{\sum\limits_{i = 1}^n {\left| {{x_i} - M} \right|} }}{n}$ to calculate the mean deviation from the median.

To find the median the data must be arranged in either ascending or descending order of their value. So the arrangement in ascending order is $20$, $33$, $39$, $40$, $50$, $53$, $59$, $65$, $69$.

The number of observations $\left( n \right)$ is odd that means the median $\left( M \right)$ is given as ${\left( {\dfrac{{n + 1}}{2}} \right)^{th}}$ term.

$\Rightarrow M = {\left( {\dfrac{{9 + 1}}{2}} \right)^{th}}{\text{ term}}$

$\Rightarrow M = {\left( 5 \right)^{th}}{\text{ term}}$

$\Rightarrow M = 50$ 

Now, the mean deviation from the median is given as $M.D. = \dfrac{{\sum\limits_{i = 1}^n {\left| {{x_i} - M} \right|} }}{n}$. So for the given data,

$\Rightarrow M.D. = \dfrac{{\sum\limits_{i = 1}^9 {\left| {{x_i} - 50} \right|} }}{9}$

$\Rightarrow M.D.=\dfrac{{\left|{20-50}\right|+\left|{33-50}\right|+\left|{39-50}\right|+ \left|{40-50}\right|+\left|{50-50} \right|+\left|{53-50}\right|+\left|{59-50}\right|+\left| {65 - 50} \right|+\left|{69-50}\right|}}{9}$

$\Rightarrow M.D. = 12.67$ 

 Correct Answer: C

28: The standard deviation of the data $6$, $5$, $9$, $13$, $12$, $8$, $10$ is

(A) $\sqrt {\dfrac{{52}}{7}}$

(B) $\dfrac{{52}}{7}$

(C) $\sqrt 6 $

(D) $6$

Ans: The correct option is (A).

Given: The data $6$, $5$, $9$, $13$, $12$, $8$, $10$.

Use the formula $\overline x  = \dfrac{{\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)}}{n}$ to calculate the mean of the given data. Here $\overline x $ is the mean.

Now, use the formula $\sigma  = \sqrt {\dfrac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \overline x } \right)}^2}} }}{n}}$, where $\sigma $ is the standard deviation.

The total number of observations is $n = 7$. Using the formula $\overline x  = \dfrac{{\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)}}{n}$, where $\overline x $ is the mean,

$\Rightarrow \overline x  = \dfrac{{\left( {\sum\limits_{i = 1}^7 {{x_i}} } \right)}}{7}$

$\Rightarrow \overline x  = \dfrac{{6 + 5 + 9 + 13 + 12 + 8 + 10}}{7}$

$\Rightarrow \overline x  = 9$ 

Now, the standard deviation is given as $\sigma  = \sqrt {\dfrac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \overline x } \right)}^2}} }}{n}}$, where $\sigma $ is the standard deviation.

$\Rightarrow \sigma  = \sqrt {\dfrac{{\sum\limits_{i = 1}^7 {{{\left( {{x_i} - 9} \right)}^2}} }}{7}} $

$\Rightarrow \sigma  = \sqrt {\dfrac{{{{\left( {6 - 9} \right)}^2} + {{\left( {5 - 9} \right)}^2} + {{\left( {9 - 9} \right)}^2} + {{\left( {13 - 9} \right)}^2} + {{\left( {12 - 9} \right)}^2} + {{\left( {8 - 9} \right)}^2} + {{\left( {10 - 9} \right)}^2}}}{7}} $

$\Rightarrow \sigma  = \sqrt {\dfrac{{52}}{7}} $ 

Correct Answer: A

29: Let ${x_1}, {x_2}, ...., {x_n}$ be $n$ observations and \[\overline x \] be their arithmetic mean. The formula for the standard deviation is given by

(A) $\sum\limits_{i = 1}^n {{{\left( {{x_i} - \overline x } \right)}^2}}$

(B) $\dfrac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \overline x } \right)}^2}} }}{n}$

(C) $\sqrt {\dfrac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \overline x } \right)}^2}} }}{n}}$

(D) $\sqrt {\dfrac{{\sum\limits_{i = 1}^n {{x_i}^2} }}{n} + {{\left( {\overline x } \right)}^2}}$

Ans: The correct option is (C).

