Question

If the mean and variance of eight numbers $3,7,9,12,13,20,x$ and $y$ be $10$ and $25$ respectively then $xy$ is equal to

Answer 1

Compute the required value.

Step $1$: Find the equation between $x$ and $y$.

Given : The mean and variance of eight numbers is $10$ and $25$.

Since, mean $\left(\overline{x}\right)$is given by, $\begin{array}{rcl}& & \frac{{\displaystyle \underset{i=1}{\overset{n}{\sum x_i}}}}{n}\end{array}$

Thus,

$\begin{array}{rcl}\left(\overline{x}\right)& =& \frac{{\displaystyle \underset{i=1}{\overset{n}{\sum x_i}}}}{n}\\ & \Rightarrow & 10=\frac{3+7+9+12+13+20+x+y}{8}\\ & \Rightarrow & 80=64+x+y\\ & \Rightarrow & 16=x+y..........................\left(1\right)\end{array}$

Step $2$: Find the value of $xy$.

We know that variance $\left({\sigma }^2\right)$ is given by,

$\begin{array}{rcl}{\sigma }^2& =& \frac{{\displaystyle \sum _{i=1}^n}{\left(x_i\right)}^2}{n}-{\left(\overline{x}\right)}^2\\ & \Rightarrow & 25=\frac{\left(3^2+7^2+9^2+{12}^2+{13}^2+{20}^2+x^2+y^2\right)}{8}-{\left(10\right)}^2\\ & \Rightarrow & 200=9+49+81+144+169+400+x^2+y^2-800\\ & \Rightarrow & 1000=852+x^2+y^2\\ & \Rightarrow & 148={(x+y)}^2-2xy\\ & \Rightarrow & 148=256-2xy\left[Fromequation1\right]\\ & \Rightarrow & 2xy=108\\ & \Rightarrow & xy=54\end{array}$

Hence the value of $xy$ is equal to $54$.