Solution

Solution

$ \begin{matrix} \text { Now, } & M_{e}=\frac{20+1}{2} \text { th item }=\frac{21}{2}=10.5 \text { th item } \\ \therefore & M_{e}=12 \\ \therefore & M D=\frac{\sum f_{i} d_{i}}{\sum f_{i}}=\frac{25}{20}=1.25 \end{matrix} $

Solution

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

$ \begin{aligned} \text { Mean } \bar{x} & =\frac{1+2+3+\ldots+n}{n}=\frac{n(n+1)}{2 n}=\frac{n+1}{2} \\ \therefore \quad \text { MD } & =\frac{|1-\frac{n+1}{2}|+|2-\frac{n+1}{2}|+|3-\frac{n+1}{2}|+\ldots+|n-\frac{n+1}{2}|}{n} \\ & =\frac{+|\frac{n+1}{2}-\frac{n+1}{2}|+|\frac{n+3}{2}-\frac{n+1}{2}|+\ldots+|\frac{n+1}{2}|+|2-\frac{n+1}{2}|+\ldots+|\frac{n-1}{2}-\frac{n+1}{2}|+|n-\frac{n+1}{2}|}{n} \\ & =\frac{2}{n} 1+2+\ldots+\frac{n-3}{2}+\frac{n-1}{2} \quad \frac{n-1}{2} \text { terms } \\ & =\frac{2}{n} \frac{\frac{n-1}{2} \frac{n-1}{2}+1}{2} \quad \because \text { sum of first } n \text { natural numbers }=\frac{n(n+1)}{2} \\ & =\frac{2}{n} \cdot \frac{1}{2} \frac{n-1}{2} \frac{n+1}{2}=\frac{1}{n} \frac{n^{2}-1}{4}=\frac{n^{2}-1}{4 n} \end{aligned} $

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}=\frac{1+2+3+\ldots+n}{n}=\frac{n(n+1)}{2 n}=\frac{n+1}{2} \\ & MD=\frac{1}{n}|1-\frac{n+1}{2}|+|2-\frac{n+1}{2}|+|3-\frac{n+1}{2}|+|\frac{n-2}{2}-\frac{n+1}{2}|+|\frac{n}{2}-\frac{n+1}{2}| \\ & +|\frac{n+2}{2}-\frac{n+1}{2}|+\ldots+|n-\frac{n+1}{2}| \\ & =\frac{1}{n}|\frac{1-n}{2}|+|\frac{3-n}{2}|+|\frac{5-n}{2}|+\ldots .+|\frac{-3}{2}|+|\frac{1}{2}|+\ldots+|\frac{n-1}{2}| \\ & =\frac{2}{n} \frac{1}{2}+\frac{3}{2}+\ldots .+\frac{n-1}{2} \quad \frac{n}{2} \text { terms } \\ & =\frac{1}{n} \cdot \frac{n^{2}}{2} \quad[\because \text { sum of first } n \text { natural numbers }=n^{2}] \\ & =\frac{1}{n} \cdot \frac{n^{2}}{4}=\frac{n}{4} \end{aligned} $

Solution

$ \text { Now, } \quad \begin{aligned} \Sigma x_{i} & =1+2+3+4+\ldots+n=\frac{n(n+1)}{2} \\ \text { and } \quad \sum x_i^{2} & =1^{2}+2^{2}+3^{2}+\ldots+n^{2}=\frac{n(n+1)(2 n+1)}{6} \\ \therefore \quad & =\sqrt{\frac{\sum x_i^{2}}{N}-\frac{\sum x_{i}{ }^{2}}{N}} \\ & =\sqrt{\frac{n(n+1)(2 n+1)}{6 n}-\frac{n^{2}(n+1)^{2}}{4 n^{2}}} \\ & =\sqrt{\frac{(n+1)(2 n+1)}{6}-\frac{(n+1)^{2}}{4}} \\ & =\sqrt{\frac{2(2 n^{2}+3 n+1)-3(n^{2}+2 n+1)}{12}} \\ & =\sqrt{\frac{4 n^{2}+6 n+2-3 n^{2}-6 n-3}{12}} \\ & =\sqrt{\frac{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} & =\frac{\sum x_i^{2}}{25}-(18.2)^{2} \\ 25^{2} & =\frac{\sum x_i^{2}}{25}-331.24 \\ 1.24 & =\frac{\sum x_i^{2}}{25} \\ \Sigma x_i^{2} & =25 \times(10.5625) \\ & =25 \times 341.8025 \\ & =8545.0625 \end{aligned} $

