College students are randomly selected and arranged in groups of three. The random variable x is the number in the group who say that they take one or more online courses. Determine whether a probability distribution is given. If a probability distribution is​ given, find its mean and standard deviation. If a probability distribution is not​ given, identify the requirements that are not satisfied.
x
​P(x)

0
0.103
1
0.354
2
0.395
3
0.148
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Part 1
Does the table show a probability​ distribution? Select all that apply.
A.
​Yes, the table shows a probability distribution.
B.
​No, the numerical values of the random variable x are not associated with probabilities.
C.
​No, the sum of all the probabilities is not equal to 1.
D.
​No, not every probability is between 0 and 1 inclusive.
E.
​No, the random variable x is categorical instead of numerical.

To determine if the table shows a probability distribution, we need to check the following requirements:

1. The sum of all probabilities must be equal to 1.
2. Each probability must be between 0 and 1 inclusive.

Given the table:

\[
\begin{array}{|c|c|}
\hline
x & P(x) \\
\hline
0 & 0.103 \\
1 & 0.354 \\
2 & 0.395 \\
3 & 0.148 \\
\hline
\end{array}
\]

1. Check if the sum of all probabilities is equal to 1:
   \[
   0.103 + 0.354 + 0.395 + 0.148 = 1.000
   \]
   The sum is 1, so this requirement is satisfied.

2. Check if each probability is between 0 and 1 inclusive:
   \[
   0.103 \quad (0 \leq 0.103 \leq 1) \\
   0.354 \quad (0 \leq 0.354 \leq 1) \\
   0.395 \quad (0 \leq 0.395 \leq 1) \\
   0.148 \quad (0 \leq 0.148 \leq 1)
   \]
   Each probability is between 0 and 1 inclusive, so this requirement is satisfied.

Since both requirements are satisfied, the table does show a probability distribution.

So, the correct answer is:
A. Yes, the table shows a probability distribution.

### Part 2: Finding the Mean and Standard Deviation

To find the mean (μ) of the probability distribution, use the formula:
\[
\mu = \sum [x \cdot P(x)]
\]

Calculating:
\[
\mu = 0 \cdot 0.103 + 1 \cdot 0.354 + 2 \cdot 0.395 + 3 \cdot 0.148
\]
\[
\mu = 0 + 0.354 + 0.79 + 0.444
\]
\[
\mu = 1.588
\]

To find the variance (σ²) and then the standard deviation (σ), use the formulas:
\[
\sigma^2 = \sum [x^2 \cdot P(x)] - \mu^2
\]
\[
\sigma = \sqrt{\sigma^2}
\]

Calculating:
\[
\sigma^2 = (0^2 \cdot 0.103) + (1^2 \cdot 0.354) + (2^2 \cdot 0.395) + (3^2 \cdot 0.148) - \mu^2
\]
\[
\sigma^2 = (0) + (0.354) + (4 \cdot 0.395) + (9 \cdot 0.148) - 1.588^2
\]
\[
\sigma^2 = 0.354 + 1.58 + 1.332 - 2.523744
\]
\[
\sigma^2 = 3.266 - 2.523744
\]
\[
\sigma^2 = 0.742256
\]
\[
\sigma = \sqrt{0.742256}
\]
\[
\sigma \approx 0.8615
\]

So, the mean (μ) is approximately 1.588, and the standard deviation (σ) is approximately 0.8615.

Do you want more details or have any questions?

Here are some related questions to expand on this topic:

1. How do you calculate the expected value for a discrete random variable?
2. What are the properties of a probability distribution?
3. How can you verify if a given set of probabilities forms a valid probability distribution?
4. Why is it important that the sum of probabilities equals 1 in a probability distribution?
5. What is the difference between variance and standard deviation in a probability distribution?
6. How do you interpret the mean of a probability distribution?
7. Can the standard deviation of a probability distribution ever be negative?
8. What is the significance of the standard deviation in the context of a probability distribution?

**Tip:** Always ensure that the sum of the probabilities in a probability distribution equals 1, and that each probability is between 0 and 1 inclusive, to validate it as a proper distribution.

Understand probability distributions and how to calculate mean and standard deviation from a given probability table. This problem involves a random variable x representing college students' responses to online courses. Learn the formulas for mean and variance calculation. Suitable for college-level students.

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