Q49 / 77 · NTA Abhyas (JEE Main) 2020 · Mathematics · Statistics

If $\sum_{i = 1}^{5} (x_{i} - 6) = 5$ and $\sum_{i = 1}^{5} {(x_{i} - 6)}^{2} = 25$ , then the standard deviation of observations $3 x_{1} + 2, 3 x_{2} + 2, 3 x_{3} + 2,$ $3 x_{4} + 2$ and $3 x_{5} + 2$ is equal to

Correct answer: 6

Solution

Variance of

(3 x_{i} + 2) = 9 v a r (x_{i})

= 9 (\frac{\sum {x_{i}}^{2}}{5} - {(\frac{\sum x_{i}}{5})}^{2})

= 9 (5 - 1)

= 36

\Rightarrow S . D = 6

Hence, the correct answer is 6.

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