Question

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

Answer 1

Let first $2n$ observations be denoted by $x_i$ and last $n$ by $y_i$
We know
$\frac{\sum x_i}{2n}=6,\frac{\sum y_i}{n}=3{\sigma }^2=4$
The combined mean is
$\overline{x}=\frac{12n+3n}{3n}=5\Rightarrow {\sigma }^2=\frac{\sum x_i^2+\sum y_i^2}{3n}-(5{)}^2\Rightarrow \sum x_i^2+\sum y_i^2=29\times 3n=87n\cdots \left(1\right)$

Now, for new set $x_i^{\prime }=x_i+1,y_i^{\prime }=y_i-1$
$\frac{\sum x_i^{\prime }}{2n}=7,\frac{\sum y_i^{\prime }}{n}=2$
The combined mean is
$\overline{x^{\prime }}=\frac{14n+2n}{3n}=\frac{16}{3}$
So, the new variance is
$({\sigma }^{\prime }{)}^2=\frac{\sum {(x_i+1)}^2+\sum (y_i-1{)}^2}{3n}-{\left(\frac{16}{3}\right)}^2=\frac{\sum x_i^2+\sum y_i^2+2\left(\sum x_i-\sum y_i\right)+3n}{3n}-{\left(\frac{16}{3}\right)}^2=36-\frac{{16}^2}{9}\therefore 9k=(9\times 36)-{16}^2=68$