Question

fill in the $p(x = x)$ values to give a legitimate probability distribution for the discrete random variable $x$, whose possible values are 2, 3, 4, 5, and 6.

value $x$ of $x$	$p(x = x)$
3	0.25
4	0.23
5	0.12
6	0.12

Explanation:

Step1: Sum given probabilities

$0.27 + 0.25 + 0.21 + 0.12 = 0.85$

Step2: Calculate missing probability

$1 - 0.85 = 0.15$

Step3: Verify total probability

$0.27 + 0.25 + 0.21 + 0.12 + 0.15 = 1$

Answer:

The missing $P(X=6)$ value is $0.15$, so the completed table is:

value $x$ of $X$	$P(X=x)$
3	0.25
4	0.21
5	0.12
6	0.15