The mean and standard deviation of 20 observa | vedclass

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Question

(B) Given $n = 20$,Incorrect Mean $\bar{x} = 10$,Incorrect Standard Deviation $\sigma = 2$.Incorrect sum of observations $\sum x_i = n 	imes \bar{x}

Vedclass · Accepted Answer

Mean $= 10.2$,Standard Deviation $= 1.9899$

Vedclass · Answer

(B) Given $n = 20$,Incorrect Mean $\bar{x} = 10$,Incorrect Standard Deviation $\sigma = 2$.Incorrect sum of observations $\sum x_i = n 	imes \bar{x} = 20 	imes 10 = 200$.Correct sum of observations $= 200 - 8 + 12 = 204$.Correct mean $= \frac{204}{20} = 10.2$.Using the formula $\sigma^2 = \frac{1}{n} \sum x_i^2 - (\bar{x})^2$,we have $2^2 = \frac{1}{20} \sum x_i^2 - 10^2$.$4 = \frac{1}{20} \sum x_i^2 - 100 \Rightarrow \sum x_i^2 = 20 	imes 104 = 2080$.Correct $\sum x_i^2 = 2080 - 8^2 + 12^2 = 2080 - 64 + 144 = 2160$.Correct standard deviation $= \sqrt{\frac{2160}{20} - (10.2)^2} = \sqrt{108 - 104.04} = \sqrt{3.96} \approx 1.9899$.

Value	$4$	$5$	$8$	$9$	$6$	$12$	$11$
Frequency	$5$	$f_1$	$f_2$	$2$	$1$	$1$	$3$

List-$I$	List-$II$
$(P) \ 7f_1+9f_2$ is equal to	$(1) \ 146$
$(Q) \ 19\alpha$ is equal to	$(2) \ 47$
$(R) \ 19\beta$ is equal to	$(3) \ 48$
$(S) \ 19\sigma^2$ is equal to	$(4) \ 145$
	$(5) \ 55$

The mean and standard deviation of $20$ observations are found to be $10$ and $2$ respectively. On rechecking,it was found that an observation $8$ was incorrect. If the incorrect observation $8$ is replaced by $12$,calculate the correct mean and standard deviation.

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