Questions: Given the table of probabilities for the random variable x, does this form a probability distribution? X 0 1 2 3 4 P(x) 0.02 0.07 0.22 0.27 0.42 Yes No

Solution Steps

We need to verify that all probabilities \( P(x) \) are non-negative. The given probabilities are:

\[ P(0) = 0.02, \quad P(1) = 0.07, \quad P(2) = 0.22, \quad P(3) = 0.27, \quad P(4) = 0.42 \]

Since all values are greater than or equal to \( 0 \), we conclude that:

\[ \text{All } P(x) \geq 0 \quad \text{(True)} \]

Next, we calculate the sum of the probabilities:

\[ \text{Sum} = P(0) + P(1) + P(2) + P(3) + P(4) = 0.02 + 0.07 + 0.22 + 0.27 + 0.42 \]

Calculating this gives:

\[ \text{Sum} = 0.02 + 0.07 + 0.22 + 0.27 + 0.42 = 1.00 \]

Since the sum equals \( 1 \), we have:

\[ \text{Sum of probabilities} = 1 \quad \text{(True)} \]

Since both conditions are satisfied (non-negativity and sum equals 1), we conclude that the given probabilities form a valid probability distribution.

Final Answer