Consider a set of 3 n numbers having variance | vedclass

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Question

Let number be $a _{1}, a _{2}, a _{3}, \ldots \ldots a _{2 n }, b _{1}, b _{2}, b _{3} \ldots b _{ n }$

$\sigma^{2}=\frac{\sum a^{2}+\sum b^{2}}{3

Vedclass · Accepted Answer

$68$

Vedclass · Answer

Let number be $a _{1}, a _{2}, a _{3}, \ldots \ldots a _{2 n }, b _{1}, b _{2}, b _{3} \ldots b _{ n }$

$\sigma^{2}=\frac{\sum a^{2}+\sum b^{2}}{3 n}-(5)^{2}$

$\Rightarrow \sum a^{2}+\sum b^{2}=87 n$

Now, distribution becomes

$a _{1}+1, a _{2}+1, a _{3}+1, \ldots \ldots a _{2 n }+1, b _{1}-1,b_{2}-1 \ldots \ldots b_{n}-1$

Variance

$=\frac{\sum(a+1)^{2}+\sum(b-1)^{2}}{3 n}-\left(\frac{12 n+2 n+3 n-n}{3 n}\right)^{2}$

$=\frac{\left(\sum a^{2}+2 n+2 \sum a\right)+\left(\sum b^{2}+n-2 \sum b\right)}{3 n}$

$=\frac{\left(\sum a^{2}+2 n+2 \sum a\right)+\left(\sum b^{2}+n-2 \sum b\right)}{3 n}-\left(\frac{16}{3}\right)^{2}$

$=\frac{87 n+3 n+2(12 n)-2(3 n)}{3 n}-\left(\frac{16}{3}\right)^{2}$

$\Rightarrow k=\frac{108}{3}-\left(\frac{16}{5}\right)^{2}$

$\Rightarrow 9 k=3(108)-(16)^{2}=324-256=68$

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