Statistics

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.

Given, E(X) = 3 and E(X²) = 11

Variable of X = E(X²) $-$ [E(X)]² = 11 $-$ 3² = 11 $-$ 9 = 2

Given, x₁ + x₂ + x₃ + .... + x₁₀ = 12

and x$_1^2$ + x$_2^2$ + ..... + x$_{10}^2$ = 18

${\sigma ^2} = {1 \over n}\sum {{x^2} - {{\left( {{1 \over n}\sum x } \right)}^2}} $

$ = {1 \over {10}} \times 18 - {\left( {{1 \over {10}} \times 12} \right)^2} = {9 \over 5} - {{36} \over {25}} = {9 \over {25}}$

$\therefore$ Standard deviation = 3/5.

Let us make the table from the given data

For player A, Mean = ${{\sum {{x_i}} } \over n} = {{610} \over {10}} = 61$

$SD = \sqrt {{{\sum {{{({x_i} - 61)}^2}} } \over N}} $

$ = \sqrt {{{526} \over {10}}} = 7.25$

For player B, Mean = ${{\sum {{y_i}} } \over n} = {{610} \over {10}} = 61$

$SD = \sqrt {{{\sum {{{({y_i} - 61)}^2}} } \over N}} $

$ = \sqrt {{{11416} \over {10}}} = 33.79$

Since, SD for player A is 7.25 < SD for player B is 33.79.

Hence, player A is more consistent player.

The variance of first $2 k$ natural numbers

$ \begin{aligned} S_1 & =\frac{2 k(2 k+1)(4 k+1)}{6 \times 2 k}-\left(\frac{2 k(2 k+1)}{2 \times 2 k}\right)^2 \\ & =(2 k+1)\left[\frac{4 k+1}{6}-\frac{2 k+1}{4}\right] \\ & =\frac{2 k+1}{12}[8 k+2-6 k-3] \\ & =\frac{4 k^2-1}{12} \end{aligned} $

and variance of first $k$ natural numbers

$ \begin{aligned} S_2= & \frac{k(k+1)(2 k+1)}{6 \times k}-\left(\frac{k(k+1)}{2 \times k}\right)^2 \\ & =(k+1)\left[\frac{2 k+1}{6}-\frac{k+1}{4}\right] \\ & =\frac{k+1}{12}[4 k+2-3 k-3]=\frac{k^2-1}{12} \\ \therefore \quad & \frac{S_1}{S_2}=\frac{4 k^2-1}{k^2-1}=4+\frac{3}{k^2-1},(k>1) \\ \therefore \quad & \frac{S_1}{S_2} \in(1,4] \end{aligned} $

From the given information, we can say that

$ \begin{aligned} & \sqrt{\frac{\sum_{i=1}^{100}\left(x_i-\bar{x}\right)^2}{100}}=5 \text { and } \sqrt{\frac{\sum_{i=1}^{100}\left(y_i-\bar{y}\right)^2}{100}}=6 \\ & \therefore \quad \sum_{i=1}^{100}\left(x_i-\bar{x}\right)^2=2500 \end{aligned} $

and $\sum_{i=1}^{100}\left(y_i-\bar{y}\right)^2=3600$

and it is also given that

$ \begin{gathered} \sum_{i=1}^{100}\left(x_i-\bar{x}\right)\left(y_i-\bar{y}\right)=600 \\ \because \quad a^2+b^2-2 a b=(a-b)^2 \\ \sum_{i=1}^{100}\left(x_i-\bar{x}\right)^2+\sum_{i=1}^{100}\left(y_i-\bar{y}\right)^2 \\ -\sum_{i=1}^{100}\left(x_i-\bar{x}\right)\left(y_i-\bar{y}\right) \end{gathered} $

$ \begin{aligned} & =\sum_{i=1}^{100}\left(\left(x_i-\bar{x}\right)-\left(y_i-\bar{y}\right)\right)^2 \\ \Rightarrow \quad & \sum_{i=1}^{100}\left[\left(x_i-\bar{x}\right)-\left(y_i-\bar{y}\right)\right]^2 \\ & =2500+3600-1200 \\ \Rightarrow \quad & \sum_{i=1}^{100}\left[\left(x_i-y_i\right)-(\bar{x}-\bar{y})\right]^2=4900 \end{aligned} $

$ \Rightarrow \sqrt{\frac{\sum_{i=1}^{100}\left[\left(x_i-y_i\right)-(\bar{x}-\bar{y})\right]^2}{100}}=7 $

$ \begin{aligned} & \text { Given, } \frac{1}{4} \text { th observation }=a \\ & \text { another } \frac{1}{4} \text { th observation }=-a \\ & \quad \frac{1}{4} \text { th observation }=b \\ & \Rightarrow \frac{1}{4} \text { th observation }=-b \\ & \therefore \quad \bar{x}=\frac{a-a+b-b}{n}=0 \end{aligned} $

