Solution

$\bar{x}=\dfrac{\sum f _ {i} x _ {i}}{\sum f _ {i}}=\dfrac{433}{20}=21.65$

$MD=\dfrac{\sum f _ {i}\left|x _ {i}-\bar{x}\right|}{\sum f _ {i}}=\dfrac{25}{20}=1.25$

Solution

$ M _ {e}=\left(\dfrac{20+1}{2}\right)^{th}\ \text{item }=\dfrac{21}{2}=10.5^ {th} \text { item } $

$ \therefore \quad M _ {e}=12 $

$ \therefore \quad M D=\dfrac{\sum f _ {i} d _ {i}}{\sum f _ {i}}=\dfrac{25}{20}=1.25$

Solution

Consider first natural number when $n$ is an odd i.e., 1, 2, 3, 4, … $n$, [odd].

$ \text { Mean } \bar{x} =\dfrac{1+2+3+\ldots+n}{n}=\dfrac{n(n+1)}{2 n}=\dfrac{n+1}{2} \qquad\left[\because \quad \text { sum of first } n \text { natural numbers }=\dfrac{n(n+1)}{2}\right] $

$\therefore \quad \mathrm{MD} =\dfrac{\left|1-\dfrac{n+1}{2}\right|+\left|2-\dfrac{n+1}{2}\right|+\left|3-\dfrac{n+1}{2}\right|+\cdots+\left|n-\dfrac{n+1}{2}\right|}{n} $

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

$=\dfrac{2}{n}\left[1+2+\ldots+\dfrac{n-3}{2}+\dfrac{n-1}{2} \right] $

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

$=\dfrac{2}{n} \cdot \dfrac{1}{2} \left[\left(\dfrac{n-1}{2}\right) \left(\dfrac{n+1}{2}\right)\right]=\dfrac{1}{n} \left(\dfrac{n^{2}-1}{4}\right)=\dfrac{n^{2}-1}{4 n} $

Solution

Consider first $n$ natural number, when $n$ is even i.e., $1,2,3,4, \ldots \ldots . n$.

$ \begin{aligned} \therefore \quad \text { Mean } \bar{x} & =\dfrac{1+2+3+\ldots+n}{n}=\dfrac{n(n+1)}{2 n}=\dfrac{n+1}{2} \\ \\ MD & =\dfrac{1}{n}\Bigg[\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|+\ldots+\left|n-\dfrac{n+1}{2}\right|\Bigg] \\ \\ & =\dfrac{1}{n}\Bigg[\left|\dfrac{1-n}{2}\right|+\left|\dfrac{3-n}{2}\right|+\left|\dfrac{5-n}{2}\right|+\ldots .+\left|\dfrac{-3}{2}\right|+\left|\dfrac{1}{2}\right|+\ldots+\left|\dfrac{n-1}{2}\right|\Bigg] \\ \\ & =\dfrac{2}{n} \Bigg[\dfrac{1}{2}+\dfrac{3}{2}+\ldots .+\dfrac{n-1}{2}\Bigg] \\ \\ & =\dfrac{1}{n} \cdot \dfrac{n^{2}}{2} \quad[\because \quad \text { sum of first } n \text { odd natural numbers }=n^{2}] \\ \\ & =\dfrac{1}{n} \cdot \dfrac{n^{2}}{4}=\dfrac{n}{4} \end{aligned} $

Solution

$ \begin{aligned} \Sigma x _ {i} & =1+2+3+4+\ldots+n=\dfrac{n(n+1)}{2} \\ \\ \text { and } \quad \sum x _ i^{2} & =1^{2}+2^{2}+3^{2}+\ldots+n^{2}=\dfrac{n(n+1)(2 n+1)}{6} \\ \\ \text { Now, }\quad SD & =\sqrt{\dfrac{\sum x _ i^{2}}{N}-(\dfrac{\sum x _ {i}{ }}{N})^{2}} \\ \\ & =\sqrt{\dfrac{n(n+1)(2 n+1)}{6 n}-\dfrac{n^{2}(n+1)^{2}}{4 n^{2}}} \\ \\ & =\sqrt{\dfrac{(n+1)(2 n+1)}{6}-\dfrac{(n+1)^{2}}{4}} \\ \\ & =\sqrt{\dfrac{2(2 n^{2}+3 n+1)-3(n^{2}+2 n+1)}{12}} \\ \\ & =\sqrt{\dfrac{4 n^{2}+6 n+2-3 n^{2}-6 n-3}{12}} \\ \\ & =\sqrt{\dfrac{n^{2}-1}{12}} \end{aligned} $

Solution

Given,

$n _ {i} =25, \bar x _ i=18.2, \sigma _ 1=3.25 \\ \\ $

$n _ 2 =15, \sum _ {i=1}^{15} x _ {i}=279 \text { and } \sum _ {i=1}^{15} x _ i^{2}=5524 $

For first set,

$ \begin{aligned} \sum x _ {i} & =25 \times 18.2=455 \\ \\ \sigma _ 1^{2} & =\dfrac{\sum x _ i^{2}}{25}-(18.2)^{2} \\ \\ 25^{2} & =\dfrac{\sum x _ i^{2}}{25}-331.24 \\ \\ 1.24 & =\dfrac{\sum x _ i^{2}}{25} \\ \\ \Sigma x _ i^{2} & =25 \times(10.5625) \\ \\ & =25 \times 341.8025 \\ \\ & =8545.0625 \end{aligned} $

