Question

Find the variance of first $10$ multiples of $3$.

• A. $70.15$
• B. $74.15$
• C. $73.15$
• D. $74.25$

Answer 1

Answer: (D)

$74.25$

Answer 2

Variance, $\sigma^{2} = \frac{\sum x^{2}_{i}}{n}-\left(\bar{x}\right)^{2}$ = \left\\{\frac{3^{2}+6^{2}+9^{2}+....+30^{2}}{10}-\left(16.5\right)^{2}\right\\} = \frac{3^{2}\times\left\\{1^{2}+2^{2}+3^{2}+... +10^{2}\right\\}}{10}-\left(16.5\right)^{2} $= \frac{9\times 10\times \left(10+1\right)\left(2\times 10+1\right)}{6\times 10}-\left(16.5\right)^{2}$ $= \left(\frac{9\times 10\times 11\times 21}{6\times 10}-272\cdot25\right)$ $= \left(346.5 - 272.25\right) = 74.25$. $\therefore$ Variance, $\sigma^{2} = 74.25$.