$ \begin{aligned} & \text { Variance }=a b \\ & \qquad a b=\frac{\Sigma x_i^2}{n}-(\bar{x})^2=\frac{\Sigma x_i^2}{n} \\ & \Rightarrow \quad a b=\frac{\frac{n}{4}\left(a^2+a^2\right)+\frac{n}{4}\left(b^2+b^2\right)}{n} \end{aligned} $

$ \begin{array}{llrl} \Rightarrow & a b=\frac{a^2+b^2}{2} & \\ \Rightarrow & a^2+b^2-2 a b =0 \\ \Rightarrow & (a-b)^2 =0 \\ \Rightarrow & a =b \end{array} $

Let variance of $x_1, x_2, x_3, \ldots, x_n$ be $\sigma^2$, then variance of $4 x_1, 4 x_2, \ldots, 4 x_n$ is $4^2 \sigma^2$ is $16 \sigma^2$ And variance dependent on change of scale.

$ \begin{aligned} &\text { Mean of first five prime numbers }\\ &\begin{aligned} & =\frac{2+3+5+7+11}{5} \\ & =\frac{28}{5}=5.6 \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { ∴ Mean deviation about mean }\\ &\begin{aligned} & =\frac{|2-5.6|+|3-5.6|+|5-5.6|+|7-5.6|}{+|11-5.6|} \\ & =\frac{3.6+2.6+0.6+1.4+5.4}{5} \\ & =\frac{13.6}{5}=2.72 \\ & \text { Variance }=\frac{\Sigma x i^2}{n}-\left(\frac{\Sigma x i}{n}\right)^2 \\ & \quad=\frac{4+9+25+49+121}{5}-(5.6)^2 \\ & \quad=41.6-31.36=10.24 \end{aligned} \end{aligned} $

$ \begin{array}{c|c|c|c|c|c} \hline \mathbf{C l} & \boldsymbol{f}_{\boldsymbol{i}} & \boldsymbol{x}_{\boldsymbol{i}} & \boldsymbol{f}_{\boldsymbol{i}} \boldsymbol{x}_{\boldsymbol{i}} & \boldsymbol{x}_{\boldsymbol{i}}^2 & \boldsymbol{f}_{\boldsymbol{i}} \boldsymbol{x}_{\boldsymbol{i}}^2 \\ \hline 0-5 & 4 & 2.5 & 10 & 6.25 & 25 \\ \hline 5-10 & 1 & 7.5 & 7.5 & 56.25 & 56.25 \\ \hline 10-15 & 10 & 12.5 & 125 & 156.25 & 1562.5 \\ \hline 15-20 & 3 & 17.5 & 52.5 & 306.25 & 918.75 \\ \hline 20-25 & 2 & 22.5 & 45 & 506.25 & 1012.50 \\ \hline \text { Total } & 20 & & 240 & & 3575 \\ \hline \end{array} $

$ \begin{gathered} \text { } \begin{aligned} Now,\,\,\, & \operatorname{var}(x)=\frac{1}{N} \Sigma f_i x_i^2-\left(\frac{1}{N} \Sigma f_i x_i\right)^2 \\ = & \frac{1}{20} \times 3575-\left(\frac{1}{20} \times 240\right)^2 \\ = & 178.75-144=34.75 \end{aligned} \end{gathered} $

We have,

$ \begin{aligned} & \text { Mean }=6 \\ & \therefore \quad \frac{8+9+6+5+x+4+6+5}{8}=6 \\ & \Rightarrow \quad 43+x=48 \Rightarrow x=5 \end{aligned} $

So, data are $8,9,6,5,5,4,6,5$

$ \begin{array}{lc} \therefore & x_i=8,9,6,5,5,4,6,5 \\ \Rightarrow & x_i^2=64,81,36,25,25,16,36,25 \\ \therefore & \Sigma x_i^2=64+81+36+25+25+16+36+25 \\ & =308 \\ \therefore & \text { SD }(x)=\sqrt{\frac{1}{N} \Sigma x_i^2-(\bar{x})^2} \\ & =\sqrt{\frac{1}{8} \times 308-(6)^2}=\sqrt{38.5-36} \\ & =\sqrt{2.5} \quad=1.58 \end{array} $

Let two observations be x and y.

$\therefore$ Mean $ = {{1 + 3 + 8 + x + y} \over 5} = {{12 + x + y} \over 5}$

$ \Rightarrow 5 = {{x + y + 12} \over {15}} \Rightarrow x + y = 13$ ..... (i)

Now, variance $ = {{\sum {x_i^2} } \over n} - {(\overline x )^2}$

$ \Rightarrow 9.20 = {{{1^2} + {3^2} + {8^2} + {x^2} + {y^2}} \over 5} - 25$

$ \Rightarrow 46 = 74 + {x^2} + {y^2} - 125$

$ \Rightarrow {x^2} + {y^2} = 97$ ...... (ii)

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

${(x + y)^2} - 2xy = 97$

$ \Rightarrow {(13)^2} - 2xy = 97$

$ \Rightarrow 2xy = 169 - 97 = 72$

$\Rightarrow$ xy = 36

$\therefore$ x : y = 4 : 9