Question

If +(x, a) mand X Ix, af na then the standard deviation of variate x.

Answer 1

Given,

$\displaystyle\longrightarrow\sum_{i=1}^n(x_i-a)=n\quad\quad\dots(1)$

$\displaystyle\longrightarrow\sum_{i=1}^nx_i-\sum_{i=1}^na=n$

$\displaystyle\longrightarrow\sum_{i=1}^nx_i-na=n$

$\displaystyle\longrightarrow\sum_{i=1}^nx_i=na+n$

$\displaystyle\longrightarrow\sum_{i=1}^nx_i=n(a+1)$

$\displaystyle\longrightarrow\dfrac{1}{n}\sum_{i=1}^nx_i=a+1$

The LHS here is the mean of the $n$ terms.

$\displaystyle\longrightarrow\bar x=a+1$

$\displaystyle\longrightarrow a=\bar x-1$

Then (1) becomes,

$\displaystyle\longrightarrow\sum_{i=1}^n(x_i-\bar x+1)=n$

$\displaystyle\longrightarrow\sum_{i=1}^n(x_i-\bar x)+\sum_{i=1}^n1=n$

$\displaystyle\longrightarrow\sum_{i=1}^n(x_i-\bar x)+n=n$

$\displaystyle\longrightarrow\sum_{i=1}^n(x_i-\bar x)=0$

Given,

$\displaystyle\longrightarrow\sum_{i=1}^n(x_i-a)^2=na$

$\displaystyle\longrightarrow\sum_{i=1}^n(x_i-\bar x+1)^2=na$

$\displaystyle\longrightarrow\sum_{i=1}^n((x_i-\bar x)+1)^2=na$

$\displaystyle\longrightarrow\sum_{i=1}^n[(x_i-\bar x)^2+2(x_i-\bar x)+1]=na$

$\displaystyle\longrightarrow\sum_{i=1}^n(x_i-\bar x)^2+2\sum_{i=1}^n(x_i-\bar x)+\sum_{i=1}^n1=na$

$\displaystyle\longrightarrow\sum_{i=1}^n(x_i-\bar x)^2+2\times0+n=na$

$\displaystyle\longrightarrow\sum_{i=1}^n(x_i-\bar x)^2+n=na$

$\displaystyle\longrightarrow\sum_{i=1}^n(x_i-\bar x)^2=na-n$

$\displaystyle\longrightarrow\sum_{i=1}^n(x_i-\bar x)^2=n(a-1)$

$\displaystyle\longrightarrow\dfrac{1}{n}\sum_{i=1}^n(x_i-\bar x)^2=a-1$

$\displaystyle\longrightarrow\underline{\underline{\sqrt{\dfrac{1}{n}\sum_{i=1}^n(x_i-\bar x)^2}=\sqrt{a-1}}}$

This is the standard deviation of the variate.

Hence (2) is the answer.