$ \begin{matrix} \therefore \quad & \sigma _ 1^{2}=\dfrac{\Sigma x _ i^{2}}{25}-(18.2)^{2} \\ \\ \Rightarrow & (3.25)^{2}=\dfrac{\Sigma x _ i^{2}}{25}-331.24 \\ \\ \Rightarrow & 10.5625+331.24=\dfrac{\Sigma x _ i^{2}}{25} \\ \\ \Rightarrow & \Sigma x _ i^{2}=25 \times(10.5625+331.24) \end{matrix} $

For combined SD of the 40 observations $n=40$,

$ \begin{aligned} \text { Now } \quad \Sigma x _ i^{2} & =5524+8545.0625=14069.0625 \\ \\ \text { and } \quad \Sigma x _ {i} & =455+279=734 \\ \\ \therefore \quad SD &=\sqrt{\dfrac{14069.0625}{40}-(\dfrac{734}{40})^{2}} \\ \\ & =\sqrt{351.726-(18.35)^{2}} \\ \\ & =\sqrt{351.726-336.7225} \\ \\ & =\sqrt{15.0035}=3.87 \end{aligned} $

Solution

Given : mean and standard deviation of a set of $ n _ {1} $ observation are $ \bar{x} _ {1} $ and $ s _ {1} $ respectively while the mean and standard deviation of another set of $ n $ observation are $ \bar{x} _ {2} $ and $ s _ {2} $ respectively.

Let $ s $ be the standard deviation and $ \bar{x} $ be the mean of combined sent of $( n _ {1}+n _ {2} )$ observations.

$ \bar{x}=\dfrac{n _ {1} \bar{x} _ {1}+n _ {2} \bar{x} _ {2}}{n _ {1}+n _ {2}} \qquad…(1) $

If $ d _ {1}=\bar{x} _ {1}-\bar{x} $ and $ d _ {2}=\bar{x} _ {2}-\bar{x} $ then

$ s^{2}=\dfrac{1}{n _ {1}+n _ {2}}\left[n _ {1}\left(s _ {1}^{2}+d _ {1}^{2}\right)+n _ {2}\left(s _ {2}^{2}+d _ {2}^{2}\right)\right] \qquad…(2) $

Now,

$d _ {1}^{2}=\left(\bar{x} _ {1}-\bar{x}\right)^{2}=\left\lbrace \bar{x} _ {1}-\dfrac{n _ {1} \bar{x} _ {1}+n _ {2} \bar{x} _ {2}}{n _ {1}+n _ {2}}\right\rbrace^{2}$

$ d _ {1}^{2}=\dfrac{n _ {1}^{2}\left(\bar{x} _ {1}-\bar{x} _ {2}\right)^{2}}{\left(n _ {1}+n _ {2}\right)^{2}}\qquad…(3) $

and $ d _ {2}^{2}=\left(\bar{x} _ {2}-\bar{x}\right)^{2} =\left\lbrace \bar{x} _ {2}-\dfrac{n _ {1} \bar{x} _ {1}+n _ {2} \bar{x} _ {2}}{n _ {1}+n _ {2}}\right\rbrace^{2}$

$ \qquad\quad=\dfrac{n _ {2}^{2}\left(\bar{x} _ {2}-\bar{x} _ {1}\right)^{2}}{\left(n _ {1}+n _ {2}\right)^{2}}=\dfrac{n _ {1}^{2}\left(\bar{x} _ {1}-\bar{x} _ {2}\right)^{2}}{\left(n _ {1}+n _ {2}\right)^{2}} \qquad…(4) $

From (2), (3) and (4) we get,

$ \begin{array}{l} s^{2}=\dfrac{1}{n _ {1}+n _ {2}} {\left[n _ {1}\left\lbrace s _ {1}^{2}+\dfrac{n _ {2}^{2}\left(\bar{x} _ {1}-\bar{x} _ {2}\right)^{2}}{\left(n _ {1}+n _ {2}\right)^{2}}\right\rbrace+n _ {2}\left\lbrace s _ {2}^{2}+\dfrac{n _ {1}^{2}\left(\bar{x} _ {1}-\bar{x} _ {2}\right)^{2}}{\left(n _ {1}+n _ {2}\right)^{2}}\right\rbrace\right]} \end{array} $

$ s^{2}=\dfrac{1}{n _ {1}+n _ {2}}\left[n _ {1} s _ {1}^{2}+n _ {2} s _ {2}^{2}+\dfrac{\left(\bar{x} _ {1}-\bar{x} _ {2}\right)^{2}}{\left(n _ {1}+n _ {2}\right)^{2}}\left(n _ {1} n _ {2}^{2}+n _ {2} n _ {1}^{2}\right)\right] $

$ s^{2}=\dfrac{1}{n _ {1} n _ {2}}\left[n _ {1} s _ {1}^{2}+n _ {2} s _ {2}^{2}+\dfrac{n _ {1} n _ {2}}{\left(n _ {1}+n _ {2}\right)}\left(\bar{x} _ {1}-\bar{x} _ {2}\right)^{2}\right] $

$ s=\sqrt{\dfrac{n _ {1} s _ {1}^{2}+n _ {2} s _ {2}^{2}}{n _ {1}+n _ {2}}+\dfrac{n _ {1} n _ {2}}{\left(n _ {1}+n _ {2}\right)^{2}}\left(\bar{x} _ {1}-\bar{x} _ {2}\right)^{2}} $ is proved.

