MHT CET · Maths · Statistics
The variance of first 10 multiples of 3 is
- A \(74.15\)
- B \(73.15\)
- C \(74.25\)
- D \(70.15\)
Answer & Solution
Correct Answer
(C) \(74.25\)
Step-by-step Solution
Detailed explanation
Variance
\(=\frac{\sum x_i^2}{n}-(\bar{x})^2=\frac{3^2+6^2+9^2+\ldots . .+30^2}{10}-\) \(\left(\frac{3+6+9+\ldots \ldots+30}{10}\right)^2 \)
\( =\frac{3^2\left(1^2+2^2+3^2 \ldots .0^2\right)}{10}-3^2\left(\frac{1+2+3+\ldots . .+10}{10}\right)^2 \)
\( =\frac{9 \times 10 \times 11 \times 21}{6 \times 10}-9 \times\left(\frac{10 \times 11}{2 \times 10}\right)^2=\frac{33 \times 21}{2}\) \(-\frac{9 \times 121}{4} \)
\( =\frac{99}{4}(14-11) \)
\( =\frac{99 \times 3}{4}=24.75 \times 3=74.25\)
\(=\frac{\sum x_i^2}{n}-(\bar{x})^2=\frac{3^2+6^2+9^2+\ldots . .+30^2}{10}-\) \(\left(\frac{3+6+9+\ldots \ldots+30}{10}\right)^2 \)
\( =\frac{3^2\left(1^2+2^2+3^2 \ldots .0^2\right)}{10}-3^2\left(\frac{1+2+3+\ldots . .+10}{10}\right)^2 \)
\( =\frac{9 \times 10 \times 11 \times 21}{6 \times 10}-9 \times\left(\frac{10 \times 11}{2 \times 10}\right)^2=\frac{33 \times 21}{2}\) \(-\frac{9 \times 121}{4} \)
\( =\frac{99}{4}(14-11) \)
\( =\frac{99 \times 3}{4}=24.75 \times 3=74.25\)
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