To solve for $E(Y)$ and $\text{Var}(Y)$ in each case, we will use the properties of expected values and variances.

Given:

• $X$ is a random variable with mean $E(X) = 5$ and standard deviation $\sigma_X = 2$.
• The variance of $X$ is $\text{Var}(X) = \sigma_X^2 = 4$.

(a) For $Y = 2X + 3$:

1. Expected Value $E(Y)$: $E(Y) = E(2X + 3) = 2 \cdot E(X) + 3 = 2 \cdot 5 + 3 = 10 + 3 = 13$

2. Variance $\text{Var}(Y)$: $\text{Var}(Y) = \text{Var}(2X + 3) = 2^2 \cdot \text{Var}(X) = 4 \cdot 4 = 16$

   Result for (a): $E(Y) = 13, \quad \text{Var}(Y) = 16$

(b) For $Y = 2X - 3$:

1. Expected Value $E(Y)$: $E(Y) = E(2X - 3) = 2 \cdot E(X) - 3 = 2 \cdot 5 - 3 = 10 - 3 = 7$

2. Variance $\text{Var}(Y)$: $\text{Var}(Y) = \text{Var}(2X - 3) = 2^2 \cdot \text{Var}(X) = 4 \cdot 4 = 16$

   Result for (b): $E(Y) = 7, \quad \text{Var}(Y) = 16$

(c) For $Y = (X - 5)^2$:

1. Expected Value $E(Y)$:

   • Here, $Y = (X - 5)^2$ represents the square of a centered random variable.
   • Note that $X - 5$ has a mean of $0$ and variance equal to $\text{Var}(X) = 4$.
   • The expected value of $Y$ is the variance of $X$: $E(Y) = \text{Var}(X) = 4$

2. Variance $\text{Var}(Y)$:

   • To find $\text{Var}(Y)$, we would need additional information about the distribution of $X$, as calculating $\text{Var}((X-5)^2)$ involves higher moments of $X$.
   • Without additional information on the distribution of $X$, we cannot determine $\text{Var}(Y)$ for this case.

   Result for (c): $E(Y) = 4, \quad \text{Var}(Y) = \text{(Cannot be determined with given information)}$

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Would you like more details on these calculations, or do you have any questions?

Related Questions

1. How would the values change if $X$ had a different standard deviation?
2. How does the linear transformation affect the variance in cases (a) and (b)?
3. Why does the variance formula for $Y = 2X + 3$ involve squaring the coefficient of $X$?
4. What assumptions are needed to calculate higher moments of $Y = (X-5)^2$?
5. How would you find $\text{Var}(Y)$ if $X$ were normally distributed?

Tip

For linear transformations $Y = aX + b$, remember that $E(Y) = aE(X) + b$ and $\text{Var}(Y) = a^2 \text{Var}(X)$.