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

0.1040.104

1

0.3530.353

2

0.3980.398

3

0.1450.145

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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, not every probability is between 0 and 1 inclusive.

D.

​No, the sum of all the probabilities is not equal to 1.

E.

​No, the random variable x is categorical instead of numerical.

To determine whether the table represents a valid probability distribution, we must verify if it satisfies the following key requirements:

1. **The sum of all probabilities must equal 1.**
2. **Each probability must be between 0 and 1 inclusive.**
3. **The random variable \( x \) must represent numerical values (not categorical).**

Let’s check each requirement with the provided dataTo determine whether the table represents a valid probability distribution, we must verify if it satisfies the following key requirements:

1. **The sum of all probabilities must equal 1.**
2. **Each probability must be between 0 and 1 inclusive.**
3. **The random variable \( x \) must represent numerical values (not categorical).**

Let’s check each requirement with the provided 

| \( x \)  | \( P(x) \) |
|----------|------------|
| 0        | 0.104      |
| 1        | 0.353      |
| 2        | 0.398      |
| 3        | 0.145      |

### Step 1: Check if the probabilities are between 0 and 1
Each of the probabilities provided is within the range [0, 1]:
- \( P(0) = 0.104 \)
- \( P(1) = 0.353 \)
- \( P(2) = 0.398 \)
- \( P(3) = 0.145 \)

Thus, the probabilities satisfy this condition.

### Step 2: Check if the sum of the probabilities equals 1
The sum of the probabilities is:
\[
0.104 + 0.353 + 0.398 + 0.145 = 1.000
\]
The sum is exactly 1, which also satisfies this requirement.

### Step 3: Check if \( x \) represents numerical values
The random variable \( x \) represents the number of students in the group who take online courses, which is a numerical value.

### Conclusion
Since all the requirements for a probability distribution are satisfied, the table does represent a valid probability distribution. Therefore, **Option A** is correct.

---

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

Now, let's calculate the **mean** \( \mu \) and **standard deviation** \( \sigma \).

#### Mean \( \mu \):
The mean of a probability distribution is calculated as:
\[
\mu = \sum x \cdot P(x)
\]
Using the values in the table:
\[
\mu = (0 \cdot 0.104) + (1 \cdot 0.353) + (2 \cdot 0.398) + (3 \cdot 0.145)
\]
\[
\mu = 0 + 0.353 + 0.796 + 0.435 = 1.584
\]

#### Standard Deviation \( \sigma \):
The standard deviation is given by:
\[
\sigma = \sqrt{\sum (x - \mu)^2 \cdot P(x)}
\]
First, calculate \( (x - \mu)^2 \cdot P(x) \) for each \( x \):

\[
(0 - 1.584)^2 \cdot 0.104 = 1.583^2 \cdot 0.104 = 2.505 \cdot 0.104 = 0.2605
\]
\[
(1 - 1.584)^2 \cdot 0.353 = (-0.584)^2 \cdot 0.353 = 0.341 \cdot 0.353 = 0.1203
\]
\[
(2 - 1.584)^2 \cdot 0.398 = 0.416^2 \cdot 0.398 = 0.173 \cdot 0.398 = 0.0688
\]
\[
(3 - 1.584)^2 \cdot 0.145 = 1.416^2 \cdot 0.145 = 2.006 \cdot 0.145 = 0.291
\]

Now sum the values and take the square root:
\[
\sigma = \sqrt{0.2605 + 0.1203 + 0.0688 + 0.291} = \sqrt{0.7406} \approx 0.8609
\]

### Final Answer:
- The **mean** is \( \mu \approx 1.584 \).
- The **standard deviation** is \( \sigma \approx 0.8609 \).

---

Do you want more details or have any questions?  
Here are 5 related questions:

1. How is the mean of a probability distribution used in practical scenarios?
2. How do you interpret standard deviation in a probability distribution?
3. What is the difference between a probability mass function and a probability density function?
4. How would you calculate the variance of this distribution?
5. What happens if the sum of the probabilities is less than or greater than 1?

**Tip:** Always check that probabilities sum to 1 before calculating the mean and standard deviation of a distribution.

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.

Determine whether the given table of probabilities represents a valid probability distribution for the number of students taking online courses. Learn how to calculate the mean and standard deviation step-by-step with key formulas and explanations. Ideal for college-level statistics students.

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