Question

If ${\sum }_{i=1}^{18}(x_i-8)=9and{\sum }_{i=1}^{18}(x_i-8{)}^2=45$ then the standard deviation of $x_1,x_2,\dots ..x_{18}$ is

• A. $\frac{1}{2}$
• B. $2$
• C. $\frac{3}{2}$
• D. $\frac{5}{2}$

Answer 1

The correct option is C

$\frac{3}{2}$

Let $d_i=x_i-8$ but

${\delta }_x^2={\delta }_d^2=\frac{1}{18}\sum d_i^2-(\frac{1}{18}\sum d_i{)}^2$

$=\frac{1}{18}\times 45-(\frac{9}{18}{)}^2=\frac{5}{2}-\frac{1}{4}=\frac{9}{4}$

Therefore ${\delta }_x=\frac{3}{2}$