JEE Main 2019 — Statistics Question with Solution

Let, the two unknown terms are $x \& y.$

The mean is defined as $\mu =\frac{\sum x_i}{N}$

$\Rightarrow 5=\frac{1+3+8+x+y}{5}$

$\Rightarrow 25=12+x+y$

$\Rightarrow x+y=13 ...\left(1\right)$

And, the variance is defined as ${\sigma }^2=\frac{\sum {\left(x_i-\mu \right)}^2}{N}=\frac{\sum x_i^2}{N}-{\mu }^2$

$\Rightarrow 9.2=\frac{1+9+64+x^2+y^2}{5}-25$

$\Rightarrow 34.2\times 5=74+x^2+y^2$

$\Rightarrow 171=74+x^2+y^2$

$\Rightarrow 97=x^2+y^2 ...\left(2\right)$

Put the value of $y$ from equation $\left(1\right)$ and put in $\left(2\right),$ to get

$97=x^2+{\left(13-x\right)}^2$

$\Rightarrow 97=x^2+169-26x+x^2$

$\Rightarrow 2x^2-26x+72=0$

$\Rightarrow x^2-13x+36=0$

$\Rightarrow \left(x-4\right)\left(x-9\right)=0$

$\Rightarrow x=4$ or $x=9$

And, $y=9$ or $y=4$ respectively.

Hence, the ratio of $x : y$ is $4 : 9$ or $9 : 4.$