QJEEM22Ma2d34: JEE Main 2022 Mathematics - Class 12 JEEM22Ma2d34 Solution: The probability distribution of X is : .tg…

We know, $0 \le P(x) \le 1$

$\therefore$ $0 \le {{1 - d} \over 4} \le 1$

$\Rightarrow 0 \le 1 - d \le 4$

$\Rightarrow - 1 \le - d \le 3$

$\Rightarrow 1 \ge d \ge - 3$

Also,

$0 \le {{1 + 2d} \over 4} \le 1$

$\Rightarrow 0 \le 1 + 2d \le 4$

$\Rightarrow - 1 \le 2d \le 3$

$\Rightarrow - {1 \over 2} \le d \le {3 \over 2}$

Also,

$0 \le {{1 - 4d} \over 4} \le 1$

$\Rightarrow 0 \le 1 - 4d \le 4$

$\Rightarrow - 1 \le - 4d \le 3$

$\Rightarrow 1 \ge 4d \ge - 3$

$\Rightarrow {1 \over 4} \ge d \ge - {3 \over 4}$

And,

$0 \le {{1 + 3d} \over 4} \le 1$

$\Rightarrow 0 \le 1 + 3d \le 4$

$\Rightarrow - 1 \le 3d \le 3$

$\Rightarrow - {1 \over 3} \le d \le 1$

Common range is $= - {1 \over 3}$ to ${1 \over 4}$

$\therefore$ $d\, \in \left[ { - {1 \over 3},{1 \over 4}} \right]$

$\therefore$ Minimum value of $d = - {1 \over 3}$

We know, mean

$E(x) = \sum {x\,.\,P(x)}$

$= 0 \times {{1 - d} \over 4} + 1 \times {{1 + 2d} \over 4} + 2 \times {{1 - 4d} \over 4} + 3 \times {{1 + 3d} \over 4}$

$= {{1 + 2d + 2 - 8d + 3 + 9d} \over 4}$

$= {{6 + 3d} \over 4}$

For $d = - {1 \over 3}$ , $E(x) = {{6 + 3 \times - {1 \over 3}} \over 4} = {5 \over 4}$

$\therefore$ $60E(x) = 60 \times {5 \over 4} = 75$