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.

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

• 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): $\frac{1}{n} \sum_{i=1}^{n}(x_{i}-\bar{x})^{2}$ 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 } & =\frac{1357+1090+1666+1494+1623}{5} \\ & =\frac{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 $=\frac{9+1}{2}$ term $=5$ th term

• Option (a) 9 is incorrect because the sum of the absolute deviations from the median is 114, and dividing this by the number of students (9) gives a mean deviation of 12.67, not 9.

• Option (b) 10.5 is incorrect because, as calculated, the mean deviation from the median is 12.67, not 10.5.

• Option (d) 14.76 is incorrect because the correct mean deviation from the median, based on the given data, is 12.67, not 14.76.

Solution

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

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

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

• Option (b) $\frac{52}{7}$ is incorrect because it represents the variance, not the standard deviation. The standard deviation is the square root of the variance.

• Option (c) $\sqrt{6}$ is incorrect because it does not match the calculated value of the standard deviation, which is $\sqrt{\frac{52}{7}}$.

• Option (d) 6 is incorrect because it is not derived from the given data set and does not match the calculated standard deviation.

Solution

(c) SD is given by

$ \sigma=\sqrt{\frac{\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) $\frac{\Sigma(x_{i}-\bar{x})^{2}}{n}$ is incorrect because it represents the variance, not the standard deviation. The standard deviation is the square root of the variance.

• Option (d) $\sqrt{\frac{\sum x_i^{2}}{n}+\bar x^{-2}}$ 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,

$ \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} $

• Option (a) 50000: This option is incorrect because it significantly underestimates the sum of the squares of the observations. Given the mean and standard deviation, the sum of the squares must be much larger.

• Option (b) 250000: This option is incorrect because it is close to the correct answer but still slightly underestimates the sum of the squares. The correct calculation shows that the sum of the squares is 252500.

• Option (d) 255000: This option is incorrect because it overestimates the sum of the squares of the observations. The correct calculation shows that the sum of the squares is 252500, not 255000.

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} $

• Option (b) ( ks ): This option is incorrect because adding a constant ( k ) to each observation does not affect the spread or dispersion of the data, which is what the standard deviation measures. The standard deviation remains the same regardless of the constant added.

• Option (c) ( s+k ): This option is incorrect because the standard deviation is a measure of the spread of the data around the mean, and adding a constant ( k ) to each observation shifts the mean but does not change the spread. Therefore, the standard deviation does not increase by ( k ).

• Option (d) ( \frac{s}{k} ): This option is incorrect because dividing the standard deviation by the constant ( k ) implies that the spread of the data decreases when a constant is added to each observation. However, adding a constant does not affect the spread of the data, so the standard deviation remains unchanged.

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} $

• Option (a) ( k + s ): This option is incorrect because the standard deviation of a set of observations scaled by a factor ( k ) is not simply the original standard deviation plus ( k ). The standard deviation measures the spread of the data, and scaling the data by ( k ) scales the spread by ( k ), not by adding ( k ).

• Option (b) ( \frac{s}{k} ): This option is incorrect because dividing the original standard deviation ( s ) by ( k ) would imply that the spread of the data decreases when the data is scaled by ( k ). However, scaling the data by ( k ) actually increases the spread by a factor of ( k ), not decreases it.

• Option (d) ( s ): This option is incorrect because it suggests that the standard deviation remains unchanged when the data is scaled by ( k ). However, scaling the data by ( k ) changes the spread of the data, and thus the standard deviation must also change by the same factor ( k ).

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 $

• Option (b) $l=-1.25, k=5$:

  • The value of $l$ should be positive because the standard deviation of $w_i$ is greater than the standard deviation of $x_i$. A negative $l$ would imply a reduction in the spread, which contradicts the given standard deviations.
  • The value of $k$ should be negative to satisfy the equation $55 = 48 + k$. Here, $k = 5$ does not satisfy this equation.