$ \begin{matrix} \therefore & \sigma_1^{2}=\frac{\Sigma x_i^{2}}{25}-(18.2)^{2} \\ \Rightarrow & (3.25)^{2}=\frac{\Sigma x_i^{2}}{25}-331.24 \\ \Rightarrow & 10.5625+331.24=\frac{\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{\frac{14069.0625}{40}-\frac{734}40^{2}} \\ & =\sqrt{351.726-(18.35)^{2}} \\ & =\sqrt{351.726-336.7225} \\ & =\sqrt{15.0035}=3.87 \end{aligned} $

Solution

Let

$ x_{i}, i=1,2,3 \ldots, n_1 \text { and } y_{j}, j=1,2,3, \ldots, n_2 $

$ \therefore \bar x_1=\frac{1}{n_1} \sum_{i=1}^{n_1} x_{i} \text { and } \bar x_2=\frac{1}{n_2} \sum_{j=1}^{n_2} y_{j} \\ $

$\Rightarrow \sigma _1^2= \frac{1}{n_1} \sum _{i=1}^{n_1} (x_i - \bar x_1)^{2} $

and

$\sigma _2^{2}= \frac{1}{n_2} \sum _{j=1}^{n}(y_j - \bar x_2)^{2} $

Now, mean $\bar{x}$ of the given series is given by

$ \bar {x}= \frac{1}{n_1+n_2} \sum _{i=1}^{n_1} x_i + \sum _{j=1}^{n_2} y_j = \frac{n_1 \bar x_1+n_2 \bar x_2}{n_1+n_2} $

The variance $\sigma^{2}$ of the combined series is given by

$ \sigma^{2} =\Big[\frac{1}{n_1+n_2} \sum_{i=1}^{n_1}(x_{i}-\bar{x})^{2}+\sum_{j=1}^{n_2}(y_{j}-\bar{x})^{2}\Big] \\ $

Now,

$\sum_{i=1}^{n_1}(x_{i}-\bar{x})^{2} = \sum _{i=1}^{n_1} (x_i - \bar x_j - \bar x)^2 \\ $

But $\quad \sum _{i=1}^{n_1}(x_i - \bar x_i)=0 $

[algebraic sum of the deviation of values of first series from their mean is zero]

Also,

$ \sum _{i=1}^{n_1} (x_i-\bar{x})^2= n_1 s_1^2+n_1(\bar x_1-\bar{x})^2=n_1 s_1^2+n_1 d_1^2 $

Where,

$ d_1=(\bar x_1-\bar{x}) $

Similarly, $\quad \sum_{j=1}^{n_2}(y_{j}-\bar{x})^{2}=\sum _{j=1}^{n_2}(y_j-\bar x_i+\bar x_i-\bar{x})^2=n_2 s_2^2+n_2 d_2^2$

where,

$ d_2=\bar x_2-\bar x $

Combined SD

$ \sigma=\sqrt{\frac{[n_1(s_1^{2}+d_1^{2})+n_2(s_2^{2}+d_2^{2})]}{n_1+n_2}} $

where,

$d_1=\bar x_1-\bar x=\bar x_1-\frac{n_1 \bar x_1+n_2 \bar x_2}{n_1+n_2}=\frac{n_2(\bar x_1-\bar x_2)}{n_1+n_2} $

and

$d_2=\bar x_2-\bar{x}=\bar x_2-\frac{n_1 \bar x_1+n_2 \bar x_2}{n_1+n_2}=\frac{n_1(\bar x_2-\bar x_1)}{n_1+n_2} $

$\therefore \quad \sigma^{2}=\frac{1}{n_1+n_2} n_1 s_1^{2}+n_2 s_2^{2}+\frac{n_1 n_2(\bar x_1-\bar x_2)^{2}}{(n_1+n_2)^{2}}+\frac{n_2 n_1(\bar x_2-\bar x_1)^{2}}{(n_1+n_2)^{2}}$

