Question

Determine whether the table represents a discrete probability distribution. Explain why or why not.
x| -2 0 2 4
p(x)| .35 .15 .25 .15
The table [blank] represent a discrete probability distribution because [blank]

Answer 1

wrong

Answer 2

The table does not represent a discrete probability distribution because all probabilities are between 0 and 1, but the probabilities do not add up to 1.

Answer 3

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

1. Each probability \( p(x) \) must be between 0 and 1 (inclusive).
2. The sum of all probabilities must equal 1.

Looking at the provided values:

\[
\begin{align*}
x & : -2 & 0 & 2 & 4 \\
p(x) & : 0.35 & 0.15 & 0.25 & 0.15 \\
\end{align*}
\]

1. **Check if each \( p(x) \) is between 0 and 1:**
- \( 0.35 \geq 0 \) and \( 0.35 \leq 1 \) ✔
- \( 0.15 \geq 0 \) and \( 0.15 \leq 1 \) ✔
- \( 0.25 \geq 0 \) and \( 0.25 \leq 1 \) ✔
- \( 0.15 \geq 0 \) and \( 0.15 \leq 1 \) ✔

All probabilities are within the range of 0 to 1.

2. **Sum of the probabilities:**
\[
0.35 + 0.15 + 0.25 + 0.15 = 1.00
\]

The sum is equal to 1.

Since both criteria are satisfied, we can conclude:

The table **does** represent a discrete probability distribution because all probabilities are between 0 and 1, and the sum of the probabilities equals 1.

Answer 4

I apologize for any confusion in my previous response. Let’s carefully analyze the table again to ensure accuracy in determining whether it represents a discrete probability distribution.

Given:

\[
\begin{array}{c|cccc}
x & -2 & 0 & 2 & 4 \\
\hline
p(x) & 0.35 & 0.15 & 0.25 & 0.15 \\
\end{array}
\]

1. **Probabilities Between 0 and 1:**
Check if each probability \( p(x) \) falls within the range of 0 and 1.
- \( 0.35 \) is between 0 and 1 ✔
- \( 0.15 \) is between 0 and 1 ✔
- \( 0.25 \) is between 0 and 1 ✔
- \( 0.15 \) is between 0 and 1 ✔

All probabilities are valid.

2. **Sum of the Probabilities:**
Calculate the sum of all probabilities to see if it equals 1.
\[
0.35 + 0.15 + 0.25 + 0.15 = 1.00
\]

The sum is equal to 1.

Since both conditions are met, I can confirm that:

The table **does** represent a discrete probability distribution because all probabilities are between 0 and 1, and the sum of the probabilities equals 1.

If this doesn't resolve your concern, please clarify further so I can assist you accurately.

Answer 5

Thank you for your patience, and I apologize for the earlier oversight in my calculations.

Let's break it down correctly regarding whether the table represents a discrete probability distribution:

### Given Table:

\[
\begin{align*}
x & : -2 & 0 & 2 & 4 \\
p(x) & : 0.35 & 0.15 & 0.25 & 0.15 \\
\end{align*}
\]

### Review the Criteria for a Discrete Probability Distribution:

1. **Each probability \( p(x) \) must be between 0 and 1 (inclusive).**
- \( p(-2) = 0.35 \) ✔
- \( p(0) = 0.15 \) ✔
- \( p(2) = 0.25 \) ✔
- \( p(4) = 0.15 \) ✔

All probabilities are valid as they are between 0 and 1.

2. **The sum of all probabilities must equal 1.**
- Let's sum the probabilities:
\[
0.35 + 0.15 + 0.25 + 0.15 = 0.90
\]

The total sum equals 0.90, which is **not equal to 1**.

### Conclusion:

The table **does not** represent a discrete probability distribution because, although all probabilities are between 0 and 1, they do not add up to 1.

Thank you for your guidance in correcting this analysis!