Solution

Given, $n _ 1=20, \sigma _ 1=5, \bar x _ 1=17$ and $n _ 2=20, \sigma _ 2=5, \bar x _ 2=22$

We know that, $\sigma=\sqrt{\dfrac{n _ 1 s _ 1^{2}+n _ 2 s _ 2^{2}}{n _ 1+n _ 2}+\dfrac{n _ 1 n _ 2(\bar x _ 1-\bar x _ 2)^{2}}{(n _ 1+n _ 2)^{2}}}$

$ \hspace {2.2cm}=\sqrt{\dfrac{20 \times(5)^{2}+20 \times(5)^{2}}{20+20}+\dfrac{20 \times 20(17-22)^{2}}{(20+20)^{2}}}$

$ \hspace {2.2cm}=\sqrt{\dfrac{1000}{40}+\dfrac{400 \times 25}{1600}}=\sqrt{25+\dfrac{25}{4}}$

$ \hspace {2.2cm}=\sqrt{\dfrac{125}{4}}=\sqrt{31.25}=5.59$

Solution

$\therefore \quad \sigma^{2}=\dfrac{\Sigma f _ {i} x _ i^{2}}{n}-\left(\dfrac{\Sigma f _ {i} x _ {i}}{n}\right)^2 $

$\Rightarrow 160=\dfrac{92 A^{2}}{7}-(\dfrac{(22 A)}{7})^{2} $

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

$\Rightarrow 160=(644-484) \dfrac{A^{2}}{49} $

$\Rightarrow 160=\dfrac{160 A^{2}}{49} $

$ \Rightarrow A^{2}=49 $

$\therefore \quad A=7$

Solution

$\therefore \quad SD$ $ =\sqrt{\dfrac{\Sigma f _ {i} d _ i^{2}}{N}-\Big(\dfrac{\Sigma f _ {i} d _ i}{N}\Big)^{2}}$

$\qquad \qquad =\sqrt{\dfrac{137}{60}-\Big(\dfrac{37}{60}\Big)^{2}} $

$\qquad \qquad=\sqrt{2.2833-(0.616)^{2}} $

$\qquad \qquad=\sqrt{2.2833-0.3794} $

$\qquad \qquad =\sqrt{1.9037}=1.38$

Solution

$\therefore \quad$ Sum of frequencies,

$\begin{aligned} & x-2+x+x^2+(x+1)^2+2 x+x+1=60 \\ \\ & \Rightarrow \quad 2 x-2+x^2+x^2+1+2 x+2 x+x+1=60 \\ \\ & \Rightarrow \quad 2 x^2+7 x=60 \\ \\ & \Rightarrow \quad 2 x^2+7 x-60=0 \\ \\ & \Rightarrow \quad 2 x^2+15 x-8 x-60=0 \\ \\ & \Rightarrow \quad x(2 x+15)-4(2 x+15)=0 \\ \\ & \Rightarrow \quad(2 x+15)(x-4)=0 \\ \\ & \Rightarrow x=-\dfrac{15}{2}, 4 \\ \\ & \Rightarrow x=-\dfrac{15}{2} \qquad \text { [inaddmisible since x is positive ]} \\ \\ & \end{aligned}$

$\begin{aligned} \text { Mean } & =A+\dfrac{\sum f _ i d _ i}{\Sigma f _ i}=3+\dfrac{-12}{60}=2.8 \\ \\ \sigma & =\sqrt{\dfrac{\sum f _ i d _ i^2}{\Sigma f _ i}-\Big(\dfrac{\sum f _ i d _ i)}{\Sigma f _ i}\Big)^2}=\sqrt{\dfrac{78}{60}-\Big(\dfrac{-12}{60}\Big)^2} \\ \\ & =\sqrt{1.3-0.04}=\sqrt{1.26}=1.12\end{aligned}$

Solution

Here, $n _ 1=60, \bar x _ 1=650, s _ 1=8$ and $n _ 2=80, \bar x _ 2=660, s _ 2=7$

$\therefore \quad \sigma =\sqrt{\dfrac{n _ 1 s _ 1{ }^{2}+n _ 2 s _ 2{ }^{2}}{n _ 1+n _ 2}+\dfrac{n _ 1 n _ 2(\bar x _ 1-\bar x _ 2)^{2}}{(n _ 1+n _ 2)^{2}}} $

$\hspace{1cm}=\sqrt{\dfrac{60 \times(8)^{2}+80 \times(7)^{2}}{60+80}+\dfrac{60 \times 80(650-660)^{2}}{(60+80)^{2}}} $

$\hspace{1cm}=\sqrt{\dfrac{6 \times 64+8 \times 49}{14}+\dfrac{60 \times 80 \times 100}{140 \times 140}} $

$\hspace{1cm}=\sqrt{\dfrac{192+196}{7}+\dfrac{1200}{49}}=\sqrt{\dfrac{388}{7}+\dfrac{1200}{49}} $

$\hspace{1cm}=\sqrt{\dfrac{2716+1200}{49}}=\sqrt{\dfrac{3916}{49}}$

$\hspace{1cm}=\dfrac{62.58}{7}=8.9 $

Solution

$\Rightarrow \quad \dfrac{\sum \mathrm{x} _ {\mathrm{i}}}{100}=50 $

$\Rightarrow \quad \sum \mathrm{x} _ {\mathrm{i}}=5000 $

$\Rightarrow \quad \dfrac{\sum\left(\mathrm{x} _ {\mathrm{i}}-50\right)^{2}}{\mathrm{ ~ N}}=16 $