Also,

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

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, $\sigma=\sqrt{\frac{n_1 s_1^{2}+n_2 s_2^{2}}{n_1+n_2}+\frac{n_1 n_2(\bar x_1-\bar x_2)^{2}}{(n_1+n_2)^{2}}}$

$ \begin{aligned} & =\sqrt{\frac{20 \times(5)^{2}+20 \times(5)^{2}}{20+20}+\frac{20 \times 20(17-22)^{2}}{(20+20)^{2}}} \\ & =\sqrt{\frac{1000}{40}+\frac{400 \times 25}{1600}}=\sqrt{25+\frac{25}{4}}=\sqrt{\frac{125}{4}}=\sqrt{31.25}=5.59 \end{aligned} $

Solution

$ \begin{matrix} \therefore & \sigma^{2}=\frac{\Sigma f_{i} x_i^{2}}{n}-\frac{\Sigma f_{i} x_{i}}{n} \\ \Rightarrow & 160=\frac{92 A^{2}}{7}-\frac{22 A^{2}}{7} \\ \Rightarrow & 160=\frac{92 A^{2}}{7}-\frac{484 A^{2}}{49} \\ \Rightarrow & 160=(644-484) \frac{A^{2}}{49} \\ \Rightarrow & 160=\frac{160 A^{2}}{49} \Rightarrow A^{2}=49 \\ \therefore & A=7 \end{matrix} $

Solution

Solution

$\therefore$ Sum of frequencies,

$\begin{aligned} & x-2+x+x^2+(x+1)^2+2 x+x+1=60 \\ & \Rightarrow \quad 2 x-2+x^2+x^2+1+2 x+2 x+x+1=60 \\ & \Rightarrow \quad 2 x^2+7 x=60 \\ & \Rightarrow \quad 2 x^2+7 x-60=0 \\ & \Rightarrow \quad 2 x^2+15 x-8 x-60=0 \\ & \Rightarrow \quad x(2 x+15)-4(2 x+15)=0 \\ & \Rightarrow \quad(2 x+15)(x-4)=0 \ & \Rightarrow \\ & \Rightarrow x=-\frac{15}{2}, 4 \\ & \Rightarrow x=-\frac{15}{2} \ & \text { [inaddmisible] }\left[\because x \in /^{+}\right] \\ & \end{aligned}$

$\begin{array}{|c|c|c|c|c|} \hline x_{i} & f_{i} & d_{i}=x_{i}-3 & f_{i} d_{i} & f_{i} d_i^2 \\ \hline 0 & 2 & -3 & -6 & 18 \\ 1 & 4 & -2 & -8 & 16 \\ 2 & 16 & -1 & -16 & 16 \\ A=3 & 25 & 0 & 0 & 0 \\ 4 & 8 & 1 & 8 & 8 \\ 5 & 5 & 2 & 10 & 20 \\ \hline \text { Total } & \Sigma f_i=60 & & \Sigma f_i d_i=-12 & \Sigma f_i d_i^2=\mathbf{7 8} \\ \hline \end{array}$

$\begin{aligned} \text { Mean } & =A+\frac{\sum f_i d_i}{\Sigma f_i}=3+\frac{-12}{60}=2.8 \\ \sigma & =\sqrt{\frac{\sum f_i d_i^2}{\Sigma f_i}-\frac{\sum f_i d_i^2}{\Sigma f_i}}=\sqrt{\frac{78}{60}-\frac{-12^2}{60}} \\ & =\sqrt{1.3-0.04}=\sqrt{1.26}=1.12\end{aligned}$

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{\frac{n_1 s_1{ }^{2}+n_2 s_2{ }^{2}}{n_1+n_2}+\frac{n_1 n_2(\bar x_1-\bar x_2)^{2}}{(n_1+n_2)^{2}}} \\ $

$=\sqrt{\frac{60 \times(8)^{2}+80 \times(7)^{2}}{60+80}+\frac{60 \times 80(650-660)^{2}}{(60+80)^{2}}} \\ $

