JEE Main 2024 (Online) 8th April Evening ShiftQuestion 1
Let $a,b,c\in\mathbf{N}$ and $a<b<c$. Let the mean, the mean deviation about the mean and the variance of the $5$ observations $9,25,a,b,c$ be $18$, $4$ and $\frac{136}{5}$, respectively. Then $2a+b-c$ is equal to $\_\_\_\_\_\_$ .
Statistics
Answer$33$
Solution
Sum $=90\Rightarrow a+b+c=56$. Deviations of $a,b,c$ from 18: $d_1,d_2,d_3$ (integers) with $\sum d=2$. $\sum(x-18)^2=5\cdot\frac{136}{5}=136$: $81+49+\sum d^2=136\Rightarrow\sum d^2=6$. MD: $9+7+\sum|d|=20\Rightarrow\sum|d|=4$. So $d=\{-1,1,2\}$ in increasing order: $a=17,b=19,c=20$. $2a+b-c=34+19-20=33$.