JEE Main 2021 — Statistics Question with Solution

We know that, mean$=\frac{\sum f_i{x}_i}{\sum f_i}$

Given, mean$=\frac{309}{22}$

$\Rightarrow \frac{3a+9 b+180+189+135}{a+b+26}=\frac{309}{22}$

$\Rightarrow \frac{504+3a+9 b}{a+b+26}=\frac{309}{22}$

$\Rightarrow 66a+198 b+11088=309a+309 b+8034$

$\Rightarrow 243a+111 b=3054$

$\Rightarrow 81a+37b=1018 ...\left(1\right)$

Now, median$=14,$ which lies in the interval $12-18,$ thus $12-18$ is the median class and we have $=l+\frac{\left({\displaystyle \frac{N}{2}}-c\right)\times h}{f},$

Where, $l$ is the lower limit of the median class i.e. $l=12, c$ is the cumulative frequency of the class above the median class i.e. $c=a+b, f$ is the frequency of the median class i.e. $f=12$ and $h$ is the length of the interval i.e. $h=6.$

Thus, we have $12+\left(\frac{\left({\displaystyle \frac{a+b+26}{2}}\right)-(a+b)}{12}\right)\times 6=14$

$\Rightarrow \frac{\left({\displaystyle \frac{a+b+26-2a-2b}{2}}\right)}{2}=14-12$

$\Rightarrow \frac{26-a-b}{4}=2$

$\Rightarrow 26-a-b=8$

$\Rightarrow a+b=18 ...\left(2\right)$

On solving the equations $\left(1\right)$ and $\left(2\right),$ we get $a=8, b=10.$

$\therefore {\left(a-b\right)}^2={\left(8-10\right)}^2=4.$