Statistics

$ \begin{array}{lcccc} \hline \begin{array}{l} \text { Class } \\ \text { interval } \end{array} & \begin{array}{l} \text { Frequency } \\ \left(\boldsymbol{f}_{\boldsymbol{i}}\right) \end{array} & \boldsymbol{x}_{\boldsymbol{i}} & \boldsymbol{f}_{\boldsymbol{i}} \boldsymbol{x}_{\boldsymbol{i}} & \boldsymbol{f}_{\boldsymbol{i}} x_{\boldsymbol{i}}^2 \\ \hline 0-4 & 1 & 2 & 2 & 4 \\ \hline 4-8 & 2 & 6 & 12 & 72 \\ \hline 8-12 & 2 & 10 & 20 & 200 \\ \hline 12-16 & 1 & 14 & 14 & 196 \\ \hline & \begin{array}{c} \sum f_i=N \\ =6 \end{array} & & \begin{array}{c} \sum f_i x_i \\ =48 \end{array} & \begin{array}{c} \sum f_i x_i^2 \\ =472 \end{array} \\ \hline \end{array} $

$ \begin{aligned} & \text { So, mean }(\bar{x})=\frac{\sum f_i x_i}{N}=\frac{48}{6}=8 \\ & \text { and variance }\left(\sigma^2\right)=\frac{\sum f_i x_i^2}{N}-(\bar{x})^2 \\ & \qquad=\frac{472}{6}-8^2=\frac{236}{3}-64 \\ & =\frac{236-192}{3}=44 / 3 \end{aligned} $

We have, $\sum_{i=1}^9 x_i-5=9$

$ \begin{aligned} & \text { and } \sum_{i=1}^9\left(x_i-5\right)^2=45 \\ & \qquad \begin{array}{l} \Sigma x_i=9+45=54 \\ \Sigma x_i^2-10 \Sigma x_i+25 \times 9=45 \\ \Sigma x_i^2-540+225=45 \\ \Sigma x_i^2=585-225=360 \end{array} \end{aligned} $

∴ Standard deviation $=\sqrt{\text { Variance }}$

$ \begin{aligned} & =\sqrt{\frac{\Sigma x_i^2}{9}-\left(\frac{\Sigma x_i}{9}\right)^2}=\sqrt{\frac{360}{9}-\left(\frac{54}{9}\right)^2} \\ & =\sqrt{40-36}=2 \end{aligned} $

$ \begin{array}{|c|c|c|c|c|} \hline \boldsymbol{x}_{\boldsymbol{i}} & \boldsymbol{f}_{\boldsymbol{i}} & \boldsymbol{c f} & \left|\boldsymbol{x}_{\boldsymbol{i}}-\mathbf{3}\right| & \boldsymbol{f}_{\boldsymbol{i}}\left|\boldsymbol{x}_{\boldsymbol{i}}-\mathbf{3}\right| \\ \hline 2 & 8 & 8 & 1 & 8 \\ \hline 3 & 6 & 14 & 0 & 0 \\ \hline 5 & 4 & 18 & 2 & 8 \\ \hline 7 & 2 & 20 & 4 & 8 \\ \hline 9 & 1 & 21 & 6 & 6 \\ \hline \end{array} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \quad N=21 \Rightarrow \frac{N}{2}=10.5 \\ & \therefore \text { Median }=M=3 \end{aligned}\\ &\text { and Mean deviation about median }\\ &=\frac{\sum f_i\left|x_i-M\right|}{\sum f_i}=\frac{30}{21}=\frac{10}{7} \end{aligned} $

$ \begin{aligned} &\begin{aligned} \sum f_i & =11, \sum f_i x_i=55 \\ \bar{X} & =\frac{\sum f_i x_i}{\sum f_i}=\frac{55}{11}=5 \\ \sum f_i\left|x_i-\bar{x}\right| & =20 \end{aligned}\\ &\therefore \text { Mean deviation about mean is }\\ &\frac{\sum f_i\left|x_i-\bar{x}\right|}{\sum f_i}=\frac{20}{11} \end{aligned} $

