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To solve part (a), we need to compute the expected value \( E(X) \), the expected value of \( X^2 \), and the variance \( V(X) \).

### 1. Expected Value \( E(X) \):
The expected value is given by the formula:
\[
E(X) = \sum x \cdot f_X(x)
\]
where \( x \) is the value of the random variable and \( f_X(x) \) is the probability associated with \( x \).

Given dataTo solve part (a), we need to compute the expected value \( E(X) \), the expected value of \( X^2 \), and the variance \( V(X) \).

### 1. Expected Value \( E(X) \):
The expected value is given by the formula:
\[
E(X) = \sum x \cdot f_X(x)
\]
where \( x \) is the value of the random variable and \( f_X(x) \) is the probability associated with \( x \).

Given 
- \( x = 2, 3, 4, 5, 6 \)
- \( f_X(x) = 0.01, 0.25, 0.40, 0.30, 0.04 \)

We calculate \( E(X) \).

### 2. Expected Value of \( X^2 \) \( E(X^2) \):
The expected value of \( X^2 \) is given by:
\[
E(X^2) = \sum x^2 \cdot f_X(x)
\]
We will calculate this value by squaring \( x \) and multiplying by \( f_X(x) \).

### 3. Variance \( V(X) \):
Variance is calculated using:
\[
V(X) = E(X^2) - [E(X)]^2
\]
Once we compute \( E(X) \) and \( E(X^2) \), we can find \( V(X) \).

Let me compute these now.The calculations yield the following results:

1. \( E(X) = 4.11 \)
2. \( E(X^2) = 17.63 \)
3. \( V(X) = 0.738 \) (rounded to three decimal places)

Would you like further clarification or additional details on any of these steps?

Here are some related questions to expand understanding:
1. What is the interpretation of \( E(X) \) in probability distributions?
2. How does \( E(X^2) \) relate to higher moments in statistics?
3. Why is variance \( V(X) \) important in understanding a random variable's spread?
4. What is the significance of calculating the second moment in probability?
5. How would the results change if the probabilities were altered?

**Tip**: To compute variance, always remember that it reflects the spread of data around the mean, while \( E(X) \) captures the central tendency.

Find E(X), E(X^2), and V(X) based on the given probability distribution table for X = {2, 3, 4, 5, 6} with probabilities {0.01, 0.25, 0.40, 0.30, 0.04}.

Learn how to compute the expected value (E(X)), E(X^2), and the variance (V(X)) for a discrete random variable X = {2, 3, 4, 5, 6} with given probabilities. This problem applies key probability distribution concepts and is suitable for Grades 11-12 or College students.

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