• Option (c) $l=2.5, k=-5$:

  • The value of $l$ should be 1.25, not 2.5, because the standard deviation of $w_i$ is 15 and the standard deviation of $x_i$ is 12. The correct $l$ is calculated as $\frac{15}{12} = 1.25$.
  • Although $k = -5$ is correct, the incorrect value of $l$ makes this option invalid.

• Option (d) $l=2.5, k=5$:

  • The value of $l$ should be 1.25, not 2.5, for the same reason as in option (c).
  • The value of $k$ should be negative to satisfy the equation $55 = 48 + k$. Here, $k = 5$ does not satisfy this equation.

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 $

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

• Option (b) 3.87 is incorrect because, when applying the formula $\sqrt{\frac{n^{2}-1}{12}}$ 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 $\sqrt{\frac{n^{2}-1}{12}}$ 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 \\ & =\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} $

• Option (a) 6.5: This option is incorrect because it represents the mean of the original numbers (1 to 10) rather than the variance. The variance calculation involves squaring the deviations from the mean, which results in a different value.

• Option (b) 2.87: This option is incorrect because it is not derived from the correct formula for variance. The variance of a set of numbers is calculated using the sum of the squared deviations from the mean, divided by the number of observations. The value 2.87 does not match the result of this calculation for the given set of numbers.

• Option (c) 3.87: This option is incorrect because it does not match the correct variance calculation for the given set of numbers. The correct variance is obtained by subtracting the square of the mean from the mean of the squares of the numbers, which results in 8.25, not 3.87.

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} $

• Option (b) 6.5: This option is incorrect because the variance calculation involves squaring the deviations from the mean. The correct variance, as derived, is 8.25, not 6.5. The steps in the solution show that the sum of the squared deviations divided by the number of observations results in 8.25.

• Option (c) 3.87: This option is incorrect because it significantly underestimates the variance. The correct variance calculation involves summing the squares of the deviations from the mean and then dividing by the number of observations, which results in 8.25, not 3.87.

• Option (d) 2.87: This option is incorrect because it also underestimates the variance. The correct variance, as shown in the solution, is derived from the sum of the squared deviations from the mean, which results in 8.25, not 2.87.

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} $

• Option (a) 6.63: This value is incorrect because it is significantly lower than the calculated variance. The variance formula used, (\frac{\Sigma x_i^{2}}{n} - \left(\frac{\Sigma x_{i}}{n}\right)^2), results in a value of 44, not 6.63. This option might be a result of a miscalculation or misunderstanding of the formula.

• Option (b) 16: This value is incorrect because it does not match the calculated variance. The correct calculation, (\frac{18000}{60} - \left(\frac{960}{60}\right)^2), results in 44. The value 16 could be a result of an incorrect intermediate step or a different formula.

• Option (c) 22: This value is incorrect because it is half of the correct variance. The correct variance calculation yields 44. This option might be a result of an error in the final subtraction step or a misinterpretation of the formula.

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 $

• Option (b) is incorrect because the calculated standard deviations for both distributions are equal (15), so their difference cannot be 1.
• Option (c) is incorrect because the calculated standard deviations for both distributions are equal (15), so their difference cannot be 1.5.
• Option (d) is incorrect because the calculated standard deviations for both distributions are equal (15), so their difference cannot be 2.5.

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 $

• Option (b) 57 is incorrect because the variance is calculated as the square of the standard deviation. Given the standard deviation in Fahrenheit is 9, the variance should be (9^2 = 81), not 57.

• Option (c) 36 is incorrect because, similar to option (b), the variance is the square of the standard deviation. Since the standard deviation in Fahrenheit is 9, the variance should be (9^2 = 81), not 36.

• Option (d) 25 is incorrect because it represents the variance if the standard deviation were 5. However, the standard deviation in Fahrenheit is 9, so the variance should be (9^2 = 81), not 25.

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.