$ \begin{aligned} & \text { } \sum_{i=1}^{11}\left(x_i-4\right)=22 \\ & \Rightarrow \quad \sum_{i=1}^{11} x_i-4 \times 11=22 \\ & \Rightarrow \quad \sum_{i=1}^{11} x_i=66 \end{aligned} $

So, mean $(\bar{x})=\frac{66}{11}=6$ and                                      ...(i)

$ \begin{aligned} \because \quad x_i-\bar{x} & =x_i-6 \\ & =\left(x_i-4\right)-2 \\ \therefore \quad\left[x_i-\bar{x}\right]^2 & =\left[\left(x_i-4\right)-2\right]^2 \\ & =\left(x_i-4\right)^2-4\left(x_i-4\right)+4 \end{aligned} $

$ \begin{aligned}So,\,\, \sum_{i=1}^{11}\left(x_i-\bar{x}\right)^2 & =\sum_{i=1}^{11}(x-4)^2 \\ - & 4 \sum_{i=1}^{11}\left(x_i-4\right)+11 \times 4 \\ =154- & 4 \times 22+44=110 \end{aligned} $

$ \text { } \begin{aligned}Variance\,\, (\sigma) & =\frac{1}{n} \sum\left(x_i-\bar{x}\right)^2 \\ & =\frac{1}{11} \times 110=10 \end{aligned} $

Thus, $\alpha=6, \beta=10$

Therefore, $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{6}{10}+\frac{10}{6}=\frac{3}{5}+\frac{5}{3}$

$ =\frac{9+25}{15}=34 / 15=\text { Sum of roots } $

and $\frac{\alpha}{\beta} \cdot \frac{\beta}{\alpha}=1=$ product of roots

Hence, the quadratic is

$ \begin{aligned} & x^2-\left(\frac{34}{15}\right) x+1=0 \\ \Rightarrow & 15 x^2-34 x+15=0 \end{aligned} $

Use formula for combined variance

Let $n_1=15, \mu_1=2, \sigma_1^2=4$

$ n_2=10, \mu_2=2, \sigma_2^2=5 $

$ \begin{aligned} \text { Combined mean } & =\mu=\frac{n_1 \mu_1+n_2 \mu_2}{n_1+n_2} \\ & =\frac{15 \times 2+10 \times 2}{25}=2 \end{aligned} $

Combined variance $=\sigma^2$

$ =\frac{n_1 \sigma_1^2+n_2 \sigma_2^2+n_1\left(\mu_1-\mu\right)^2+n_2\left(\mu_2-\mu\right)^2}{n_1+n_2} $

Since, $\mu_1=\mu_2=$ the last two terms becomes zero.

So, $\sigma^2=\frac{15 \times 4+10 \times 5}{25}=\frac{60+50}{25}=\frac{110}{25}$

$ =4.4 $

$ \begin{array}{cccc} \hline \text { Class interval } & \text { Frequency } & \boldsymbol{x}_{\boldsymbol{i}} & \boldsymbol{f}_{\boldsymbol{i}} \boldsymbol{x}_{\boldsymbol{i}} \\ \hline 0-2 & 2 & 1 & 2 \\ \hline 2-4 & 3 & 3 & 9 \\ \hline 4-6 & 5 & 5 & 25 \\ \hline 6-8 & 3 & 7 & 21 \\ \hline 8-10 & 2 & 9 & 18 \\ \hline & \Sigma \boldsymbol{f}_{\boldsymbol{i}}=15 & & 75 \\ \hline \end{array} $

$ \begin{aligned} & \text { Mean }(\mu)=\frac{75}{15}=5 \\ & \qquad \begin{aligned} \sigma^2 & =\frac{\Sigma\left(x_i-\mu\right)^2 \cdot f_i}{N} \\ & =\frac{32+12+0+12+32}{15}=\frac{88}{15} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { We have, } \Sigma f=20\\ &\begin{aligned} \therefore \quad(\alpha+2)+(\alpha-1)^2+4 & +(\alpha-1)+2+\alpha=20 \end{aligned} \end{aligned} $

