The mean and S.D. of the marks of 200 candida | vedclass

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Question

(B) Given $n = 200$,incorrect mean $\bar{x} = 40$,incorrect $S.D. \sigma = 15$.Incorrect $\Sigma x = n 	imes \bar{x} = 200 	imes 40 = 8000$.Correcte

Vedclass · Accepted Answer

$39.95, 14.98$

Vedclass · Answer

(B) Given $n = 200$,incorrect mean $\bar{x} = 40$,incorrect $S.D. \sigma = 15$.Incorrect $\Sigma x = n 	imes \bar{x} = 200 	imes 40 = 8000$.Corrected $\Sigma x = 8000 - 50 + 40 = 7990$.Corrected mean $\bar{x}_{new} = \frac{7990}{200} = 39.95$.Incorrect $\Sigma x^2 = n(\sigma^2 + \bar{x}^2) = 200(15^2 + 40^2) = 200(225 + 1600) = 200(1825) = 365000$.Corrected $\Sigma x^2 = 365000 - 50^2 + 40^2 = 365000 - 2500 + 1600 = 364100$.Corrected $\sigma_{new} = \sqrt{\frac{\Sigma x^2}{n} - (\bar{x}_{new})^2} = \sqrt{\frac{364100}{200} - (39.95)^2}$.$\sigma_{new} = \sqrt{1820.5 - 1596.0025} = \sqrt{224.4975} \approx 14.98$.

List-$I$	List-$II$
$(P)$ The mean of the above data is	$(1) 2.5$
$(Q)$ The median of the above data is	$(2) 5$
$(R)$ The mean deviation about the mean of the above data is	$(3) 6$
$(S)$ The mean deviation about the median of the above data is	$(4) 2.7$
	$(5) 2.4$

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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