The mean and variance of the marks obtained b | vedclass

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(C) Let the number of students be $n$. The initial mean $\bar{x} = 10$ and variance $\sigma^2 = 4$.$\bar{x} = \frac{\sum x_i}{n} = 10 \implies \sum x_

Vedclass · Accepted Answer

$3.96$

Vedclass · Answer

(C) Let the number of students be $n$. The initial mean $\bar{x} = 10$ and variance $\sigma^2 = 4$.$\bar{x} = \frac{\sum x_i}{n} = 10 \implies \sum x_i = 10n$.$\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2 = 4 \implies \frac{\sum x_i^2}{n} - 100 = 4 \implies \sum x_i^2 = 104n$.When one student's mark changes from $8$ to $12$,the new sum of marks is $\sum x_i' = 10n - 8 + 12 = 10n + 4$.The new mean is $\frac{10n + 4}{n} = 10.2 \implies 10n + 4 = 10.2n \implies 0.2n = 4 \implies n = 20$.Now,the new sum of squares is $\sum x_i'^2 = \sum x_i^2 - 8^2 + 12^2 = 104(20) - 64 + 144 = 2080 + 80 = 2160$.The new variance is $\sigma'^2 = \frac{\sum x_i'^2}{n} - (\bar{x}')^2 = \frac{2160}{20} - (10.2)^2 = 108 - 104.04 = 3.96$.

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 variance of the marks obtained by $n$ students in a test are $10$ and $4$ respectively. Later,the marks of one of the students is increased from $8$ to $12$. If the new mean of the marks is $10.2$,then their new variance is equal to:

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