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The mean and the standard deviation(s.d.)of five observations are 9 and 0, respectively. If one of the observations is changed such that the mean of the new set of five observations becomes 10, then their s.d. is :

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a

0

b

1

c

4

d

2

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detailed solution

Correct option is D

Let a, b, c, d, e be five observations.

$\begin{array}{ll} ∴ S.D. = \sqrt{\frac{Σ (a - \bar{x})^{2}}{5}} = 0 \Rightarrow Σ (a - \bar{x})^{2} = 0 \\ \Rightarrow (a - \bar{x})^{2} + (b - \bar{x})^{2} + (c - \bar{x})^{2} + (d - \bar{x})^{2} + (e - \bar{x})^{2} = 0 \\ \Rightarrow a - \bar{x} = b - \bar{x} = c - \bar{x} = d - \bar{x} = e - \bar{x} = 0 \\ \Rightarrow a = b = c = d = e = \bar{x} = 9 \\ ∴ a + b + c + d + e = 9 \times 5 = 45 \end{array}$

Let $x$ be the change in term. Then

New sum $a + b + c + (d + x) + e = 5 \times 10 = 50$

$\Rightarrow x = 5$

$∴$ New observations becomes 9, 9, 9, 14, 9

New S.D.= $\sqrt{\frac{Σ (a - \bar{x})^{2}}{5}} = \sqrt{\frac{1^{2} + 1^{2} + 1^{2} + 4^{2} + 1^{2}}{5}} = \sqrt{\frac{20}{5}} = 2$

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