The mean and standard deviation of 100 observ | vedclass

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(A) Given that number of observations $(n) = 100$.Incorrect mean $(\bar{x}) = 40$.Incorrect standard deviation $(\sigma) = 5.1$.We know that $\bar{x}

Vedclass · Accepted Answer

$39.9, 5$

Vedclass · Answer

(A) Given that number of observations $(n) = 100$.Incorrect mean $(\bar{x}) = 40$.Incorrect standard deviation $(\sigma) = 5.1$.We know that $\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i$.$40 = \frac{1}{100} \sum_{i=1}^{100} x_i \implies \sum_{i=1}^{100} x_i = 4000$.Incorrect sum of observations $= 4000$.Correct sum of observations $= 4000 - 50 + 40 = 3990$.Correct mean $= \frac{3990}{100} = 39.9$.Also,standard deviation $\sigma = \sqrt{\frac{1}{n} \sum x_i^2 - (\bar{x})^2}$.$5.1 = \sqrt{\frac{1}{100} \sum x_i^2 - (40)^2}$.$26.01 = \frac{1}{100} \sum x_i^2 - 1600$.Incorrect $\sum x_i^2 = 100(26.01 + 1600) = 162601$.Correct $\sum x_i^2 = 162601 - (50)^2 + (40)^2 = 162601 - 2500 + 1600 = 161701$.Correct standard deviation $= \sqrt{\frac{161701}{100} - (39.9)^2} = \sqrt{1617.01 - 1592.01} = \sqrt{25} = 5$.

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	$(4) 2.7$
	$(5) 2.4$

Income group	No. of firms
$150-300$	$300$
$300-500$	$500$
$500-800$	$900$
$800-1200$	$1000$
$1200-1800$	$1200$

$X$	$1$	$3$	$5$	$7$	$9$
$f$	$4$	$24$	$28$	$\alpha$	$8$

The mean and standard deviation of $100$ observations were calculated as $40$ and $5.1$,respectively,by a student who took by mistake $50$ instead of $40$ for one observation. What are the correct mean and standard deviation?

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