The probability distribution of X is: X 0 1 2 3 P ( X ) 1 - d 4 1 + 2 d 4 1 - 4 d 4 1 + 3 d 4 For the minimum possible value of d , sixty times the mean of X is equal to

Question

The probability distribution of $X$ is:

$X$	$0$	$1$	$2$	$3$
$P (X)$	$\frac{1 - d}{4}$	$\frac{1 + 2 d}{4}$	$\frac{1 - 4 d}{4}$	$\frac{1 + 3 d}{4}$

For the minimum possible value of $d$ , sixty times the mean of $X$ is equal to _______.

Smart Achievers · Accepted Answer

(75)

$0 \leq \frac{1 - d}{4} \leq 1 \Rightarrow - 3 \leq d \leq 1$ $\dots (i)$

$0 \leq \frac{1 + 2 d}{4} \leq 1 \Rightarrow - \frac{1}{2} \leq d \leq \frac{3}{2}$ $\dots (i i)$

$0 \leq \frac{1 - 4 d}{4} \leq 1 \Rightarrow - \frac{3}{4} \leq d \leq \frac{1}{4}$ $\dots (i i i)$

$0 \leq \frac{1 + 3 d}{4} \leq 1 \Rightarrow - \frac{1}{3} \leq d \leq 1$ $\dots (i v)$

$From (i), (i i), (i i i) and (i v)$

$- \frac{1}{3} \leq d \leq \frac{1}{4}$

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

$Mean = 0 + \frac{1 + 2 d}{4} + \frac{2 (1 - 4 d)}{4} + \frac{3 (1 + 3 d)}{4}$

$X = \frac{6 + 3 d}{4} = \frac{1}{4} (6 - 3 \times \frac{1}{3}) = \frac{5}{4}$

$60 \bar{X} = 60 \times \frac{5}{4} = 75$

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