# Consider a set of 3 n numbers having

Question:

Consider a set of $3 \mathrm{n}$ numbers having variance $4 .$ In this set, the mean of first $2 \mathrm{n}$ numbers is 6 and the mean of the remaining numbers is 3 . A new - set is constructed by adding 1 into each of first $2 \mathrm{n}$ numbers, and subtracting 1 from each of the remaining $\mathrm{n}$ numbers. If the variance of the new set is $\mathrm{k}$, then $9 \mathrm{k}$ is equal to__________.

Solution:

Let number be $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots \ldots \mathrm{a}_{2 \mathrm{n}}, \mathrm{b}_{1}, \mathrm{~b}_{2}, \mathrm{~b}_{3} \ldots \mathrm{b}_{\mathrm{n}}$

$\sigma^{2}=\frac{\sum a^{2}+\sum b^{2}}{3 n}-(5)^{2}$

$\Rightarrow \sum a^{2}+\sum b^{2}=87 n$

Now, distribution becomes $\mathrm{a}_{1}+1, \mathrm{a}_{2}+1, \mathrm{a}_{3}+1, \ldots \ldots \mathrm{a}_{2 \mathrm{n}}+1, \mathrm{~b}_{1}-1$

$\mathrm{b}_{2}-1 \ldots \ldots \mathrm{b}_{\mathrm{n}}-1$

Variance

$=\frac{\sum(a+1)^{2}+\sum(b-1)^{2}}{3 n}-\left(\frac{12 n+2 n+3 n-n}{3 n}\right)^{2}$

$=\frac{\left(\sum a^{2}+2 n+2 \sum a\right)+\left(\sum b^{2}+n-2 \sum b\right)}{3 n}$

$=\frac{\left(\sum a^{2}+2 n+2 \sum a\right)+\left(\sum b^{2}+n-2 \sum b\right)}{3 n}-\left(\frac{16}{3}\right)^{2}$

$=\frac{87 n+3 n+2(12 n)-2(3 n)}{3 n}-\left(\frac{16}{3}\right)^{2}$

$\Rightarrow k=\frac{108}{3}-\left(\frac{16}{5}\right)^{2}$

$\Rightarrow 9 k=3(108)-(16)^{2}=324-256=68$