$ \begin{array}{ll} \Rightarrow & \alpha^2+\alpha-12=0 \\ & (\alpha+4)(\alpha-3)=0 \Rightarrow \alpha=+3,-4 \\ \therefore & \alpha=3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,[\because \alpha \neq-4]\end{array} $

$ \begin{array}{ccccc} \hline x_i & f_i & f_i x_i & \left|x_i-\bar{x}\right| & f \cdot\left|x_i-\bar{x}\right| \\ \hline 3 & 5 & 15 & 3 & 15 \\ \hline 4 & 4 & 16 & 2 & 8 \\ \hline 5 & 4 & 20 & 1 & 4 \\ \hline 8 & 2 & 16 & 2 & 4 \\ \hline 10 & 2 & 20 & 4 & 8 \\ \hline 11 & 3 & 33 & 5 & 15 \\ \hline & 20 & 120 & & \sum\left|x_i-\bar{x}\right|=54 \\ \hline \end{array} $

$ \begin{aligned} \bar{x} & =\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{120}{20}=6 \\ \mathrm{MD} & =\frac{\Sigma f_i\left|x_i-\bar{x}\right|}{\Sigma f}=\frac{54}{20}=27 \end{aligned} $

Variance of Ist $n$ natural number is

$ =\frac{n^2-1}{12}=10 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

and variance of Ist $m$ even natural number is

$ \frac{m^2-1}{3}=16 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

On solving Eqs. (i) and (ii), we get

$ \begin{aligned} & n^2=121 \text { and } m^2=49 \\ & n: m=11: 7 \end{aligned} $

$ \begin{aligned} &\text { Given, } 2,12,3,11,5,10,6,7\\ &\begin{aligned} \text { Mean } \bar{x} & =\frac{56}{8}=7 \\ \Sigma\left(x_i-\bar{x}\right)^2 & =25+25+16+16+4 +9+1+0 \\ & =96 \\ \operatorname{var}(x) & =\frac{\left(x_i-\bar{x}\right)^2}{n}=\frac{96}{8}=12 \end{aligned} \end{aligned} $

To find the variance of the data set $7, 8, 9, 7, 8, 7, \lambda, 8$ with a mean of 8, follow these steps:

First, solve for $\lambda$:

$ \frac{7 + 8 + 9 + 7 + 8 + 7 + \lambda + 8}{8} = 8 $

Simplify the equation:

$ \frac{54 + \lambda}{8} = 8 $

Multiply both sides by 8:

$ 54 + \lambda = 64 $

Solve for $\lambda$:

$ \lambda = 64 - 54 = 10 $

Next, calculate the variance using the formula:

$ \text{Variance} = \frac{\Sigma x_i^2}{8} - \bar{x}^2 $

Where $\Sigma x_i^2$ is the sum of squares of each data point:

$ \begin{aligned} & \Sigma x_i^2 = 7^2 + 8^2 + 9^2 + 7^2 + 8^2 + 7^2 + 10^2 + 8^2 \\ & = 49 + 64 + 81 + 49 + 64 + 49 + 100 + 64 \\ & = 520 \end{aligned} $

Calculate the variance:

$ \frac{520}{8} - 8^2 = 65 - 64 = 1 $

Thus, the variance of the data set is 1.