$\Rightarrow \quad \dfrac{\sum \mathrm{x} _ {\mathrm{i}}^{2}-100 \sum \mathrm{x} _ {\mathrm{i}}+2500}{100}=16 $

$\Rightarrow \quad \sum \mathrm{x} _ {\mathrm{i}}^{2}-100 \times 5000+2500=16 \times 100 $

$\Rightarrow \quad \sum \mathrm{x} _ {\mathrm{i}}^{2}=500000-900 $

$\Rightarrow \quad \therefore \quad \sum \mathrm{x} _ {\mathrm{i}}^{2}=499100 $

Solution

Given,

$ \begin{aligned} n & =18, \Sigma(x-5)=3 \text { and } \Sigma(x-5)^{2}=43 \\ \\ & =5+\dfrac{3}{18}=5+0.1666=5.1666=5.17 \\ \\ \text { SD } & =\sqrt{\dfrac{\sum(x-5)^{2}}{n}-\Big(\dfrac{\sum(x-5)}{n}\Big)^{2}} \\ \\ & =\sqrt{\dfrac{43}{18}-\Big(\dfrac{3}{18}\Big)^2} \\ \\ & =\sqrt{2.3944-(0.166)^{2}}=\sqrt{2.3944-0.2755}=1.59 \end{aligned} $

$ \begin{aligned} & \therefore \quad \text { Mean }=A+\dfrac{\sum(x-5)}{18} \end{aligned} $

Solution

$\therefore \quad $ Mean $=\dfrac{\Sigma f _ {i} x _ {i}}{\Sigma f _ {i}}=\dfrac{66.5}{16}=4.15$

$ \begin{aligned} \text { variance } & =\sigma^{2}=\dfrac{\Sigma f _ {i} x _ i^{2}}{\Sigma f _ {i}}-\Big(\dfrac{\Sigma f _ {i} x _ {i}}{\Sigma f _ {i}}\Big)^2 \\ \\ & =\dfrac{340.25}{16}-(4.15)^{2} \\ \\ & =21.2656-17.2225=4.043 \end{aligned} $

Solution

$ \begin{array}{|c|c|c|c|c|c|} \hline \text{Class internal} & f _ i & x _ i & f _ i x _ i & d _ i=x _ i- \overline x \mid & f _ i d _ i \\ \hline 0-4 & 4 & 2 & 8 & 7.2 & 28.8 \\ 4-8 & 6 & 6 & 36 & 3.2 & 19.2 \\ 8-12 & 8 & 10 & 80 & 0.8 & 6.4 \\ 12-16 & 5 & 14 & 70 & 4.8 & 24.0 \\ 16-20 & 2 & 18 & 36 & 8.8 & 17.6 \\ \hline Total & \Sigma f _ i=25 & & \Sigma f _ i x _ i=230 & & \Sigma f _ i d _ i=96 \\ \hline \end{array} $

$\begin{aligned} & \therefore \quad \text { Mean }=\dfrac{\Sigma f _ i x _ i}{\Sigma f _ i}=\dfrac{230}{25}=9.2 \\ \\ & \text { and } \quad \text { mean deviation }=\dfrac{\Sigma f d _ i}{\Sigma f _ i}=\dfrac{96}{25}=3.84 \\ \\ & \end{aligned}$

Solution

$N=20$ , $\dfrac{N}{2}=\dfrac{20}{2}=10$

So, the median class is $12-18$.

$ \begin{aligned} \quad \text { Median } & =l+\dfrac{\dfrac{N}{2}-c f}{f} \times h \\ \\ & =12+\dfrac{6}{3}(10-9) \\ \\ & =12+2=14 \\ \\ MD & =\dfrac{\Sigma f _ {i} d _ {i}}{\Sigma f _ {i}}=\dfrac{140}{20}=7 \end{aligned} $

Solution

$ \begin{aligned} & \therefore \quad \text { Mean } \bar{x}=\dfrac{\Sigma f _ {i} x _ {i}}{\Sigma f _ {i}}=\dfrac{239}{40}=5.975 \approx 6 \\ \\ & \text { and } \\ \\ & \begin{aligned} \sigma & =\sqrt{\dfrac{\Sigma f _ {i} d _ i^{2}}{\Sigma f _ {i}}-\Big(\dfrac{\Sigma f _ {i} d _ i}{\Sigma f _ {i}}\Big)^{2}}=\sqrt{\dfrac{325}{40}-\Big(\dfrac{-1}{40}\Big)^{2}} \\ \\ & =\sqrt{8.125-0.000625}=\sqrt{8.124375}=2.85 \end{aligned} \end{aligned} $

Solution

Now,

$\sigma^2 ={\dfrac{\Sigma f _ {i} d _ i^{2}}{\Sigma f _ {i}}-\Big(\dfrac{\Sigma f _ {i} d _ i}{\Sigma f _ {i}}\Big)^{2}}$

$\quad =1.4571-0.2916=1.1655 \\ \\ \therefore \quad SD, \sigma =\sqrt{1.1655}=1.08 $

Solution

$\sum x _ {i}=\dfrac{n}{2}[2 a+(n-1)]$

Mean $=\dfrac{\sum x _ {i}}{n}=\dfrac{1}{n} \dfrac{n}{2}(2 a)+(n-1) d$ $=$ $a+\dfrac{(n-1)}{2} d$

