Question

Let \( Y = Z^2 \), \( Z = \frac{X - \mu}{\sigma} \), where \( X \) is a normal random variable with mean \( \mu \) and variance \( \sigma^2 \). The variance of \( Y \) is

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

When squaring a standard normal variable, the result follows a chi-squared distribution with 1 degree of freedom. The variance of a chi-squared distribution with 1 degree of freedom is always 2.

Answer 1

The Correct Option is B

The variable \( Z \) is a standard normal random variable, i.e., \( Z \sim N(0, 1) \), because: \[ Z = \frac{X - \mu}{\sigma} \] Since \( X \) has a normal distribution with mean \( \mu \) and variance \( \sigma^2 \), the transformation \( Z = \frac{X - \mu}{\sigma} \) gives \( Z \) a mean of 0 and variance of 1. The variable \( Y = Z^2 \) follows a chi-squared distribution with 1 degree of freedom, i.e., \( Y \sim \chi^2_1 \). The properties of a chi-squared distribution with 1 degree of freedom are: The mean of \( Y \) is \( \mu_Y = 1 \),
The variance of \( Y \) is \( \sigma_Y^2 = 2 \).
Thus, the variance of \( Y \) is \( \boxed{2} \).