Question

The standard deviation of 8 observations is 25. If each value is diminished by 3

then the new standard deviation is​

Answer 1

Let us do it by general.

Consider the formula for finding standard deviation given below.

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

Now each value is increased by $k,$ then new mean becomes,

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

$\displaystyle\longrightarrow (\bar x)'=\dfrac{1}{n}\sum_{i=1}^nx_i+\dfrac{1}{n}\sum_{i=1}^nk$

$\displaystyle\longrightarrow (\bar x)'=\bar x+k\quad\quad\left[\because\bar x=\dfrac{1}{n}\sum_{i=1}^nx_i\right]$

And so, new standard deviation,

$\displaystyle\longrightarrow\sigma'=\sqrt{\dfrac{1}{n}\sum_{i=1}^n(x_i+k)^2-(\bar x+k)^2}$

$\displaystyle\longrightarrow\sigma'=\sqrt{\dfrac{1}{n}\sum_{i=1}^n\left[(x_i)^2+2kx_i+k^2\right]-\left[(\bar x)^2+2k\bar x+k^2\right]}$

$\displaystyle\longrightarrow\sigma'=\sqrt{\dfrac{1}{n}\sum_{i=1}^n(x_i)^2+\dfrac{2k}{n}\sum_{i=1}^nx_i+\dfrac{1}{n}\sum_{i=1}^nk^2-(\bar x)^2-2k\bar x-k^2}$

$\displaystyle\longrightarrow\sigma'=\sqrt{\dfrac{1}{n}\sum_{i=1}^n(x_i)^2-(\bar x)^2+2k\cdot\dfrac{1}{n}\sum_{i=1}^nx_i+\dfrac{1}{n}\sum_{i=1}^nk^2-2k\bar x-k^2}$

$\displaystyle\longrightarrow\sigma'=\sqrt{\sigma^2+2k\bar x+k^2-2k\bar x-k^2}$

$\displaystyle\longrightarrow\underline{\underline{\sigma'=\sigma}}$

So the standard deviation does not change if each observation is increased or decreased by the same constant.

Hence the answer to our question is undoubtedly 25.