$ \begin{aligned} \Sigma(x _ {i}-a) & =d[1+2+3+\ldots+(n-1) d] \\ \\ & =d \dfrac{(n-1) n}{2} \\ \\ \text { and } \quad \sum(x _ {i}-a)^{2} & =d^{2}[1^{2}+2^{2}+3^{2}+\ldots+(n-1)^{2}] \\ \\ & =\dfrac{d^{2}(n-1) n(2 n-1)}{6} \\ \\ \sigma & =\sqrt{\dfrac{\sum(x _ {i}-a)^{2}}{n}-\Big(\sum\dfrac{x _ {i}-a}{n}\Big)^{2}} \\ \\ & =\sqrt{\dfrac{d^{2}(n-1)(n)(2 n-1)}{6 n}-\Big(\dfrac{d(n-1) n{ }}{2n }\Big)^{2}} \\ \\ & =\sqrt{\dfrac{d^{2}(n-1)(2 n-1)}{6}-\dfrac{d^{2}{n^2}(n-1)^{2}}{4n^2}} \\ \\ & =d \sqrt{\dfrac{(n-1)(2 n-1)}{6}-\dfrac{(n-1)^{2}}{4}} \\ \\ & =d \sqrt{\dfrac{(n-1)}{2} \Big(\dfrac{2 n-1}{3}-\dfrac{n-1}{2}\Big)} \\ \\ & =d \sqrt{\dfrac{(n-1)}{2} \Big(\dfrac{4 n-2-3 n+3}{6}\Big)} \\ \\ & =d \sqrt{\dfrac{(n-1)(n+1)}{12}}=d \sqrt{\dfrac{(n^{2}-1)}{12}} \end{aligned} $

Solution

For Ravi,

Now,

$ \begin{aligned} \sigma & =\sqrt{\dfrac{\Sigma d^{2} _ i}{n}-\Big(\dfrac{\Sigma d _ {i}{ }}{n}\Big)^{2}} \\ \\ & =\sqrt{\dfrac{1699}{10}-\Big(\dfrac{-14}{10}\Big)^{2}}=\sqrt{169.9-1.96} \\ \\ & =\sqrt{167.94}=12.96 \\ \\ \bar{x} & =A+\dfrac{\Sigma d _ {i}}{\Sigma f _ {i}}=45-\dfrac{14}{10}=43.6 \end{aligned} $

For Hashina,

$ \begin{aligned} & \because \quad \quad \text { Mean }=55 \\ \\ & \therefore \quad \sigma=\sqrt{\dfrac{6368}{10}}=\sqrt{636.8}=25.2 \\ \\ & CV=\dfrac{\sigma}{\bar{x}} \times 100=\dfrac{12.96}{43.6} \times 100=29.72 \\ \\ & CV=\dfrac{\sigma}{\bar{x}} \times 100=\dfrac{25.2}{55} \times 100=45.89 \end{aligned} $

Hence, Hashina is more consistent and intelligent.

Solution

Given, $ \quad n =100, \bar{x}=40, \sigma=10 \text { and } \bar{x}=40 $

$\therefore \quad \dfrac{\sum x _ {i}}{n} =40$

$\Rightarrow \quad \dfrac{\Sigma x _ {i}}{100}=40$

$\Rightarrow\quad \Sigma x _ {i} =4000$

$ \begin{aligned} \text { Corrected } \Sigma x _ {i} & =4000-30-70+3+27 \\ \\ & =4030-100=3930 \\ \\ \text { Corrected mean } & =\dfrac{2930}{100}=39.3 \end{aligned} $

$ \begin{aligned} \sigma^{2} & =\dfrac{\Sigma x _ i^{2}}{n}-(40)^{2} \\ \\ 100 & =\dfrac{\Sigma x _ i^{2}}{100}-1600 \\ \\ \Sigma x _ i^{2} & =170000 \end{aligned} $

Corrected $\Sigma x _ i^{2}=170000-(30)^{2}-(70)^{2}+3^{2}+(27)^{2}$

$ \begin{aligned} & =164939 \\ \\ \text { Corrected } \sigma & =\sqrt{\dfrac{164939}{100}-(39.3)^{2}} \\ \\ & =\sqrt{1649.39-39.3 \times 39.3} \\ \\ & =\sqrt{1649.39-1544.49} \\ \\ & =\sqrt{104.9}=10.24 \end{aligned} $

Solution

$ \begin{array}{l} \operatorname{mean}(\bar{x})=45 \\ \\ \text { variance }=\left(\sigma^{2}\right)=16 \\ \\ n=10 \end{array} $

As we know, $ \operatorname{mean}(\bar{x})=\dfrac{\sum x}{n} $

$ \Rightarrow \Sigma x=45 x _ {10}=450 $

Now, replacing the urlong rading 52 by 25 , we get,

$ \sum x(\text { Corrected })=450-52+25=423 $

$ \therefore \quad \text { Corrected mean }(\bar{x})=\dfrac{\sum x(\text { Corrected })}{n}=\dfrac{423}{10} $

$ \text { Corrected mean }(\bar{x})=42.3 $

Now, given variance $ =16 $

$ \begin{array}{l} \therefore \quad \dfrac{\sum x^{2}}{n}-\left(\dfrac{\sum x}{n}\right)^{2}=\sigma^{2}=16 \\ \\ \Rightarrow\quad \dfrac{\sum x^{2}}{10}-\left(\dfrac{450}{10}\right)^{2}=16 \\ \\ \Rightarrow\quad \sum x^{2}=20410 \end{array} $

