Question

If standard deviation of $x_i,i=1,2,......,n$ is ${\sigma }_x$, then the standard deviation of $\frac{a{x}_i+b}{p}\forall a,b,p\in R,i=1,2,3,.....,n$, is

• A. $\left|\frac{a}{p}\right|{\sigma }_x$
• B. $\frac{p}{a}{\sigma }_x$
• C. $\frac{p}{2a}{\sigma }_x$
• D. $\left|\frac{p}{a}\right|{\sigma }_x$

Answer 1

The correct option is A $\left|\frac{a}{p}\right|{\sigma }_x$

Let $u=\frac{ax+b}{p}\therefore \overline{u}=\frac{a\overline{x}+b}{p}$
$\therefore SD=\sqrt{\frac{\sum (u-\overline{u}{)}^2}{\sum f}}=\sqrt{\frac{\sum {(\frac{ax+b}{p}-\frac{a\overline{x}+b}{p})}^2}{\sum f}}$
$=\sqrt{\frac{\sum {\left(\frac{ax+b-a\overline{x}-b}{p}\right)}^2}{\sum f}}=\sqrt{\frac{a^2}{p^2}\frac{\sum (x-\overline{x}{)}^2}{\sum f}}$
$=\sqrt{\frac{a^2}{p^2}{\sigma }_x^2}=\left|\frac{a}{p}\right|{\sigma }_x$