Questions: Does the following table represent a valid discrete probability distribution? x 1 2 3 4 5 P(X=x) 0.11 0.06 0.18 0.06 0.96 yes no

Transcript text: Does the following table represent a valid discrete probability distribution? $x$ & 1 & 2 & 3 & 4 & 5 $P(X=x)$ & 0.11 & 0.06 & 0.18 & 0.06 & 0.96 yes no

Solution

Solution Steps

To determine if the table represents a valid discrete probability distribution, we need to check two conditions:

Each probability $ P(X=x) $ must be between 0 and 1, inclusive.
The sum of all probabilities must equal 1.

Solution Approach

Verify that each probability value is between 0 and 1.
Calculate the sum of all probability values and check if it equals 1.

Step 1: Verify Probability Values

To determine if the table represents a valid discrete probability distribution, we first check that each probability $ P(X=x) $ is between 0 and 1. The given probabilities are $ 0.11, 0.06, 0.18, 0.06, $ and $ 0.96 $. All these values satisfy the condition $ 0 \leq P(X=x) \leq 1 $.

Step 2: Calculate the Sum of Probabilities

Next, we calculate the sum of all the probabilities: \[ 0.11 + 0.06 + 0.18 + 0.06 + 0.96 = 1.370 \] For a valid discrete probability distribution, the sum of probabilities must be exactly 1. Here, the sum is $ 1.370 $, which is not equal to 1.

Final Answer

Since the sum of the probabilities is not equal to 1, the table does not represent a valid discrete probability distribution. Therefore, the answer is $\boxed{\text{no}}$.

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