Now, replacing wrong reading 52 by 25 we get,

$ \begin{array}{l} \Sigma x^{2}=20410-(52)^{2}+(25)^{2} \\ \\ \Sigma x^{2}=18331 \end{array} $

Hence, Corrected variance $ \sigma^{2}=\dfrac{\sum x^{2} \text { (Corrected) }}{n}-\left(\dfrac{\sum x}{n}\right)^{2} $

$ \begin{aligned} & \qquad =\dfrac{18331}{10}-\left(\dfrac{423}{10}\right)^{2} \\ \\ \text { Corrected variance } & =43.81 \end{aligned} $

Solution

(b) Given, observations are $3,10,10,4,7,10$ and $5.$

Here, $n=7,\bar x=\dfrac{3+10+10+4+7+10+5}{7}=7$

Now, $MD=\dfrac{\sum d _ {i}}{N}=\dfrac{18}{7}=2.57$

• Option (a) is incorrect because the mean deviation calculated from the given data is not 2. The correct mean deviation is 2.57.
• Option (c) is incorrect because the mean deviation calculated from the given data is not 3. The correct mean deviation is 2.57.
• Option (d) is incorrect because the mean deviation calculated from the given data is not 3.75. The correct mean deviation is 2.57.

Solution

(b) $MD=\dfrac{1}{n} \sum _ {i=1}^{n}|x _ {i}-\bar{x}|$

• Option (a): $\sum _ {i=1}^{n}(x _ {i}-\bar{x})$ is incorrect because the sum of deviations from the mean is always zero. This is due to the property of the mean, which balances the data points such that the positive and negative deviations cancel each other out.

• Option (c): $\sum _ {i=1}^{n}(x _ {i}-\bar{x})^{2}$ is incorrect because this represents the sum of squared deviations from the mean, which is used to calculate the variance, not the mean deviation.

• Option (d): $\dfrac{1}{n} \sum _ {i=1}^{n}(x _ {i}-\bar{x})^{2}$ is incorrect because this represents the mean of the squared deviations from the mean, which is the definition of variance, not the mean deviation.

Solution

(a) Since, the lives of 5 bulbs are 1357, 1090, 1666, 1494 and 1623.

$ \begin{aligned} \therefore \quad \text { Mean } & =\dfrac{1357+1090+1666+1494+1623}{5} \\ \\ & =\dfrac{7230}{5}=1446 \end{aligned} $

• Option (b) 179 is incorrect because the correct mean deviation calculated from the given data is 178, not 179. The sum of the absolute deviations from the mean is 890, and dividing this by the number of bulbs (5) gives 178.

• Option (c) 220 is incorrect because it represents the absolute deviation of one of the bulbs (1666) from the mean, not the mean deviation of all bulbs.

• Option (d) 356 is incorrect because it represents the absolute deviation of one of the bulbs (1090) from the mean, not the mean deviation of all bulbs.

Solution

(c) Since, marks obtained by 9 students in Mathematics are 50, 69, 20, 33, 53, 39, 40, 65 and 59.

Rewrite the given data in ascending order.

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

Here, $n=9 \ [odd]$

$\therefore \quad $ Median $=\dfrac{9+1}{2}$ term $=5$ th term

$M _ e=50$

$\therefore \quad \text{Option (c) is correct.}$

Solution

(a) Given, data are 6, 5, 9, 13, 12, 8, and 10.

$\begin{array}{|c|c|} \hline x _ i & x _ i^2 \\ \hline 6 & 36 \\ 5 & 25 \\ 9 & 81 \\ 13 & 169 \\ 12 & 144 \\ 8 & 64 \\ 10 & 100 \\ \hline \Sigma x _ i=6 3 & \Sigma x _ i^ 2=619 \\ \hline \end{array}$

$\begin{aligned} \therefore \quad \mathrm{SD} & =\sigma=\sqrt{\dfrac{\Sigma x _ i^2}{N}-{\Big(\dfrac{\Sigma x _ i}{N}}\Big)^2}=\sqrt{\dfrac{619}{7}-\Big(\dfrac{63}{7}\Big)^2} \\ \\ & =\sqrt{\dfrac{7 \times 619-3969}{49}} \\ \\ & =\sqrt{\dfrac{4333-3969}{49}} \\ \\ & =\sqrt{\dfrac{364}{49}}=\sqrt{\dfrac{52}{7}}\end{aligned}$

$\therefore \quad \text{Option (a) is correct.}$

Solution

(c) SD is given by $ \sigma=\sqrt{\dfrac{\sum(x _ {i}-\bar{x})^{2}}{n}} $

• Option (a) $\Sigma(x _ {i}-\bar{x})^{2}$ is incorrect because it represents the sum of squared deviations from the mean, not the standard deviation. The standard deviation requires taking the square root of the average of these squared deviations.

• Option (b) $\dfrac{\Sigma(x _ {i}-\bar{x})^{2}}{n}$ is incorrect because it represents the variance, not the standard deviation. The standard deviation is the square root of the variance.

• Option (d) $\sqrt{\dfrac{\sum x _ i^{2}}{n}+\bar x^{-2}}$ is incorrect because it does not correctly represent the formula for standard deviation. The term $\bar x^{-2}$ is not part of the standard deviation formula and the correct formula involves the squared deviations from the mean, not the sum of squares of the observations divided by the number of observations.

