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The mean and standard deviation of a sample of size 20 are found 10 and 2 respectively. If one item 8 is replaced by 12 then

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standard deviation increases

standard deviation decreases

standard deviation will remain same

mean increases by 4%

answer is B.

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Detailed Solution

$\begin{array}{l} Σ x_{i} = 20 \times 10 = 200 so new sum = 200 - 8 + 12 \\ = 204 . \end{array}$

New mean = 10.2 so increase of 2%.

Also $\frac{1}{20} Σ x_{i}^{2} - 10^{2} = 4 . so Σ x_{i}^{2} = 2080$

New sum of squares $= 2080 - 64 ∣ + 144$

= 2160

New variance $\begin{array}{l} = \frac{1}{20} \times 2160 - {(\frac{204}{20})}^{2} \\ = 108 - 104.04 \\ = 3.96 < 4 \end{array}$

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