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

Statistics

Solution:

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, $\sigma^{2} = 74.25$.