Solution

(c) Given,

$\bar{x}=50, n =100 \text { and } \sigma=5 $

$\Sigma x _ i^{2} =? $

$\bar{x} =\dfrac{\Sigma x _ {i}}{n}$

$\Rightarrow \quad 50=\dfrac{\Sigma x _ {i}}{100} $

$\therefore \quad \Sigma x _ {i}=50 \times 100=5000$

$ \begin{aligned} & \text { Now, } \\ \\ & \Rightarrow \sigma=\sqrt{\dfrac{\Sigma x _ i^{2}}{n}-\dfrac{\sum x _ {i}{ }^{2}}{n}} \\ \\ & \Rightarrow \sigma^{2}=\dfrac{\sum x _ i^{2}}{n}-(\bar{x})^{2} \\ \\ & \Rightarrow \quad 25=\dfrac{\Sigma x _ i^{2}}{100}-(50)^{2} \\ \\ & \Rightarrow 25=\dfrac{\Sigma x _ i^{2}}{100}-2500 \\ \\ & \Rightarrow \quad 2525=\dfrac{\Sigma x _ i^{2}}{100} \\ \\ & \therefore \quad \Sigma x _ i^{2}=252500 \end{aligned} $

$\therefore \quad \text{Option (c) is correct.}$

Solution

(a) Given observations are $a, b, c, d$ and $e$.

$ \begin{aligned} \text { Mean } & =m=\dfrac{a+b+c+d+e}{5} \\ \\ \Sigma x _ {i} & =a+b+c+d+e=5 m \\ \\ \text { Now, } & \\ \\ \text { mean } & =\dfrac{a+k+b+k+c+k+d+k+e+k}{5} \\ \\ & =\dfrac{(a+b+c+d+e)+5 k}{5}=m+k \\ \\ \therefore \quad SD & =\sqrt{\dfrac{\sum(x _ {i}+k)^{2}}{n}-(m+k)^{2}} \\ \\ & =\sqrt{\dfrac{\sum(x _ i^{2}+k^{2}+2 k x _ {i})}{n}-(m^{2}+k^{2}+2 m k)} \\ \\ & =\sqrt{\dfrac{\Sigma x _ i^{2}}{n}-m^{2}+\dfrac{2 k \Sigma x _ {i}}{n}-2 m k} \\ \\ & =\sqrt{\dfrac{\Sigma x _ i^{2}}{n}-m^{2}+2 k m-2 m k} \quad \left[\because \quad \dfrac{\Sigma x _ {i}}{n}=m\right] \\ \\ & =\sqrt{\dfrac{\Sigma x _ i^{2}}{n}-m^{2}} \\ \\ & =s \end{aligned} $

$\therefore \quad \text{Option (a) is correct.}$

Solution

(c) Here,

$ \begin{aligned} m & =\dfrac{\Sigma x _ {i}}{5}, s=\sqrt{\dfrac{\Sigma x _ i^{2}}{5}-\Big(\dfrac{\Sigma x _ {i}}{5}\Big)^2} \\ \\ SD & =\sqrt{\dfrac{k^{2} \Sigma x _ i^{2}}{5}-\Big(\dfrac{k \Sigma x _ {i}{ }}{5}\Big)^{2}} \\ \\ & =\sqrt{\dfrac{k^{2} \Sigma x _ i^{2}}{5}-k^{2} \Big(\dfrac{\Sigma x _ i}{5})^{2}} \\ \\ & =k\sqrt{\dfrac{\Sigma x _ i^{2}}{5}-{\left(\dfrac{\Sigma x _ {i}}{5}\right)^{2}}}=k s \end{aligned} $

$\therefore \quad \text{Option (c) is correct.}$

Solution

(a) Given, $w _ {i}=x _ {i}+k, \bar x _ i=48, s x _ {i}=12, w _ {i}=55$ and $s w _ {i}=15$

Then, $\quad \bar w _ i=\bar x _ i+k $

[where, $\bar w _ i$ is mean $w _ {i}$ ’s and $\bar x _ i$ is mean of $x _ {i}{ }^{\prime} s$ ]

$\Rightarrow \quad 55=48+k \qquad…(i)$

Now, $\quad \text{SD of } w _ {i}= \text{ SD of } x _ {i}$

$\Rightarrow \quad 15=12$

$\Rightarrow \quad l=\dfrac{15}{12} =1.25\qquad…(ii) $

From Eqs. (i) and (ii),

$k=55-1.25 \times 48=-5$

$\therefore \quad \text{Option (a) is correct.}$

Solution

(d) We know that, SD of first $n$ natural number $=\sqrt{\dfrac{n^{2}-1}{12}}$

$\therefore \quad $ SD of first 10 natural numbers $=\sqrt{\dfrac{(10)^{2}-1}{12}}$

$ \hspace{4.6cm}=\sqrt{\dfrac{100-1}{12}}=\sqrt{\dfrac{99}{12}}$

$\hspace{4.6cm}=\sqrt{8.25}=2.87 $

• Option (a) 5.5 is incorrect because the standard deviation of the first 10 natural numbers is calculated using the formula $\sqrt{\dfrac{n^{2}-1}{12}}$, which does not yield 5.5.

• Option (b) 3.87 is incorrect because, when applying the formula $\sqrt{\dfrac{n^{2}-1}{12}}$ to the first 10 natural numbers, the result is not 3.87.