Sum of first $n$ even natural number

$ \begin{aligned} = & 2+4+6+\ldots . \\ = & 2(1+2+3+\ldots .) \\ = & 2 \times \frac{n(n+1)}{2}=n(n+1) \\ \therefore \quad \text { Mean } & =\frac{n(n+1)}{n}=(n+1) \end{aligned} $

We know that variance $=\frac{1}{n}\left(\Sigma x_1\right)^2-(\bar{x})^2$

$ \begin{aligned} & =\frac{1}{n}\left(2^2+4^2+6^2+\ldots\right)-(n+1)^2 \\ & =\frac{4}{n}\left(1^2+2^2+3^2+\ldots . .\right)-(n+1)^2 \\ & =\frac{4}{n} \times \frac{n(n+1)(2 n+1)}{6}-(n+1)^2 \\ & =\frac{n+1}{3}[2(2 n+1)-2(n+1)]=\frac{n^2-1}{3} \end{aligned} $

$\therefore$ Difference between the variance of the first $n$ even natural numbers any arithmetic mean $=\frac{399}{3}-21=112$

So, statement I is false and II is true.

Given the equations:

$ \Sigma(x_i + 2)^2 = 28n $

$ \Sigma(x_i - 2)^2 = 12n $

Let's start by expanding and simplifying each equation.

First equation:

$ \begin{aligned} \Sigma(x_i + 2)^2 & = \Sigma(x_i^2 + 4x_i + 4) \\ & = \Sigma x_i^2 + 4\Sigma x_i + 4n \\ & = 28n \end{aligned} $

Second equation:

$ \begin{aligned} \Sigma(x_i - 2)^2 & = \Sigma(x_i^2 - 4x_i + 4) \\ & = \Sigma x_i^2 - 4\Sigma x_i + 4n \\ & = 12n \end{aligned} $

Let's introduce the following notation:

$ S = \Sigma x_i $

$ S_2 = \Sigma x_i^2 $

Using these, we can set up the system of equations:

$ S_2 + 4S + 4n = 28n $

$ S_2 - 4S + 4n = 12n $

Subtract the second equation from the first:

$ \begin{aligned} (S_2 + 4S + 4n) - (S_2 - 4S + 4n) &= 28n - 12n \\ 8S &= 16n \\ S &= 2n \end{aligned} $

Substituting $ S = 2n $ back into the first equation:

$ \begin{aligned} S_2 + 4(2n) + 4n &= 28n \\ S_2 + 8n + 4n &= 28n \\ S_2 &= 28n - 12n \\ S_2 &= 16n \end{aligned} $

Now, to find the variance $\sigma^2$, use the formula:

$ \begin{aligned} \sigma^2 &= \frac{1}{n} \Sigma(x_i - \mu)^2 \\ &= \frac{1}{n} \Sigma x_i^2 - \mu^2 \end{aligned} $

Since $ \Sigma x_i = S = 2n $, the mean $\mu$ is:

$ \mu = \frac{\Sigma x_i}{n} = \frac{2n}{n} = 2 $

Therefore, the variance is:

$ \begin{aligned} \sigma^2 &= \frac{1}{n} \times 16n - 2^2 \\ &= 16 - 4 \\ &= 12 \end{aligned} $

$ \begin{aligned} &\text { Given frequency distribution }\\ &\begin{array}{llll} \hline \boldsymbol{x}_{\boldsymbol{i}} & 4 & 3 & 1 \\ \hline \boldsymbol{f}_{\boldsymbol{L}} & 1 & 3 & 5 \\ \hline \end{array} \end{aligned} $

$ \begin{array}{c|c|c|c|c|c} \hline \boldsymbol{x}_i & \boldsymbol{f}_i & \boldsymbol{f}_i \boldsymbol{x}_i & \left(x_i-\bar{x}\right) & \left(x_i-\bar{x}\right)^2 & f_i\left(x_i-\bar{x}\right)^2 \\ \hline 4 & 1 & 4 & 2 & 4 & 4 \\ \hline 3 & 3 & 9 & 1 & 1 & 3 \\ \hline 1 & 5 & 5 & -1 & 1 & 5 \\ \hline & \Sigma_i=9 & \Sigma \sum_i x_i=18 & & & 12 \\ \hline \end{array} $

$ \therefore \quad \text { Mean } \bar{x}=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{18}{9}=2 $

