Let's solve this step by step.

Step 1: Find the missing probability.

The sum of all probabilities in a probability distribution must equal 1. We are missing the probability for $X = 26$. So, we add up the probabilities we know and subtract from 1:

$P(23) + P(24) + P(25) + P(26) + P(27) = 1$

From the table: $P(23) = 0.09, \quad P(24) = 0.13, \quad P(25) = 0.16, \quad P(27) = 0.11$

So: $0.09 + 0.13 + 0.16 + P(26) + 0.11 = 1$

Add the known probabilities: $0.09 + 0.13 + 0.16 + 0.11 = 0.49$

Now solve for $P(26)$: $P(26) = 1 - 0.49 = 0.51$

Step 2: Calculate the mean (expected value).

The formula for the mean (expected value) of a probability distribution is:

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

Using the probabilities and values of $X$:

$\mu = (23 \cdot 0.09) + (24 \cdot 0.13) + (25 \cdot 0.16) + (26 \cdot 0.51) + (27 \cdot 0.11)$

Let's calculate each term:

$23 \cdot 0.09 = 2.07, \quad 24 \cdot 0.13 = 3.12, \quad 25 \cdot 0.16 = 4.00, \quad 26 \cdot 0.51 = 13.26, \quad 27 \cdot 0.11 = 2.97$

Now sum these up: $\mu = 2.07 + 3.12 + 4.00 + 13.26 + 2.97 = 25.42$

So, the mean number of students is approximately 25.42.

Step 3: Calculate the standard deviation.

The formula for the standard deviation is:

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

We already know $\mu = 25.42$. Now calculate $(X - \mu)^2 \cdot P(X)$ for each $X$:

1. For $X = 23$: $(23 - 25.42)^2 \cdot 0.09 = (-2.42)^2 \cdot 0.09 = 5.8564 \cdot 0.09 = 0.527076$

2. For $X = 24$: $(24 - 25.42)^2 \cdot 0.13 = (-1.42)^2 \cdot 0.13 = 2.0164 \cdot 0.13 = 0.262132$

3. For $X = 25$: $(25 - 25.42)^2 \cdot 0.16 = (-0.42)^2 \cdot 0.16 = 0.1764 \cdot 0.16 = 0.028224$

4. For $X = 26$: $(26 - 25.42)^2 \cdot 0.51 = (0.58)^2 \cdot 0.51 = 0.3364 \cdot 0.51 = 0.171564$

5. For $X = 27$: $(27 - 25.42)^2 \cdot 0.11 = (1.58)^2 \cdot 0.11 = 2.4964 \cdot 0.11 = 0.274604$

Now sum all these values: $0.527076 + 0.262132 + 0.028224 + 0.171564 + 0.274604 = 1.2636$

Finally, take the square root: $\sigma = \sqrt{1.2636} \approx 1.12$

So, the standard deviation is approximately 1.12.

Final Answers:

• Missing probability $P(26) = 0.51$
• Mean number of students $\mu \approx 25.42$
• Standard deviation $\sigma \approx 1.12$

Would you like further details or have any questions?

Here are 5 related questions to expand your understanding:

1. What is the significance of the mean in a probability distribution?
2. How would the mean change if the probability for $X = 26$ were smaller?
3. What does the standard deviation tell you about the distribution of students?
4. How do we interpret a probability distribution in terms of expected outcomes?
5. How can the variance of this distribution be used to understand data spread?

Tip: The variance is just the square of the standard deviation, and it's a key measure to understand how much values differ from the mean.