$=\sqrt{\frac{6 \times 64+8 \times 49}{14}+\frac{60 \times 80 \times 100}{140 \times 140}} \\ $

$=\sqrt{\frac{192+196}{7}+\frac{1200}{49}}=\sqrt{\frac{388}{7}+\frac{1200}{49}} \\ $

$=\sqrt{\frac{2716+1200}{49}}=\sqrt{\frac{3916}{49}}=\frac{62.58}{7}=8.9 $

Solution

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

$ \begin{aligned} & \therefore \quad \frac{\Sigma x_{i}}{100}=50 \\ & \Rightarrow \quad \Sigma x_{i}=5000 \\ & \text { and } \quad \sigma^{2}=\frac{\Sigma f_{i} x_i^{2}}{\Sigma f_{i}}-\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}} \\ & \Rightarrow \quad(4)^{2}=\frac{\Sigma f_{i} x_i^{2}}{100}-(50)^{2} \\ & \Rightarrow \quad 16=\frac{\Sigma f_{i} x_i^{2}}{100}-2500 \\ & \Rightarrow \quad \frac{\Sigma f_{i} x_i^{2}}{100}=16+2500 \Rightarrow 2516 \\ & \therefore \quad \sum f_{i} x_i^{2}=251600 \end{aligned} $

Solution

Given,

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

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

Solution

$\therefore \quad$ Mean $=\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}=\frac{66.5}{16}=4.15$

$ \begin{aligned} \text { variance } & =\sigma^{2}=\frac{\Sigma f_{i} x_i^{2}}{\Sigma f_{i}}-\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}} \\ & =\frac{340.25}{16}-(4.15)^{2} \\ & =21.2656-17.2225=4.043 \end{aligned} $

Solution

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

$\begin{aligned} & \therefore \quad \text { Mean }=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{230}{25}=9.2 \\ & \text { and } \quad \text { mean deviation }=\frac{\Sigma f d_i}{\Sigma f_i}=\frac{96}{25}=3.84 \\ & \end{aligned}$

Solution

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

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

Solution

$ \begin{aligned} & \therefore \quad \text { Mean } \bar{x}=\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}=\frac{239}{40}=5.975 \approx 6 \\ & \text { and } \\ & \begin{aligned} \sigma & =\sqrt{\frac{\Sigma f_{i} d_i^{2}}{\Sigma f_{i}}-\frac{\Sigma f_{i} d_i^{2}}{\Sigma f_{i}}}=\sqrt{\frac{325}{40}-\frac{-1}40^{2}} \\ & =\sqrt{8.125-0.000625}=\sqrt{8.124375}=2.85 \end{aligned} \end{aligned} $

Solution

$\Sigma f_{i}=70$ $\Sigma f_{i} d_{i}=$
-38 $\Sigma f_{i} d_i^{2}=102$ | $\frac{T f_{i}}{\sum f_{i}}$

Now,
$=1.4571-0.2916=1.1655 \\ \sigma =\sqrt{1.1655}=1.08 g$

Solution

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

Solution

For Ravi,

Now,

$ \begin{aligned} \sigma & =\sqrt{\frac{\Sigma d^{2} i}{n}-\frac{\Sigma d_{i}{ }^{2}}{n}} \\ & =\sqrt{\frac{1699}{10}-\frac{-14^{2}}{10}}=\sqrt{169.9-0.0196} \\ & =\sqrt{169.88}=13.03 \\ \bar{x} & =A+\frac{\Sigma d_{i}}{\Sigma f_{i}}=45-\frac{14}{10}=43.6 \end{aligned} $

For Hashina,

$ \begin{aligned} & \because \quad \text { Mean }=55 \\ & \therefore \quad \sigma=\sqrt{\frac{6368}{10}}=\sqrt{636.8}=25.2 \\ & CV=\frac{\sigma}{\bar{x}} \times 100=\frac{13.03}{43.6} \times 100=29.88 \\ & CV=\frac{\sigma}{\bar{x}} \times 100=\frac{25.2}{55} \times 100=45.89 \end{aligned} $

Hence, Hashina is more consistent and intelligent.