Given: $n$ observations ${x_1},{x_2},....,{x_n}$ and their arithmetic mean as \[\overline x \].

Use the definition of the variance which is ‘the ratio of the summation of the squares of the deviations of each observation from their mean and the total number of observations’.

Take the square root of this mathematical expression of the variance to get the standard deviation.

The arithmetic means is generally referred to as the mean $\overline x $ of the data.

According to the definition of variance it is the ratio of the summation of the squares of the deviations of each observation from their mean and the total number of observations. Mathematically it is given as,

${\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \overline x } \right)}^2}} }}{n}$

Now, the standard deviation is the square root of the variance, so it is given by the relation,

$\sigma  = \sqrt {\dfrac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \overline x } \right)}^2}} }}{n}}$

Correct Answer: C

30: 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) $255000$

Ans: The correct option is (C).

Given: Total number of observations $ = 100$

Mean of the observations $ = 50$

Standard deviation $ = 5$

Use the formula $\overline x  = \dfrac{{\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)}}{n}$ to calculate the sum of all the observations, i.e.  $\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)$. Here $\overline x $ is the mean.

Now, use the formula $\sigma  = \sqrt {\dfrac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \overline x } \right)}^2}} }}{n}}$, where $\sigma $ is the standard deviation, and calculate the variance. Break the summation into several parts and find the value of $\left( {\sum\limits_{i = 1}^n {{{\left( {{x_i}} \right)}^2}} } \right)$.

Since there are $100$ observations and the mean is $50$, so using the formula $\overline x  = \dfrac{{\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)}}{n}$,

$ \Rightarrow 50 = \dfrac{{\left( {\sum\limits_{i = 1}^{100} {{x_i}} } \right)}}{{100}}$

$ \Rightarrow \left( {\sum\limits_{i = 1}^{100} {{x_i}} } \right) = 5000$        …… $\left( 1 \right)$

Now, the standard deviation is given as $\sigma  = \sqrt {\dfrac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \overline x } \right)}^2}} }}{n}}$, where $\sigma $ is the standard deviation. On squaring both the sides the variance is obtained. The provided value of $\sigma $ is $5$.

$\Rightarrow \sqrt {\dfrac{{\sum\limits_{i = 1}^{100} {{{\left( {{x_i} - \overline x } \right)}^2}} }}{{100}}}  = 5$

$\Rightarrow \dfrac{{\sum\limits_{i = 1}^{100} {{{\left( {{x_i} - \overline x } \right)}^2}} }}{{100}} = 25$

$\Rightarrow \sum\limits_{i = 1}^{100} {{{\left( {{x_i} - \overline x } \right)}^2}}  = 2500$ 

Breaking the summation into several parts,

\[ \Rightarrow \sum\limits_{i = 1}^{100} {{{\left( {{x_i}} \right)}^2}}  + \sum\limits_{i = 1}^{100} {{{\left( {\overline x } \right)}^2}}  - \sum\limits_{i = 1}^{100} {2\overline x {x_i}}  = 2500\]

Using the value from equation $\left( 1 \right)$,

$\Rightarrow \sum\limits_{i = 1}^{100} {{{\left( {{x_i}} \right)}^2}}  + \sum\limits_{i = 1}^{100} {{{\left( {50} \right)}^2}}  - \sum\limits_{i = 1}^{100} {2 \times 50 \times {x_i}}  = 2500$

$\Rightarrow \sum\limits_{i = 1}^{100} {{{\left( {{x_i}} \right)}^2}}  + 100 \times {\left( {50} \right)^2} - 100\sum\limits_{i = 1}^{100} {{x_i}}  = 2500$

$\Rightarrow \sum\limits_{i = 1}^{100} {{{\left( {{x_i}} \right)}^2}}  + 250000 - 500000 = 2500$

$\Rightarrow \sum\limits_{i = 1}^{100} {{{\left( {{x_i}} \right)}^2}}  = 252500$ 

Correct Answer: C

31: Let \[a, b, c, d, e \] be the observations with mean \[{\text{m}}\] and standard deviation \[{\text{s }}{\text{.}}\] the standard deviation of the observations \[a  +  k, b  +  k, c  +  k, d  +  k, e  +  k\]  is

(A) \[s\]

(B) \[s . k\]

(C) \[s  +  k\]

(D) \[\dfrac{s}{k}\]

Ans: The correct option is (A).

