Question

initial knowledge check question 14 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 $x$ | $p(x = x)$---|---2 | 0.273 | 0.254 | 0.235 | 0.126 | 0.15

Explanation:

Step1: Sum given probabilities

$0.27 + 0.25 + 0.23 + 0.12 + 0.15 = 1.02$

Step2: Identify required total

A valid distribution needs $\sum P(X=x) = 1$

Step3: Calculate adjustment amount

$1.02 - 1 = 0.02$
We need to reduce the total by 0.02. We can subtract this from one of the probabilities; the most logical choice is the largest or the last value, here we adjust $P(X=6)$:
$0.15 - 0.02 = 0.13$

Step4: Verify new total

$0.27 + 0.25 + 0.23 + 0.12 + 0.13 = 1$

Answer:

Value $x$ of $X$	$P(X=x)$
3	0.25
4	0.23
5	0.12
6	0.13

(Note: Alternatively, the 0.02 could be distributed across multiple probabilities as long as all $P(X=x) \geq 0$ and the total equals 1; adjusting the final value is a simple valid solution.)