Solution

Given,

$ \begin{aligned} & n=100, \bar{x}=40, \sigma=10 \text { and } \bar{x}=40 \\ & \therefore \frac{\sum x_{i}}{n}=40 \\ \Rightarrow & \frac{\Sigma x_{i}}{100}=40 \\ & \Rightarrow \Sigma x_{i}=4000 \\ & \text { Corrected } \Sigma x_{i}=4000-30-70+3+27 \\ & =4030-100=3930 \\ & \text { Corrected mean }=\frac{2930}{100}=39.3 \end{aligned} $

$ \begin{aligned} \sigma^{2} & =\frac{\Sigma x_i^{2}}{n}-(40)^{2} \\ 100 & =\frac{\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{\frac{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

Given,

$ \begin{aligned} & n=10, \bar{x}=45 \text { and } \sigma^{2}=16 \\ & \bar{x}=45 \Rightarrow \frac{\Sigma x_{i}}{n}=45 \\ & \frac{\Sigma x_{i}}{10}=45 \Rightarrow \Sigma x_{i}=450 \end{aligned} $

$ \therefore $

Corrected $\Sigma x_{i}=450-52+25=423$

$\therefore$

$\bar{x}=\frac{423}{10}=42.3$

$\Rightarrow$

$\sigma^{2}=\frac{\Sigma x_i^{2}}{n}-\frac{\Sigma x_{i}}{n}$

$\Rightarrow$

$16=\frac{\Sigma x_i^{2}}{10}-(45)^{2}$

$\Rightarrow$

$\Sigma x_i^{2}=10(2025+16)$

$\Rightarrow$

$\Sigma x_i^{2}=20410$

$\therefore$

Corrected $\Sigma x_i^{2}=20410-(52)^{2}+(25)^{2}=18331$

and

corrected $\sigma^{2}=\frac{18331}{10}-(42.3)^{2}=43.81$

Solution

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

Solution

(b) $MD=\frac{1}{n} \sum_{i=1}^{n}|x_{i}-\bar{x}|$

Solution

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

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

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 $=\frac{9+1}{2}$ term $=5$ th term

Solution

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

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

$\begin{aligned} \therefore \quad \mathrm{SD} & =\sigma=\sqrt{\frac{\Sigma x_i^2}{N}-{\frac{\Sigma x_i}{N}}^2}=\sqrt{\frac{619}{7}-\frac{63}{7}^2} \\ & =\sqrt{\frac{7 \times 619-3969}{49}} \\ & =\sqrt{\frac{4333-3969}{49}} \\ & =\sqrt{\frac{364}{49}}=\sqrt{\frac{52}{7}}\end{aligned}$

Solution

(c) SD is given by

$ \sigma=\sqrt{\frac{\sum(x_{i}-\bar{x})^{2}}{n}} $

Solution

(c) Given,

$ \begin{aligned} \bar{x}=50, n & =100 \text { and } \sigma=5 \\ \Sigma x_i^{2} & =? \\ \bar{x} & =\frac{\Sigma x_{i}}{n} \end{aligned} $

$ \begin{matrix} \Rightarrow & 50=\frac{\Sigma x_{i}}{100} \\ \therefore & \Sigma x_{i}=50 \times 100=5000 \end{matrix} $

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

Solution

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

$ \begin{aligned} & \text { Mean }=m=\frac{a+b+c+d+e}{5} \\ & \Sigma x_{i}=a+b+c+d+e=5 m \\ & \text { Now, } \\ & \text { mean }=\frac{a+k+b+k+c+k+d+k+e+k}{5} \\ & =\frac{(a+b+c+d+e)+5 k}{5}=m+k \\ & \therefore \quad SD=\sqrt{\frac{\sum(x_{i}+k)^{2}}{n}-(m+k)^{2}} \\ & =\sqrt{\frac{\sum(x_i^{2}+k^{2}+2 k x_{i})}{n}-(m^{2}+k^{2}+2 m k)} \\ & =\sqrt{\frac{\Sigma x_i^{2}}{n}-m^{2}+\frac{2 k \Sigma x_{i}}{n}-2 m k} \\ & =\sqrt{\frac{\Sigma x_i^{2}}{n}-m^{2}+2 k m-2 m k} \quad \because \frac{\Sigma x_{i}}{n}=m \\ & =\sqrt{\frac{\Sigma x_i^{2}}{n}-m^{2}} \\ & =s \end{aligned} $

