Question

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

Step1. Consider the observation :

Let the first $2n$observations be $x_{1,}{x}_2,..........,x_{2n}$and the last $n$observation be $y_1,y_2,........y_n$.

Since the mean of the first $2n$ numbers is $6$ and the mean of the remaining $n$ numbers is $3$.

$\Rightarrow \frac{{\displaystyle \underset{i=1}{\overset{2n}{\sum x_i}}}}{2n}=6,\frac{{\displaystyle \underset{i=1}{\overset{n}{\sum y_i}}}}{n}=3$

$$\begin{gathered} \underset{i=1}{\overset{2n}{\sum x_i}}=12n,\underset{i=1}{\overset{n}{\sum y_i}}=3n \\ \therefore \frac{\underset{i=1}{\overset{2n}{\sum x_i}}+\underset{i=1}{\overset{n}{\sum y_i}}}{3n}=\frac{15n}{3n}=5 \end{gathered}$$

Also, the set of $3n$ numbers has variance $4$.

$$\begin{gathered} \frac{\underset{i=1}{\overset{2n}{\sum {x_i}^2}}+\underset{i=1}{\overset{n}{\sum {y_i}^2}}}{3n}-5^2=4 \\ \Rightarrow \underset{i=1}{\overset{2n}{\sum {x_i}^2}}+\underset{i=1}{\overset{n}{\sum {y_i}^2}}=29\times 3n=87n \end{gathered}$$

Step 2. Apply the condition on mean and variance :

Now, mean is $\frac{\underset{i=1}{\overset{2n}{\sum (x_i+1)}}+\underset{i=1}{\overset{n}{\sum (y_i-1)}}}{3n}=\frac{15n+2n-n}{3n}=\frac{16}{3}$ ,& variance is

$$\begin{gathered} \frac{\underset{i=1}{\overset{2n}{\sum {(x_i+1)}^2}}+\underset{i=1}{\overset{n}{\sum {(y_i-1)}^2}}}{3n}-{\left(\frac{16}{3}\right)}^2 \\ =\frac{\underset{i=1}{\overset{2n}{\sum {x_i}^2}}+\underset{i=1}{\overset{n}{\sum {y_i}^2}}+2\left({\displaystyle \underset{i=1}{\overset{2n}{\sum x_i}}}-{\displaystyle \underset{i=1}{\overset{n}{\sum y_i}}}\right)+3n}{3n}-{\left(\frac{16}{3}\right)}^2 \\ =\frac{87n+2\left(9n\right)+3n}{3n}-{\left(\frac{16}{3}\right)}^2 \\ =\frac{108}{3}-{\left(\frac{16}{3}\right)}^2 \\ =\frac{\{324-256\}}{9} \\ =\frac{68}{9} \end{gathered}$$

Since the variance of a new set is $k$.

$$\begin{gathered} \Rightarrow k=\frac{68}{9} \\ \Rightarrow 9k=68 \end{gathered}$$

Therefore the value of $9k$ is $68$.