Chapter 15: Statistics

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.

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

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

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

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

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

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

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)\]

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

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

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

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

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

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.

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

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

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

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

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

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

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

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}}} $ 

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} $

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

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}} $ 

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

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

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

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

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

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

${\text{Similarly,}}$

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

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

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

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

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

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

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

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

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

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

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}} $

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