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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...
$14.98, 39.95$
$39.95, 14.98$
$39.95, 224.5$
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$.
Similar Questions
The diameters of circles (in mm) drawn in a design are given below:
| Diameters | $33-36$ | $37-40$ | $41-44$ | $45-48$ | $49-52$ |
| No. of circles | $15$ | $17$ | $21$ | $22$ | $25$ |
Calculate the standard deviation and mean diameter of the circles.
[ Hint : First make the data continuous by making the classes as $32.5-36.5,36.5-40.5,$ $40.5-44.5,44.5-48.5,48.5-52.5 $ and then proceed.]
