The image contains a discrete random variable $X$ with the associated probabilities $P(X)$ for each value of $X$. The task is to calculate the mean, variance, and standard deviation of $X$. Let's break down the information in the table:

The expected value (mean) $\mu$, variance $\sigma^2$, and standard deviation $\sigma$ are already calculated:

• $\mu = 8.3400$
• $\sigma^2 = 7.1244$
• $\sigma = 2.6683$

Verification of the Results:

1. Mean $\mu$: $\mu = \sum X \cdot P(X) = (3 \cdot 0.11) + (5 \cdot 0.10) + (9 \cdot 0.71) + (14 \cdot 0.08)$

2. Variance $\sigma^2$: Variance is calculated as: $\sigma^2 = \sum P(X) \cdot (X - \mu)^2$ where $\mu = 8.3400$.

3. Standard Deviation $\sigma$: Standard deviation is the square root of the variance: $\sigma = \sqrt{\sigma^2}$

Let me calculate and verify the given values.The calculations are as follows:

• Mean (Expected value) $\mu = 8.3400$, which matches the given value.
• Variance $\sigma^2 = 7.1244$, which also matches the given value.
• Standard deviation $\sigma = 2.6692$, which is very close to the provided value of $2.6683$ (slight rounding difference).

The provided values are accurate based on the given data.

Would you like further details or have any questions?

Here are 5 related questions you might find interesting:

1. How would the variance change if one of the probabilities was increased?
2. Can you compute the cumulative distribution function (CDF) from this data?
3. How is the standard deviation useful in interpreting data spread in a real-world scenario?
4. What is the role of the expected value in decision-making under uncertainty?
5. How would adding a new value to $X$ with its own probability affect the mean?

Tip: When calculating variance, always subtract the mean before squaring the deviations to get accurate results.