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Q.

The mean and S.D. of the marks of 200 candidates were found to be 40 and 15 respectively. Later, it was discovered that a score of 40 was wrongly read as 50. The correct mean and S.D. respectively are

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a

14.98, 39.95

b

39.95, 24. 5

c

39.95, 14. 98

d

39.00, 14.00

answer is B.

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Detailed Solution

Corrected $\sum x = 40 \times 200 - 50 + 40 = 7990$
incorrect $\bar{x} = \frac{7990}{200} = 39.95$
correct $\sum x^{2} = n (σ^{2} + {\bar{x}}^{2}) = 200 (5^{2} + 40^{2}) = 365000$
corrected $σ = \sqrt{\frac{364100}{200} - {(39.95)}^{2}} = \sqrt{(1820.5 - 1596)}$
$= \sqrt{224.5} = 14.98$

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