To help you solve this problem, let’s go through each required calculation step-by-step:

1. Expected Value, $E(X)$

You’ve already found $E(X) = 8.94$, which is the mean of the distribution.

2. Variance, $\sigma^2$

The variance $\sigma^2$ is calculated by the formula: $\sigma^2 = \sum \left( X - \mu \right)^2 P(X)$ where $\mu = E(X) = 8.94$.

For each value of $X$, calculate $(X - \mu)^2 P(X)$:

1. $(7 - 8.94)^2 \times 0.23$
2. $(8 - 8.94)^2 \times 0.12$
3. $(9 - 8.94)^2 \times 0.19$
4. $(10 - 8.94)^2 \times 0.42$
5. $(11 - 8.94)^2 \times 0.02$
6. $(12 - 8.94)^2 \times 0.02$

After calculating each term, add them together to get $\sigma^2$.

3. Standard Deviation, $\sigma$

The standard deviation $\sigma$ is simply the square root of the variance: $\sigma = \sqrt{\sigma^2}$

4. Probability Calculations

For the probabilities:

• $P(X \geq 11) = 0.04$
• $P(X \leq 9) = 0.54$

These values seem to be directly provided in the table or through earlier calculations.

Would you like me to perform the detailed calculations for variance and standard deviation, or is there a specific part you’d like help with?

Here are some related questions that build upon this problem:

1. How would the expected value change if the probabilities were adjusted?
2. What is the interpretation of standard deviation in this context?
3. How would you calculate $P(X = 9)$ directly from this data?
4. How does variance reflect the spread of values around the mean?
5. How might changing one $P(X)$ value impact the overall mean and variance?

Tip: Remember, variance is influenced heavily by values farther from the mean, since it squares the deviation $(X - \mu)$.