Question

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

Answer 1

Hint: Now we know that the mean of n values is given by $\dfrac{\text{sum of values }}{\text{number of values}}$ hence using this formula we can find the value of x + y. Now we also know that the formula for variance of ${{x}_{1}},{{x}_{2}},.....{{x}_{n}}$ is given by the formula $\dfrac{{{x}_{1}}^{2}+{{x}_{2}}^{2}+...+{{x}_{n}}^{2}}{n}-{{\mu }^{2}}$ where $\mu $ is the mean of ${{x}_{1}},{{x}_{2}},.....{{x}_{n}}$ Hence using this we will find the value of ${{x}^{2}}+{{y}^{2}}$ . Now we will use the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and then substitute the value of x + y to find the value of x.y

Complete step by step answer:
Now first we know that the mean of eight numbers 3, 7, 9, 12, 13, 20, x and y is 10.
The mean of n values is given by $\dfrac{\text{sum of values }}{\text{number of values}}$ . Hence using this we get
$\dfrac{3+7+9+12+13+20+x+y}{8}=10$
Multiplying the equation by 8 we get
64 + x + y = 80.
Hence x + y = 80 – 64 = 16
Hence we get x + y = 16 ……………………………… (1)
Now we are also given that the variance if data is 25.
Variance of data ${{x}_{1}},{{x}_{2}},.....{{x}_{n}}$ is given by the formula $\dfrac{{{x}_{1}}^{2}+{{x}_{2}}^{2}+...+{{x}_{n}}^{2}}{n}-{{\mu }^{2}}$ where $\mu $ is the mean of ${{x}_{1}},{{x}_{2}},.....{{x}_{n}}$
Now consider the given values
3, 7, 9, 12, 13, 20, x and y
Their variance is given as 25 hence we get
$\begin{align}
  & \dfrac{{{3}^{2}}+{{7}^{2}}+{{9}^{2}}+{{12}^{2}}+{{13}^{2}}+{{20}^{2}}+{{x}^{2}}+{{y}^{2}}}{8}-{{10}^{2}}=25 \\
 & \Rightarrow \dfrac{9+49+81+144+169+400+{{x}^{2}}+{{y}^{2}}}{8}=125 \\
 & \Rightarrow 852+{{x}^{2}}+{{y}^{2}}=1000 \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}=148 \\
\end{align}$
Now we know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
Hence we get ${{\left( x+y \right)}^{2}}-2xy=148$
Now substituting the value from equation (1) we get.
${{16}^{2}}-2xy=148$
$\begin{align}
  & \Rightarrow 256-148=2xy \\
 & \Rightarrow xy=\dfrac{108}{2} \\
\end{align}$

Hence we get x.y = 54.

Note: Note that in this solution we get two equations in x and y, Here we have made use of formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ to simplify the equation but we can also solve them by substituting the values of x or y from equation (1) and then solving the quadratic equation.