If the S.D. of $y_{1}, y_{2}, y_{3}, . . . . . y_{n}$ is $6$ , then the variance of $y_{1} - 3, y_{2} - 3, y_{3} - 3, . . . . . y_{n} - 3$ , is

6

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36

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3

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27

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Solution

The correct option is B $36$
If the S.D. of variate Y is .
$σ (Y) = \sqrt{\frac{\sum_{i = 1}^{N} (Y_{i} - ¯ ¯¯ ¯ Y)^{2}}{N}} = 6$
$σ (a Y + b) = \sqrt{\frac{\sum_{i = 1}^{N} (y_{i} - ¯ y)^{2}}{N}}$
= $\sqrt{\frac{{[(a Y_{1} + b) - (a ¯ ¯¯ ¯ Y + b)]}^{2} + {[(a Y_{2} + b) - (a ¯ ¯¯ ¯ Y + b)]}^{2} + . . . . + {[(a Y_{N} + b) - (a ¯ ¯¯ ¯ Y + b)]}^{2}}{N}}$
= $\sqrt{\frac{a^{2} [{(Y_{1} - ¯ ¯¯ ¯ Y)}^{2} + {(Y_{2} - ¯ ¯¯ ¯ Y)}^{2} + . . . + {(Y_{N} - ¯ ¯¯ ¯ Y)}^{2}]}{N}}$
= $| a | \sqrt{\frac{\sum_{i = 1}^{N} {(Y_{i} - ¯ ¯¯ ¯ Y)}^{2}}{N}}$
= $| a | σ (Y)$
$σ (Y - 3) = | 1 | \times 6$
= 6
Variance [ $(σ (Y - 3))^{2}] = 6^{2} = 36$

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