Consider a set of $3 n$ numbers having variance 4 . In this set, the mean of first $2 n$ numbers is

Question

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

Professor PC Thomas Classes · Answer

(68) : We have, $x_{1}+x_{2}+\ldots \ldots \ldots .$.\ $+x_{2 n}=12 n$\ and $x_{2 n+1}+x_{2 n+2}+\ldots \ldots \ldots . .+x_{3 n}=3 n$\ $\Rightarrow$ Combined mean, $\bar{x}=\frac{12 n+3 n}{3 n}=5$\ $\Rightarrow \quad x_{1}+x_{2}+\ldots \ldots \ldots \ldots .+x_{3 n}=15 n$\ Also, $4=\frac{\sum x_{i}^{2}}{3 n}-25 \Rightarrow \frac{\sum x_{i}^{2}}{3 n}=29$\ Now, new mean, $\bar{y}$ $$ \begin{aligned} & \frac{\left(x_{1}+x_{2}+\ldots+x_{2 n}+2 night)+\left(\begin{array}{l} x_{2 n+1}+x_{2 n+2} \ \left.+\ldots+x_{3 n}-night) \end{array}ight.}{3 n} \ \Rightarrow \quad & \bar{y}=\frac{15 n+n}{3 n}=\frac{16}{3} \end{aligned} $$ New variance, $\sigma_{y}^{2}=\frac{\sum y_{i}^{2}}{3 n}-\left(\frac{16}{3}ight)^{2}$\ $\Rightarrow k=\frac{\sum_{i=1}^{3 n} x_{i}^{2}}{3 n}+2\left(\frac{\sum_{i=1}^{2 n} x_{i}}{3 n}ight)-2\left(\frac{\sum_{i=2 n+1}^{3 n} x_{i}}{3 n}ight)$\ $+1-\frac{256}{9}$\ $\Rightarrow \quad k=29+8-2+1-\frac{256}{9}=\frac{68}{9}$\ $\Rightarrow \quad 9 k=68$\