Given the problem, we have the following information:

• $Y = aX + 5$ where $a$ is a positive integer.
• $E(aX + 5) = \text{Var}(aX + 5)$.
• The mean of $X$, $E(X) = 9$.
• The variance of $X$, $\text{Var}(X) = 2$.

Part (a): Determine the value of the constant $a$

Let's first calculate the expected value and variance of $Y = aX + 5$:

1. Expected value of $Y$: $E(Y) = E(aX + 5) = aE(X) + 5 = a \times 9 + 5 = 9a + 5$

2. Variance of $Y$: $\text{Var}(Y) = \text{Var}(aX + 5)$ Since adding a constant does not affect the variance: $\text{Var}(Y) = \text{Var}(aX) = a^2 \text{Var}(X) = a^2 \times 2 = 2a^2$

Given that $E(Y) = \text{Var}(Y)$, we set up the equation:

$9a + 5 = 2a^2$

Rearranging this into a standard quadratic equation form:

$2a^2 - 9a - 5 = 0$

We solve this quadratic equation using the quadratic formula:

$a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 2$, $b = -9$, and $c = -5$:

$a = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(2)(-5)}}{2(2)} = \frac{9 \pm \sqrt{81 + 40}}{4} = \frac{9 \pm \sqrt{121}}{4} = \frac{9 \pm 11}{4}$

This gives us two solutions:

$a = \frac{20}{4} = 5 \quad \text{or} \quad a = \frac{-2}{4} = -0.5$

Since $a$ is a positive integer, $a = 5$ is the valid solution.

Part (b): Calculate $E(Y)$ and $\text{Var}(Y)$

Now that we know $a = 5$, we can find $E(Y)$ and $\text{Var}(Y)$.

1. Expected value of $Y$: $E(Y) = 9a + 5 = 9(5) + 5 = 45 + 5 = 50$

2. Variance of $Y$: $\text{Var}(Y) = 2a^2 = 2(5^2) = 2 \times 25 = 50$

Thus, the answers are:

• (a) $a = 5$
• (b) $E(Y) = 50$ and $\text{Var}(Y) = 50$

Would you like more details or have any questions? Here are some related questions you might be interested in:

1. How do you find the expected value and variance of a linear transformation of a random variable?
2. What are some other types of problems involving continuous random variables?
3. How does adding or multiplying by constants affect the mean and variance of a random variable?
4. Can this method be applied to other forms of linear relationships between random variables?
5. What are the implications if $a$ were not a positive integer?

Tip: Always double-check the assumptions and given conditions in probability problems, as they can significantly affect the solution approach.