Given: \[a,b,c,d,e \] be the observations of mean m and standard deviations.

We know that,

$Var(ax + b) = {a^{2}}\cdot Var(x)$

or

$S.D(ax + b) = a \cdot S.D(x)$ 

Then,

$Var(ax + b) = {a^{2}}.Var(x)$

$S.D(ax + b) = a \cdot S.D(x)$

$a = 1,b = k$

$\therefore S.D(x + k) = S.D(a + k,b + k,c + k,d + k,e + k)$

$= 1\cdot S.D(a,b,c,d,e) = s$ 

Correct Answer: A

32: Let \[{x_1} , {x_2} , {x_3} , {x_{4 }}, {x_{5}}\] be the observations with mean \[m\] and standard deviation \[{\text{s}}\]. 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)  \[\dfrac{s}{k}\]

(C)  \[k.s\]

(D)  \[s\]

Ans: The correct option is (C).

Given: \[{x_1},{x_2},{x_3},{x_{4}},{x_{5}}\] be the observations of mean \[{\text{m}}\] and standard deviation S. 

We know that,

 S.D = $\sqrt {\dfrac{{\sum {{x_i}^2} }}{n} - {{\left( {\dfrac{{\sum {{x_i}} }}{n}} \right)}^2} } $

 $\sum {{x_i}^2}  = {\left( {k{x_1}} \right)^2}{\text{ + }}.........{\text{ + }}{\left( {k{x_5}} \right)^2}$

$= {k^2}\sum {{x_i}^2} $

$\sum {{x_i}^{}}  = \left( {k{x_1}} \right){\text{ + }}.........{\text{ + }}\left( {k{x_5}} \right)$

$= k\sum {{x_i}} $ 

Then,

${\text{S}}{\text{.D}} = \sqrt {\dfrac{{\sum {{x_i}^2} }}{n} - {{\left( {\dfrac{{\sum {{x_i}}}}{n}} \right)}^2} } $

$\sum {{x_i}^2}  = {\left( {k{x_1}} \right)^2}{\text{  +  }}.........{\text{ + }}{\left( {k{x_5}} \right)^2}$

$= {k^2}\sum {{x_i}^2} $

$\sum {{x_i}^{}}  = \left( {k{x_1}} \right){\text{  +  }}.........{\text{  + }}\left( {k{x_5}} \right)$

$= k\sum {{x_i}} $

S.D $= \sqrt {\dfrac{{\sum {{x_i}^2} }}{5} - {{\left( {\dfrac{{\sum {{x_i}} }}{5}} \right)}^2} }$       Since, $\overline x  = {\dfrac{{\sum {x_i}}}{n}} ;\sigma  = S = {\text{ S}}{\text{.D}}$

$= \left| k \right|\sqrt {\dfrac{{\sum {{x_i}^2} }}{5} - {{\left( {\dfrac{{\sum {{x_i}^2} }}{5}} \right)}^2} } $

$= \left| k \right|s$ 

Correct Answer:  C

33: Let \[{x_1} , {x_2} , ... {x_n}\] be \[{\text{n}}\] observations. Let \[{w_i}  =  l{x_i}  +  k \] for \[l =  1, 2, ...n,\] where \[l \text{ and } k\] are constants. If the mean of $x_i$’s is 48 and their standard deviation is \[12\], the mean of $w_i$’s is 55 and standard deviation of $w_i$’s is 15, the mean of the values of \[{\text{l}}\] and \[{\text{k}}\] should be 

 (A) \[l  =  1.25, k  = -5 \]

(B) \[l  = -1.25, k  =  5\]

(C) \[l  =  2.5, k  = -5 \]

(D) \[l  =  2.5, k  =  5\]

Ans: The correct option is (A).

Given: \[{x_1},{x_2},...{x_n}\] Be \[{\text{n}}\] observations

\[{w_i} = l{x_i} + k\]\[\text{ for } l = 1,2,...n,\]

\[landk\] are constants. The mean of $x_i$ is 48 and their Standard deviation is \[12\],

The mean of $w_i$’s is 55 and standard deviation of $w_i$’s is 15.

