JEE Main 2021 (Online) 22nd July Evening ShiftQuestion 1
Consider the following frequency distribution :
Class : $0-6$, $6-12$, $12-18$, $18-24$, $24-30$
Frequency : $a$, $b$, $12$, $9$, $5$
If mean $=\frac{309}{22}$ and median $=14$, then the value $(a-b)^{2}$ is equal to $\_\_\_\_\_\_$ .
Statistics
Answer$4$
Solution
$N=a+b+26$. Median class $12$-$18$: $12+\frac{N/2-(a+b)}{12}\cdot6=14$ gives $a+b=18$. Mean: $\frac{3a+9b+504}{44}=\frac{309}{22}$ gives $a+3b=38$. So $b=10$, $a=8$, $(a-b)^2=4$.