Question

Let \( X_1, X_2, X_3, X_4, X_5 \) be independent random variables with \[ X_1 \sim N(200, 8), \, X_2 \sim N(104, 8), \, X_3 \sim N(108, 15), \, X_4 \sim N(120, 15), \, X_5 \sim N(210, 15). \] Let \[ U = \frac{X_1 + X_2}{2}, \quad V = \frac{X_3 + X_4 + X_5}{3}. \] Then \( P(U>V) \) equals

Hint:

When dealing with the probability of linear combinations of normal variables, remember that the sum or difference of independent normal variables is also normally distributed. The mean and variance are additive, and you can standardize to compute probabilities.

Answer 1

Correct Answer: 0.97 - 0.98

Step 1: Understanding the distribution.
We are given independent normal random variables \( X_1, X_2, X_3, X_4, X_5 \) with specified means and variances. The variables \( U \) and \( V \) are linear combinations of these random variables. Since the sum of independent normal variables is also normally distributed, both \( U \) and \( V \) will follow normal distributions.

Step 2: Distribution of \( U \) and \( V \).
The mean and variance of \( U \) are calculated as: \[ \mu_U = \frac{\mu_1 + \mu_2}{2} = \frac{200 + 104}{2} = 152, \quad \sigma_U^2 = \frac{\sigma_1^2 + \sigma_2^2}{4} = \frac{8 + 8}{4} = 4. \] For \( V \), the mean and variance are: \[ \mu_V = \frac{\mu_3 + \mu_4 + \mu_5}{3} = \frac{108 + 120 + 210}{3} = 146, \quad \sigma_V^2 = \frac{\sigma_3^2 + \sigma_4^2 + \sigma_5^2}{9} = \frac{15 + 15 + 15}{9} = 5. \]
Step 3: Finding \( P(U>V) \).
We now need to calculate \( P(U>V) \). This can be done by standardizing the difference \( D = U - V \), where: \[ \mu_D = \mu_U - \mu_V = 152 - 146 = 6, \quad \sigma_D^2 = \sigma_U^2 + \sigma_V^2 = 4 + 5 = 9. \] Thus, \( D \sim N(6, 9) \). Standardizing \( D \): \[ Z = \frac{D - 6}{\sqrt{9}} = \frac{D - 6}{3}. \] We now compute \( P(Z>0) \), which corresponds to a probability between 0.97 and 0.98.

Step 4: Conclusion.
The value of \( P(U>V) \) is approximately between 0.97 and 0.98.