Fill in the P=Xx values to give a legitimate probability distribution for the discrete random variable X, whose possible values are −3, 1, 4, 5, and 6.ValuexofX P=Xx−3 0.171 4 0.165 6 0.18

The sum of all probabilities in a probability distribution must equal 1. Given that the probabilities for X=-3, X=4, and X=6 are 0.17, 0.165, and 0.18 respectively, we can find the remaining probabilities by subtracting these given probabilities from 1.

Step 1: Add the given probabilities 0.17 + 0.165 + 0.18 = 0.515

Step 2: Subtract the sum from 1 1 - 0.515 = 0.485

The sum of the probabilities for X=1 and X=5 must equal 0.485. Without additional information, we cannot determine the individual probabilities for X=1 and X=5. However, if we assume that they are equal, we can divide 0.485 by 2 to find each probability.

Step 3: Divide the remaining probability by 2 0.485 / 2 = 0.2425

So, the probability distribution could be:

X=-3, P(X) = 0.17 X=1, P(X) = 0.2425 X=4, P(X) = 0.165 X=5, P(X) = 0.2425 X=6, P(X) = 0.18

This distribution is legitimate because the sum of all probabilities is 1.