We know that,

$\sigma (ax + b) = \left| a \right|\sigma (x)$

$\sigma ({w_i}) = \sigma (l{x_i} + k)$ 

Then,

$\overline x  = 48$

$\sigma (x) = 12$

${w_i} = l{x_i} + k \ldots ..\left( 1 \right)$

${\text{for equation (1)}}$

$\sigma (ax + b) = \left| a \right|\sigma (x)$

$\Rightarrow \sigma ({w_i}) = \sigma (l{x_i} + k)$

$\Rightarrow \sigma (w) = (l)(12) \Rightarrow l = \dfrac{{15}}{{12}}$

$\Rightarrow l = 1.25$

Now, $\overline w  = l\overline x  + k$

$\Rightarrow 55 = (1.25 \times 48) + k$

$\Rightarrow k =  - 5$ 

Correct Answer: A

34: Standard deviations for the first \[{\text{10}}\] natural numbers is 

(A) \[5.5 \]

(B) \[3.87 \]

(C) \[2.97 \]

(D) \[2.87\]

Ans: The correct option is (D).

Given: First 10 natural numbers, \[n = 10\]

To find: Standard deviations

We know that,

$\sigma  = \sqrt {\dfrac{{{n^2} - 1}}{{12}}}$

$\Rightarrow \sigma  = \sqrt {\dfrac{{{n^2} - 1}}{{12}}} $

$n = 10$

$\Rightarrow \sigma  = \sqrt {\dfrac{{{{10}^2} - 1}}{{12}}} $

$= \sqrt {\dfrac{{99}}{{12}}} $

$= 2.87$ 

Correct Answer: D

35: Consider the numbers \[1, 2, 3, 4, 5, 6, 7, 8, 9, 10 .\] If \[{\text{1}}\] is added to each number, the variance of the numbers so obtained is

(A) \[6.5 \]

(B) \[2.87 \]

(C) \[3.87 \]

(D) \[8.25\]

Ans: The correct option is (D).

Given: Let us consider the numbers  \[1,2,3,4,5,6,7,8,9,10.\]

\[{\text{1}}\] is added to each number.

We know that,

\[{\sigma ^2} = \sum {\dfrac{{{x_i}^2}}{n}}  - {\left( {\sum {\dfrac{{{x_i}}}{n}} } \right)^2}\]

$\sum\limits_{i = 1}^{10} {X_i^2}  = {2^2} + {3^2} + ....... + {11^2}$

$= \dfrac{{11 \times 12 \times 23}}{6} - 1 = 505$

 ${X_1}, \ldots ,{X_n} = \sum\limits_{i = 1}^n {{X_i}}  = 2 + 3 + .... + 11 = 65$ 

Then,

  ${\sigma ^2} = \sum {\dfrac{{{x_i}^2}}{n}}  - {\left( {\sum {\dfrac{{{x_i}}}{n}} } \right)^2}$

$= \dfrac{{505}}{{10}} - {\left( {6.5} \right)^2}$

$= 8.25$ 

Correct Answer: D

36: Consider the first \[{\text{10}}\] 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) \[2.87\]

Ans: The correct option is (A).

Given: Let’s consider the first 10 positive numbers

Multiply by \[ - 1\] and add \[{\text{1}}\] to each number 

For Example,

$1 \times ( - 1) =  - 1 + 1 = 0$

We know that, \[{\sigma ^2} = {\dfrac{\sum {{x_i}^2}}{n}}  - {\left( { {\dfrac{{\sum{x_i}}}{n}} } \right)^2}\]

Since the first 10 positive integers are \[1,2,3,4,5,6,7,8,9{\text{ and }}10\].