• Option (c) 2.97 is incorrect because the correct calculation using the formula $\sqrt{\dfrac{n^{2}-1}{12}}$ for the first 10 natural numbers results in 2.87, not 2.97.

Solution

(d) Given numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10.

If 1 is added to each number, then observations will be 2, 3, 4, 5, 6, 7, 8, 9, 10 and 11 .

$ \begin{aligned} \therefore \quad \sum x _ {i}& =2+3+4+\ldots+11 \\ \\ & =\dfrac{10}{2}[2 \times 2+9 \times 1]=5[4+9]=65 \\ \\ \Sigma x _ i^{2}&=2^{2}+3^{2}+4^{2}+5^{2}+\ldots+11^{2} \\ \\ & =(1^{2}+2^{2}+3^{2}+\ldots+11^{2})-(1^{2}) \\ \\ & =\dfrac{11 \times 12 \times 23}{6}-1 \\ \\ & =\dfrac{11 \times 12 \times 23-6}{6}=505 \end{aligned} $

$ \begin{aligned} \therefore \quad s^{2} & =\dfrac{\Sigma x _ i^{2}}{n}-{\Big(\dfrac{\Sigma x _ {i}}{n}\Big)}^{2}=\dfrac{505}{10}-\Big(\dfrac{65}{10}\Big)^{2} \\ \\ & =50.5-(6.5)^{2} \\ \\ & =50.5-42.25 \\ \\ & =8.25 \end{aligned} $

$\therefore \quad \text{Option (d) is correct.}$

Solution

(a) Since, the first 10 positive integers are1, 2, 3, 4, 5, 6, 7, 8, 9 and 10.

On multiplying each number by -1 , we get

$ -1,-2,-3,-4,-5,-6,-7,-8,-9,-10 $

On adding 1 in each number, we get

$\therefore \quad 0,-1,-2,-3,-4,-5,-6,-7,-8,-9 $

$0+(-1)+(-2)+(-3)………+(-8)+(-9)=-\dfrac{9 \times 10}{2}=-45 $

$\text { and } \quad \Sigma x _ i^{2}=0^{2}+(-1)^{2}+(-2)^{2}+\ldots+(-9)^{2}$

$\hspace{1.7cm}=\dfrac{9 \times 10 \times 19}{6}=285 $

$\therefore \quad SD =\sqrt{\dfrac{285}{10}-\Big(\dfrac{-45}{10}\Big)^2}=\sqrt{\dfrac{285}{10}-\dfrac{2025}{100}} $

$\hspace{1.3cm}=\sqrt{\dfrac{2850-2025}{100}}=\sqrt{8.25}$

$\text { Now, } \quad \text { variance } =(SD)^{2}=(\sqrt{8.25})^{2}=8.25$

$\therefore \quad \text{Option (a) is correct.}$

Solution

(d)

$ \text { Variance }=\dfrac{\Sigma x _ i^{2}}{n}-\Big(\dfrac{\Sigma x _ {i}}{n}\Big)^2 $

$\hspace{1.7cm}=\dfrac{18000}{60}-\Big(\dfrac{960}{60}\Big)^{2}=300-256=44$

$\therefore \quad \text{Option (d) is correct.}$

Solution

(a) $\text{Here},\quad CV _ 1=50, CV _ 2=60, \bar x _ 1=30 \text { and } \bar x _ 2=25$

$ \therefore \quad CV _ 1=\dfrac{\sigma _ 1}{\bar x _ 1} \times 100 $

$\Rightarrow\quad 50=\dfrac{\sigma _ 1}{30} \times 100 $

$\therefore \quad \sigma _ 1=\dfrac{30 \times 50}{100}=15 \text { and } CV _ 2=\dfrac{\sigma _ 2}{\bar x _ 2} \times 100 $

$\Rightarrow \quad 60=\dfrac{\sigma _ 2}{25} \times 100 $

$\therefore \quad \sigma _ 2=\dfrac{60 \times 25}{100}=15 $

$\text { Now, }\quad \sigma _ 1-\sigma _ 2=15-15=0 $

$\therefore \quad \text{Option (a) is correct.}$

Solution

(a) Given, $\quad\sigma _ {C}=5 $

$\Rightarrow\quad \dfrac{5}{9}(F-32)=C$

$ \begin{aligned} F & =\dfrac{9 C}{5}+32 \\ \\ \sigma _ {F} & =\dfrac{9}{5} \sigma _ {C}=\dfrac{9}{5} \times 5=9 \end{aligned} $

Here, $\quad \sigma _ F^{2}=(9)^{2}=81 $

$\therefore \quad \text{Option (a) is correct.}$

Solution

$C V=\dfrac{S D}{\text { Mean }} \times 100$

Solution

If $\bar{x}$ is the mean of $n$ values of $x$, then $\sum _ {i=1}^{n}(x _ {i}-\bar{x})=0$ and if $a$ has any value other than $\bar{x}$, then $\sum _ {i=1}^{n}(x _ {i}-\bar{x})^{2}$ is less than $\sum(x _ {i}-a)^{2}$

Solution

If the variance of a data is 121 .

Then,

$ \begin{aligned} S D & =\sqrt{\text { Variance }} \\ \\ & =\sqrt{121}=11 \end{aligned} $

Solution

The standard deviation of a data is independent of any change in origin but is dependent of change of scale.

Solution

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

Solution

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

Solution

The SD is greater than or equal to the mean deviation taken from the arithmetic mean.