Determine whether the table represents a discrete probability distribution. Explain why or why not.

Question

x| -2 0 2 4
p(x)| .35 .15 .25 .15
The table [blank] represent a discrete probability distribution because [blank]

Answer 1

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:

$$\begin{array}{rlrlrlrlr}x& :-2& 0& 2& 4p(x)& :0.35& 0.15& 0.25& 0.15\end{array}$$

1. Check if each $p(x)$ is between 0 and 1:
   • $0.35\ge 0$ and $0.35\le 1$ ✔
   • $0.15\ge 0$ and $0.15\le 1$ ✔
   • $0.25\ge 0$ and $0.25\le 1$ ✔
   • $0.15\ge 0$ and $0.15\le 1$ ✔

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 2

wrong

Answer 3

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:

$$\text{Misplaced hline}$$

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 4

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 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:

$$\begin{array}{rlrlrlrlr}x& :-2& 0& 2& 4p(x)& :0.35& 0.15& 0.25& 0.15\end{array}$$

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!