On multiplying each number by \[ - 1,\]

We get \[ - 1, - 2, - 3, - 4, - 5, - 6, - 7, - 8, - 9, - 10\]

On adding \[{\text{1}}\] in each number,

We get \[ - 1, - 2, - 3, - 4, - 5, - 6, - 7, - 8, - 9, - 10\]

${\sigma ^2} = \sum {\dfrac{{{x_i}^2}}{n}}  - {\left( {\sum {\dfrac{{{x_i}^2}}{n}} } \right)^2}$

$\sum {{x_i} = \dfrac{{ - 9 \times 10}}{2} =  - 45} $

$\sum {{x_i}^2 = {0^2} +  - {1^2} + {{( - 2)}^2} + {{( - 3)}^2} + ....... + {{( - 9)}^2} = \dfrac{{9 \times 10 \times 19}}{6} = 285} $

${\sigma ^2} = {\dfrac{{285}}{{10}}}  - {\left( {{\dfrac{{ - 45}}{{10}}} } \right)^2} = {\dfrac{{285}}{{10}}}  - {\left( {{\dfrac{{2025}}{{10}}} } \right)}$ 

\[ = \sqrt {\dfrac{{2850 - 2025}}{{100}}}  = {\left( {\sqrt {8.25} } \right)^2} = 8.25\]

Correct Answer: A

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

(A) \[6.63 \]

(B) \[16 \]

(C) \[22 \]

(D) \[44\]

Ans: The correct option is (D).

Given: $\sum{x^2} = 18000,  \sum{x} =960$

We know that,

\[{\sigma ^2} = {\dfrac{\sum {{x_i}^2}}{N}}  - {{{\left( {\dfrac{{\sum {x_i}}}{n}} \right)}^2}} \]

Then,

\[{\sigma ^2} = {\dfrac{\sum {{x_i}^2}}{N}}  - {{{\left( {\dfrac{{\sum {x_i}}}{n}} \right)}^2}} \]

$N = 60$

${\sigma ^2} = {\dfrac{{18000}}{{60}}}  - {{{\left( {\dfrac{{960}}{{60}}} \right)}^2}} $

$= 300 - {(16)^2}$

$= 44$ 

Correct Answer: D

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

(A) 0

(B) 1

(C) \[1.5 \]

(D) \[2.5\]

Ans: The correct option is (A).

Given: Coefficients of variations are $50$ and $60$

Arithmetic means are $30$ and $25$

We know that,\[{\text{C}}{{\text{V}}_1} = \dfrac{{{\sigma _1}}}{{\overline {{x_1}} }} \times 100\]

Then,

$C{V_1} = 50,   C{V_2} = 60$

$\overline {{x_1}}  = 30, \overline {{x_2}}  = 25$

$c{v_1} = \dfrac{{{\sigma _1}}}{{\overline {{x_1}} }} \times 100$

${\sigma _1} = \dfrac{{50 \times 30}}{{100}} = 15$

$C{V_2} = \dfrac{{{\sigma _2}}}{{\overline {{x_2}} }} \times 100$

${\sigma _2} = \dfrac{{60 \times 25}}{{100}} = 15$

${\text{now }}$

${\sigma _1} = {\sigma _2} = 15$ 

The difference in standard deviation is \[0.\]

Correct Answer: A

39: The standard deviation of some temperature data in \[{}^{\circ}C  \text{ is }  5\]. If the data were converted into ${}^{\circ}F$ then the  variance would be 

(A) 81

(B) 57

(C) 36

(D) 25

Ans: The correct option is (A).

Given: Some data are given in \[{}^{\circ}C  \text{ is }  5\]

We know that,\[F = \dfrac{9}{5}C + 32  \]

Then,

${\sigma _c} \text{ and } {\sigma _f} = ?$

$F = \dfrac{9}{5}c + 32$       

As We know that on adding and subtracting any number in data the standard deviation doesn’t change but on multiplying and dividing the data by the same number the standard deviation changes according to the number.

$x \to s$

$kx \to ks$

$x \to s$

$k + x \to s$ 

${\sigma _f} = \dfrac{9}{5}{\sigma _c}$

$\Rightarrow {\sigma _f} = 9$ ∵ Given ${\sigma _c}=5$

$\Rightarrow {\text{variance}}  = {(SD)^2} = {9^2} = 81$ 

Correct Answer: A

Fill in the blanks in Exercises from 40 to 46.

40. Coefficient of variation = $\dfrac{{.....}}{{Mean}} \times 100$

Ans: The formula for computing variation is 

Coefficient of variation C.V $ = \dfrac{\sigma }{\mu } \times 100$

Where, $\sigma  = {\text{standard deviation}}$

$\mu  = {\text{Mean}}$

Therefore,

${\text{C}}{\text{.V}}$$ = $$\dfrac{{{\text{Standard deviation}}}}{{{\text{Mean}}}} \times 100$

41. If $x$ is the mean of $n$ values of $x$, then $\sum\limits_{i = 1}^n{\left( {{x_i} - \bar x}\right)}$ is always equal to _______.