Solution

(c) Here,

$ \begin{aligned} m & =\frac{\Sigma x_{i}}{5}, s=\sqrt{\frac{\Sigma x_i^{2}}{5}-\frac{\Sigma x_{i}}{5}} \\ SD & =\sqrt{\frac{k^{2} \Sigma x_i^{2}}{5}-\frac{k \Sigma x_{i}{ }^{2}}{5}} \\ & =\sqrt{\frac{k^{2} \Sigma x_i^{2}}{5}-k^{2} \frac{\Sigma x_i^{2}}{5}}=\sqrt{\frac{\Sigma x_i^{2}}{5}-{\frac{\Sigma x_{i}}{5}}^{2}}=k s \end{aligned} $

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 $x_{i}{ }^{\prime} s$ ]

$\Rightarrow \quad 55=48+k$

Now, $\quad$ SD of $w_{i}=$ SD of $x_{i}$

$\Rightarrow \quad 15=12$

$\Rightarrow \quad l=\frac{15}{12}$

$ =1.25 $

From Eqs. (i) and (ii),

$ k=55-1.25 \times 48 $

$ =-5 $

Solution

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

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

$ =\sqrt{\frac{100-1}{12}}=\sqrt{\frac{99}{12}}=\sqrt{8.25}=2.87 $

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 \\ & =\frac{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}) \\ & =\frac{11 \times 12 \times 23}{6}-1 \\ & =\frac{11 \times 12 \times 23-6}{6}=505 \end{aligned} $

$ \begin{aligned} \therefore \quad s^{2} & =\frac{\Sigma x_i^{2}}{n}-{\frac{\Sigma x_{i}}{n}}^{2}=\frac{505}{10}-\frac{65}10^{2} \\ & =50.5-(6.5)^{2} \\ & =50.5-42.25 \\ & =8.25 \end{aligned} $

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

$ \begin{matrix} \therefore \quad & 0,-1,-2,-3,-4,-5,-6,-7,-8,-9 \\ & =-\frac{9 \times 10}{2} \\ & =-45 \\ \text { and } \quad & \Sigma x_i^{2}=0^{2}+(-1)^{2}+(-2)^{2}+\ldots+(-9)^{2} \\ & =\frac{9 \times 10 \times 19}{6} \\ & =285 \\ \therefore \quad SD & =\sqrt{\frac{285}{10}-\frac{-45}{10}}=\sqrt{\frac{285}{10}-\frac{2025}{100}} \\ & =\sqrt{\frac{2850-2025}{100}}=\sqrt{8.25} \\ \text { Now, } \quad \text { variance } & =(SD)^{2}=(\sqrt{8.25})^{2}=8.25 \end{matrix} $

Solution

(d)

$ \begin{aligned} & \text { Variance }=\frac{\Sigma x_i^{2}}{n}-\frac{\Sigma x_{i}}{n} \\ & =\frac{18000}{60}-\frac{960}60^{2}=300-256=44 \end{aligned} $

Solution

(a) Here

$ CV_1=50, CV_2=60, \bar x_1=30 \text { and } \bar x_2=25 $

$ \therefore CV_1=\frac{\sigma _1}{\bar x_1} \times 100 \Rightarrow 50=\frac{\sigma _1}{30} \times 100 \\ $

$\therefore \sigma _1=\frac{30 \times 50}{100}=15 \text { and } CV_2=\frac{\sigma _2}{\bar x_2} \times 100 \\ $

$\Rightarrow 60=\frac{\sigma_2}{25} \times 100 \\ $

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

$\text { Now, } \sigma _1-\sigma _2=15-15=0 $

Solution

(a) Given,

$ \sigma_{C}=5 \Rightarrow \frac{5}{9}(F-32)=C $

$ \begin{aligned} F & =\frac{9 C}{5}+32 \\ \sigma_{F} & =\frac{9}{5} \sigma_{C}=\frac{9}{5} \times 5=9 \end{aligned} $

Here,

$ \sigma_F^{2}=(9)^{2}=81 $

Solution

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

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.