Now, standard deviation

$ \sigma=\sqrt{\frac{\Sigma f_i\left(x_i-\bar{x}\right)^2}{\Sigma f_i}}=\sqrt{\frac{12}{9}}=\frac{2}{\sqrt{3}} $

So, coefficient of variance $=\frac{\sigma}{\bar{x}} \times 100$

$ =\frac{2}{\sqrt{3}} \times \frac{1}{2} \times 100=\frac{100}{\sqrt{3}} $

$ \text { Mean } A=15, h=6 $

$ \begin{array}{c|c|c|c|c|c|c} \hline \begin{array}{l} \text { Class } \\ \text { Interval } \end{array} & \begin{array}{l} \text { Mid } \\ \text { value } x_i \end{array} & \text { Frequency } & d_{\boldsymbol{i}}=\frac{x_{\boldsymbol{i}}-15}{\mathbf{6}} & f \boldsymbol{d}_{\boldsymbol{i}} & x_{\boldsymbol{i}}-\bar{x} & \frac{\boldsymbol{f}_{\boldsymbol{i}}}{\boldsymbol{x}_{\boldsymbol{i}}-\boldsymbol{x}} \\ \hline 0-6 & 3 & 1 & -2 & -2 & 12 & 12 \\ \hline 6-12 & 9 & 2 & -1 & -2 & 6 & 12 \\ \hline 12-18 & 15 & 3 & 0 & 0 & 0 & 0 \\ \hline 18-24 & 21 & 2 & 1 & 2 & 6 & 12 \\ \hline 24-30 & 27 & 1 & 2 & 2 & 12 & 12 \\ \hline & & N=9 & & 0 & 36 & 48 \\ \hline \end{array} $

Mean, $\bar{X}=A+h\left(\frac{1}{N} \Sigma f_i d_i\right)=15+6\left[\frac{1}{9}(0)\right]$

$ \bar{X}=15 $

Mean-deviation about mean

$ \begin{aligned} & =\frac{1}{N} \sum f_i\left|x_i-\bar{X}\right|=\frac{1}{9} \times 48 \\ & =\frac{1}{9} \times 48=\frac{16}{3} \end{aligned} $

Given data is

Now, $\bar{x}=\frac{4+72+140+112+72}{80}$

$ =\frac{400}{80}=5 $

Thus, mean deviation about mean is $m=\frac{4 \times 4+24 \times 2+28 \times 0+16 \times 2+8 \times 4}{80}$ $=\frac{8}{5}$

and $\sigma^2=\frac{4 \times 4^2+24 \times 2^2+28 \times 0^2+16 \times(-2)^2+8 \times(-4)^2}{80}=\frac{22}{5}$

Now, $m+\sigma^2=\frac{8}{5}+\frac{22}{5}=\frac{30}{5}=6$

Given that $\bar{x}$ and $\bar{y}$ are the arithmetic means of the runs for batsmen $A$ and $B$, respectively, and $\sigma_A$ and $\sigma_B$ represent the standard deviations of their runs.

Since Batsman $A$ is more consistent than Batsman $B$, we have:

$ \sigma_A < \sigma_B \Rightarrow \frac{\sigma_A}{\sigma_B} < 1 $

Additionally, we know that if Batsman $A$ is a higher run scorer than Batsman $B$, then:

$ \bar{x} > \bar{y} \Rightarrow \frac{\bar{x}}{\bar{y}} > 1 $

Combining both conditions, we get:

$ 0 < \frac{\sigma_A}{\sigma_B} < \frac{\bar{x}}{\bar{y}} \quad \text{and} \quad \frac{\bar{x}}{\bar{y}} > 1 $

This indicates that Batsman $A$ scores more runs only when the ratio of the standard deviations is less than the ratio of the arithmetic means, with the latter being greater than 1.

We have, $20,5,15,2,7,3,11$.

