Question

If the mean and variance of $7$ variates are $8$ and $16$ respectively and five of them are $2,4,10,12,14$ then the product of the remaining two varieties is:

• A. $49$
• B. $48$
• C. $45$
• D. $40$

Answer 1

The correct option is B

$48$

Explanation for the correct answer:

Step 1: Consider the mean of the variates

The given five variates are $2,4,10,12,14$.

Let us assume that the remaining two variates are $x$ and $y$.

The mean of the $7$ variates are $m=\frac{2+4+10+12+14+x+y}{7}$.

$$\begin{gathered} \Rightarrow 8=\frac{2+4+10+12+14+x+y}{7} \\ \Rightarrow 42+x+y=56 \\ \Rightarrow x+y=14 \end{gathered}$$

Step 2: Consider the variance of the variates, and calculating the product

The variance of $7$ variates can be given by, ${\sigma }^2=\frac{{\left(2\right)}^2+{\left(4\right)}^2+{\left(10\right)}^2+{\left(12\right)}^2+{\left(14\right)}^2+x^2+y^2}{7}-m^2$.

It is given that, the variance of the variates, ${\sigma }^2=16$.

$$\begin{gathered} \Rightarrow 16=\frac{4+16+100+144+196+x^2+y^2}{7}-{\left(8\right)}^2 \\ \Rightarrow 16=\frac{460+x^2+y^2}{7}-64 \\ \Rightarrow \frac{460+x^2+y^2}{7}=80 \\ \Rightarrow x^2+y^2+460=560 \\ \Rightarrow x^2+y^2=100 \\ \Rightarrow {\left(x+y\right)}^2-2xy=100 \\ \Rightarrow {\left(14\right)}^2-2xy=100\left[\because x+y=14\right] \\ \Rightarrow 196-2xy=100 \\ \Rightarrow -2xy=-96 \\ \Rightarrow xy=48 \end{gathered}$$

Therefore, the product of the remaining two varieties is $48$.

Hence, option B is the correct answer.