Question

The standard deviation of $9,16,23,30,37,44,51$ is

• A. $7$
• B. $9$
• C. $12$
• D. $14$
• E. $16$

Answer 1

The correct option is D

$14$

Explanation for the correct option

Mean$=\frac{\left[9+16+23+30+37+44+51\right]}{7}$

$$\begin{gathered} =\frac{210}{7} \\ =30 \end{gathered}$$

Standard deviation$=\sqrt{\frac{\sum _{}{(x_i-\mu )}^2}{7}}$

$$\begin{gathered} =\sqrt{\frac{\left[{\left(9-30\right)}^2+{\left(16-30\right)}^2+{\left(23-30\right)}^2+{\left(30-30\right)}^2+{\left(37-30\right)}^2+{\left(44-30\right)}^2+{\left(51-30\right)}^2\right]}{7}} \\ =\sqrt{\frac{\left[{\left(-21\right)}^2+{\left(-14\right)}^2+{\left(-7\right)}^2+{\left(0\right)}^2+{\left(7\right)}^2+{\left(14\right)}^2+{\left(21\right)}^2\right]}{7}} \\ =\sqrt{\frac{\left[441+196+49+49+196+441\right]}{7}} \\ =\sqrt{\frac{1372}{7}} \\ =\sqrt{196} \\ =14 \end{gathered}$$

Hence the correct option is option(d) i.e. $14$