Mean and S.D. of marks of 200 candidates were found to be 40 and 15 respectively. Later, it was disc

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13.Statistics

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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...

A

$14.98, 39.95$

B

$39.95, 14.98$

C

$39.95, 224.5$

D

None of these

Solution

(b) Corrected $\Sigma x = 40 \times 200 – 50 + 40 = 7990$

Corrected $\bar x = 7990/200$$ = 39.95$

Incorrect $\Sigma {x^2} = n\,[{\sigma ^2} + {\bar x^2}] = 200[{15^2} + {40^2}] = 365000$

Correct $\Sigma {x^2} = 365000 – 2500 + 1600$$ = 364100$

Corrected $\sigma = \sqrt {\frac{{364100}}{{200}} – {{(39.95)}^2}} $

$ = \sqrt {(1820.5 – 1596)} $$ = \sqrt {224.5} = 14.98$.

Standard 11

Mathematics

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