If $a$ has any value other than $x$, then $\sum\limits_{i = 1}^n{{\left( {{x_i} - \bar x} \right)^2}}$  is _________ than ${\left( {{x_i} - a} \right)^2}$.

Ans: Mean of $n$ observations ${x_1},{x_2},.....,{x_n}$$ = \bar x$

Therefore, $\bar x = \dfrac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}$

$ \Rightarrow \sum\limits_{i = 1}^n {{x_i} = n\bar x}$ then,

$\sum\limits_{i = 1}^n {({x_i} - \bar x)}$$ = $$\sum\limits_{i = 1}^n {{x_i} - \sum\limits_{i = 1}^n {\bar x} }$

$= n\bar x - \bar x\sum\limits_{i = 1}^n {1\left[ {{\text{and}}\sum\limits_{i = 1}^n 1  = n} \right]} $

$= n\bar x - \bar x.n$

$= 0$ 

Hence, the value of $\sum\limits_{i = 1}^n {({x_i} - \bar x)}$$ = 0$.

Then, $\sum\limits_{i = 1}^n{\left( {{x_i} - \bar x} \right)^2}$$ < $${\left( {{x_i} - a} \right)^2}$.

42. If the variance of a data is \[121\], then the standard deviation of the data is _______.

Ans: Square root of variance is standard deviation.

${\text{i}}{\text{.e, standard deviation }} = \sqrt {{\text{variance}}} $ 

Therefore, given

variance = 121

${\sigma ^2} = 121$

$\Rightarrow \sigma  = \sqrt {121} $

$\Rightarrow \sigma  = 11$ 

Hence, standard deviation is $11$

43. The standard deviation of a data is ___________ of any change in orgin, but is_____ on the change of scale.

Ans: Standard deviation is not affected by the addition or subtraction of any value from the observation therefore independent of change in origin. But change in scale affects the standard deviation i.e., any value multiplied or divided to the observation changes standard deviation too.

44. The sum of the squares of the deviations of the values of the variable is _______ when taken about their arithmetic mean.

Ans: Mean of the number of observations converts the series of observations into a minimum value and mean deviation is the arithmetic average of deviations of the observation from an observed value. Sum of squares of deviation is minimum when deviations are taken from mean.

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

Ans: Sum of all the deviation of the data about the data mean is $ = $ mean deviation and the median lies between mean and mode. Therefore, mean deviation of the data is least when measured from the median.

46. The standard deviation is _______ to the mean deviation taken from the arithmetic mean.

Ans: Difference between square of deviation of data about mean and square of mean is standard deviation. And mean deviation is the sum of all deviations of a series of data about its mean. Hence, its greater or equal to mean deviation.

Ncert Exemplar Class 11 Chapter 15 Statistics

The Class 11 Maths is important for the students as it is the foundation stone for class 12, which ultimately defines their career goals. The NCERT Solutions Exemplar for Chapter 15 Statistics is provided by Vedantu and is considered to be one of the most important study materials for the students of Class 11. The solutions provided by Vedantu are formulated in such a way that every step is explained easily and clearly in detail. The Solutions for Class 11 Maths NCERT Exemplar have been prepared by the subject experts at Vedantu to help the students in their board exam preparation. The solutions will be helpful not only for their exam preparations but also in solving homework and assignments. The NCERT Exemplar Solutions provided have been explained in detail, which will ensure the students make the exam preparation effective for them. The NCERT Exemplar for Class 11 Chapter 15 is mainly focused on the measures of dispersion. It will help the students with an effective understanding of the topic and its concepts.

Important Topics In Ncert Exemplar Class 11 Chapter-15

The important topics are as follows:

• Measures of dispersion

• Definition and meaning of range

• Mean Deviation

• Mean deviation for ungrouped data

• Mean deviation for discrete frequency distribution

• Mean deviation for continuous frequency distribution (Grouped data)

• Variance

• Standard Deviation

• The standard deviation for a discrete frequency distribution

• The standard deviation of a continuous frequency distribution (grouped data)

• Coefficient of variation