Question

Find the standard deviation of $40,42$ and $48$. If each value is multiplied by $3$, find the standard deviation of the new data.

Answer 1

Let us consider the given data $40,42,48$ and find $\sigma$.
Let the assumed mean A be $44$

$x$	$d=x-44$	$d^2$
$40$	$-4$	$16$
$42$	$-2$	$4$
$48$	$4$	$16$
	$\sum d=-2$	$\sum d^2=36$

Standard deviation, $\sigma =\sqrt{\frac{\sum d^2}{n}-{\left(\frac{\sum d}{n}\right)}^2}$
$=\sqrt{\frac{36}{3}-{\left(\frac{-2}{3}\right)}^2}$
$\sigma =\frac{\sqrt{104}}{3}$
When the values are multiplied by $3$, we get $120,126,144$. Let the assumed mean A be $132$.
Let ${\sigma }_1$ be the S.D. of the new data.

$x$	$d=x-132$	$d^2$
$120$	$-12$	$144$
$126$	$-6$	$36$
$144$	$12$	$144$
	$\sum d=-6$	$\sum d^2=324$

Standard deviation, ${\sigma }_1=\sqrt{\frac{\sum d^2}{n}-{\left(\frac{\sum d}{n}\right)}^2}$
$=\sqrt{\frac{324}{3}-{\left(\frac{-6}{3}\right)}^2}$
${\sigma }_1=\sqrt{\frac{312}{3}}=\sqrt{104}$
In the example, when each value is multiplied by $3$, the standard deviation also gets multiplied by $3$.