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.