The question provides a probability distribution for the number of students in statistics classes, and you're asked to:

1. Fill in the missing probability value in the table.
2. Find the mean number of students in a class.
3. Calculate the standard deviation of the number of students in a class.

Let's walk through each part:

1. Fill in the Missing Probability Value:

We know that the sum of probabilities for a probability distribution must be 1. The given probabilities are:

• $P(X = 21) = 0.13$
• $P(X = 22) = 0.17$
• $P(X = 23) = 0.2$
• $P(X = 24) = 0.35$
• $P(X = 25) = 0.15$

Summing the given probabilities:

$0.13 + 0.17 + 0.2 + 0.35 + 0.15 = 1.00$

Thus, the distribution is complete and correct; no missing value here.

2. Mean Number of Students:

The formula for the mean $\mu$ of a probability distribution is:

$\mu = \sum X \cdot P(X)$

Using the given dataThe question provides a probability distribution for the number of students in statistics classes, and you're asked to:

1. Fill in the missing probability value in the table.
2. Find the mean number of students in a class.
3. Calculate the standard deviation of the number of students in a class.

Let's walk through each part:

1. Fill in the Missing Probability Value:

We know that the sum of probabilities for a probability distribution must be 1. The given probabilities are:

• $P(X = 21) = 0.13$
• $P(X = 22) = 0.17$
• $P(X = 23) = 0.2$
• $P(X = 24) = 0.35$
• $P(X = 25) = 0.15$

Summing the given probabilities:

$0.13 + 0.17 + 0.2 + 0.35 + 0.15 = 1.00$

Thus, the distribution is complete and correct; no missing value here.

2. Mean Number of Students:

The formula for the mean $\mu$ of a probability distribution is:

$\mu = \sum X \cdot P(X)$

Using the given

$\mu = (21 \times 0.13) + (22 \times 0.17) + (23 \times 0.20) + (24 \times 0.35) + (25 \times 0.15)$

This yields the value of 23.22 (as shown in your input), so the mean is correct.

3. Standard Deviation:

The standard deviation $\sigma$ is calculated using:

$\sigma = \sqrt{\sum (X - \mu)^2 \cdot P(X)}$

Let's calculate the variance first, which is the squared deviation from the mean:

$\text{Variance} = \sum (X - \mu)^2 \cdot P(X)$

Once we compute the variance, the standard deviation is the square root of that value.

I will calculate this now.The standard deviation of the number of students in a statistics class is approximately 1.26.

Would you like more details on this solution or have any other questions?

Here are five related questions for further exploration:

1. How do you interpret the mean in the context of this probability distribution?
2. What does the standard deviation tell us about the spread of the data?
3. Can the probabilities in a distribution ever sum to more than 1? Why or why not?
4. How would the mean change if we added more students (e.g., 26) with a certain probability?
5. What is the difference between population and sample standard deviation in this context?

Tip: The standard deviation measures how spread out the values are around the mean. A smaller standard deviation means the data points are closer to the mean.