$ \begin{aligned} \text { Mean } & =\frac{20+5+15+2+7+3+11}{7} \\ & =\frac{63}{7}=9 \end{aligned} $

Data in ascending order $=2,3,5,7,11$,

15, 20

Median = 7

$m=$ Mean deviations about means

$ \begin{aligned} & |20-9|+|5-9|+|15-9|+|2-9| \\ = & \frac{+|7-9|+|3-9|+|11-9|}{7} \\ = & \frac{11+4+6+7+2+6+2}{7}=\frac{38}{7} \end{aligned} $

$M=$ Mean deviation about median

$ \begin{aligned} & M=\frac{|20-7|+|5-7|+|15-7|+|2-7|}{+|7-7|+|3-7|+|11-7|} \\ & =\frac{13+2+8+5+0+4+4}{7}=\frac{36}{7} \end{aligned} $

Mean of $m$ and $M=\left(\frac{38}{7}+\frac{36}{7}\right) \frac{1}{2}=\frac{37}{7}$

MD (Mean of $m$ and $M$ )

$ \begin{aligned} & =\frac{1}{2}\left\{\left|\frac{36}{7}-\frac{37}{7}\right|+\left|\frac{38}{7}-\frac{37}{7}\right|\right\} \\ & =\frac{1}{2} \times \frac{2}{7}=\frac{1}{7} \end{aligned} $

To find the variance given the coefficient of variation (CV) and the mean, we can use the following relationship:

The formula for the coefficient of variation (CV) is given by:

$ \text{CV} = \frac{\sigma}{\mu} \times 100 $

where $\sigma$ is the standard deviation and $\mu$ is the mean.

In this problem, the coefficient of variation is 25, and the mean ($\mu$) is 44. Substituting these values into the equation, we have:

$ \frac{\sigma}{44} \times 100 = 25 $

Solving for $\sigma$, we get:

$ \sigma = \frac{25 \times 44}{100} = 11 $

The variance ($\sigma^2$) is the square of the standard deviation:

$ \sigma^2 = 11^2 = 121 $

Therefore, the variance is 121.

$\text { Mean } \bar{x}=\frac{\text { Sum of quantities }}{n}$

$=\frac{\frac{n}{2}(a+l)}{n}$ [$l$ = last team]

$\begin{aligned} & =\frac{1}{2}(a+l)=\frac{1}{2}(1+1+100 d) \\ & =1+50 d \end{aligned}$

$\begin{aligned} & \mathrm{MD}=\frac{1}{n} \Sigma\left|x_i-\bar{x}\right| \\ & \Rightarrow 255=\frac{1}{101}[50 d+49 d+48 d+ .... +d+0+d + .... + 50d]\\ & \Rightarrow 255=\frac{2 d}{101}\left[\frac{50 \times 51}{2}\right] \\ & d=\frac{255 \times 101}{50 \times 51}=10.1 \end{aligned}$

Given, data of number are $p, 6,6,7,8,11,15$ and 16.

$\begin{aligned} \text { Mean } \bar{x} & =\frac{\text { Sum of all observations }}{\text { Total number of observations }} \\ & =\frac{p+6+6+7+8+11+15+16}{8} \end{aligned}$

$\begin{aligned} & \Rightarrow \quad \frac{69+p}{8}=3 p \quad \text{(given)}\\ & \Rightarrow \quad 24 p-p=69 \Rightarrow 23 p=69 \\ & \Rightarrow \quad p=3 \\ \end{aligned}$

So given data is 3, 6, 6, 7, 8, 11, 15, 16 and mean is 9.

