The variance of first 10 multiples of 3 is
The variance of first 10 multiples of 3 is
- $74.15$
- $73.15$
- $74.25$
- $70.15$
Solution
Variance
$\begin{aligned} & =\frac{\sum x_i^2}{n}-(\bar{x})^2=\frac{3^2+6^2+9^2+\ldots . .+30^2}{10}-\left(\frac{3+6+9+\ldots \ldots+30}{10}\right)^2 \\ & =\frac{3^2\left(1^2+2^2+3^2 \ldots .0^2\right)}{10}-3^2\left(\frac{1+2+3+\ldots . .+10}{10}\right)^2 \\ & =\frac{9 \times 10 \times 11 \times 21}{6 \times 10}-9 \times\left(\frac{10 \times 11}{2 \times 10}\right)^2=\frac{33 \times 21}{2}-\frac{9 \times 121}{4} \\ & =\frac{99}{4}(14-11) \\ & =\frac{99 \times 3}{4}=24.75 \times 3=74.25\end{aligned}$
Asked in: MHT CET 2022 (07 Aug Shift 2)
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