MHT CET · Maths · Statistics
If 1 is added to first 10 natural numbers, then variance of the numbers so obtained is
- A 8.25
- B 3.87
- C 6.5
- D 2.87
Answer & Solution
Correct Answer
(A) 8.25
Step-by-step Solution
Detailed explanation
We have to find variance of numbers \(2,3,4, \ldots ., 11\).
Here \(\overline{\mathrm{x}}=\frac{2+3+4+\ldots+11}{10}=\frac{65}{10}=6.5\)
Variance \(=\frac{\sum\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^2}{\mathrm{n}}\)
\( \begin{aligned} & =\frac{\left[(-4.5)^2+(-3.5)^2+(-2.5)^2+(-1.5)^2+(-0.5)^2\\+(0.5)^2+ (1.5)^2+(2.5)^2+(3.5)^2+(4.5)^2\right]}{10}\end{aligned}\)
\(=\frac{2(20.25+12.25+6.25+2.25+0.25)}{10}=\frac{41.25}{5}=8.25 \)
Here \(\overline{\mathrm{x}}=\frac{2+3+4+\ldots+11}{10}=\frac{65}{10}=6.5\)
Variance \(=\frac{\sum\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^2}{\mathrm{n}}\)
\( \begin{aligned} & =\frac{\left[(-4.5)^2+(-3.5)^2+(-2.5)^2+(-1.5)^2+(-0.5)^2\\+(0.5)^2+ (1.5)^2+(2.5)^2+(3.5)^2+(4.5)^2\right]}{10}\end{aligned}\)
\(=\frac{2(20.25+12.25+6.25+2.25+0.25)}{10}=\frac{41.25}{5}=8.25 \)
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