So, the mean deviation

$\begin{aligned} & |3-9|+|6-9|+|6-9|+|7-9|+|8-9| \\ = & \frac{+|11-9|+|15-9|+|16-9|}{8} \\ = & \frac{6+3+3+2+1+2+6+7}{8}=\frac{30}{8}=3.75 \end{aligned}$

Mean,

$\begin{aligned} \bar{x} & =\frac{5+6+7+8+6+9+13+12+15}{9} \\ & =\frac{81}{9}=9 \end{aligned}$

Deviation, $x_i-\bar{x}$ :

$\begin{aligned} & 5-9,6-9,7-9,8-9,6-9,9-9,13-9,12-9,15-9 \\ & \Rightarrow-4,-3,-2,-1,-3,0,4,3,6 \\ & \left|x_i-\bar{x}\right|: 4,3,2,1,3,0,4,3,6 \end{aligned}$

$\begin{aligned} \mathrm{MD} & =\frac{\sum_\limits{i=1}^9\left|x_i-\bar{x}\right|}{9} \\ & =\frac{4+3+2+1+3+0+4+3+6}{9} \\ & =\frac{26}{9}=2.88 \end{aligned}$

Let data be

$x_1, x_2, x_3, \ldots, x_n$

Given that,

$\begin{aligned} \frac{x_1+x_2+x_3+\ldots+x_n}{n} & =10 \\ \Rightarrow \quad x_1+\ldots+x_n & =10 n \end{aligned}$

When all observation multiplied by 2, we obtain new data as $2 x_1, 2 x_2, \ldots, 2 x_n$ then

$\begin{aligned} \text { Mean } & =\frac{2 x_1+2 x_2+\ldots+2 x_n}{n} \\ & =\frac{2\left(x_1+\ldots+x_n\right)}{n}=\frac{2(10 n)}{n}=20 \end{aligned}$

Mean $ = {{ - 1 + 0 + 4} \over 3} = 1$

Deviation $ = | - 1 - 1|,|0 - 1|,|4 - 1| = 2,1,3$

Mean deviation $ = {{2 + 1 + 3} \over 3} = 2$

Variance $ = {{\sum {x_i^2} } \over n} - \sum {{{\left( {{{{x_i}} \over n}} \right)}^2}} $

$1014 = {{{{60}^2} + {a^2} + {b^2}} \over 3} - {60^2}$

$10242 = {a^2} + {b^2}$ ..... (i)

$a + b = 120$ .... (ii)

From Eqs. (i) and (ii), we get

a = 21 and b = 99

$\begin{aligned} & \Sigma P(X=r)=1 \\ & K+K+\frac{1}{7}+2 K+\frac{2}{5}=1 \Rightarrow 4 K=\frac{3}{5}-\frac{1}{7} \\ & 4 K=\frac{16}{35} \Rightarrow K=\frac{4}{35} \\ & \propto=\frac{\Sigma P_i X_i}{\Sigma P_i}=\Sigma P_i X_i=K+\frac{1}{7}+4 K+\frac{6}{5} \\ & =5 K+\frac{47}{35}=5\left(\frac{4}{35}\right)+\frac{47}{35}=\frac{67}{35}\end{aligned}$

Mean $\bar{x}=5$ and $\Sigma x^2=400$

$\begin{aligned} \because & \operatorname{var}(x)=0 \\ \because \operatorname{var}(x) & =\frac{1}{n} \sum_{i=1}^n\left(x_i-\bar{x}\right)^2 \\ 0 & =\frac{1}{n}\left[\sum_{i=1}^i x_i^2+\sum_{i=1}^i \bar{x}^2-2 \bar{x} \sum_{i=1}^i x_i\right] \\ 0 & =\frac{1}{n}[40+25 n-2 \times 5 \times n \bar{x}] \\ 0 & =\frac{1}{n}[400+25 n-50 n] \\ \Rightarrow \quad \frac{400}{n} & =25 \\ \Rightarrow \quad n & =16 \end{aligned}$

$\begin{aligned} & \text { Mean }=\bar{x}=\frac{605}{5}=121 \\ & \text { Variance }=\frac{\Sigma(x-\bar{x})^2}{n}=\frac{244}{5}=48.8\end{aligned}$

Clearly, difference of values from arithmetic mean is least in option (d).

$\Rightarrow$ Option (d) has least standard deviation.