Question

Suppose that the lifetimes (in months) of bulbs manufactured by a company have an \( \text{Exp}(\lambda) \) distribution, where \( \lambda > 0 \). A random sample of size 10 taken from the bulbs manufactured by the company yields the sample mean lifetime \( \bar{x} = 3.52 \) months. Then the uniformly minimum variance unbiased estimate of \( \frac{1}{\lambda} \) based on this sample is equal to __________ months (round off to 2 decimal places).

Answer 1

Correct Answer: 3.5 - 3.55

1. Parameter Estimation for \( \text{Exp}(\lambda) \): - For \( X_i \sim \text{Exp}(\lambda) \), the MLE of \( \lambda \) is: \[ \hat{\lambda} = \frac{1}{\bar{X}}, \] where \( \bar{X} \) is the sample mean. 2. UMVUE for \( \frac{1}{\lambda} \): - The uniformly minimum variance unbiased estimator (UMVUE) of \( \frac{1}{\lambda} \) is the sample mean \( \bar{X} \). 3. Computation: - Given \( \bar{X} = 3.52 \), the UMVUE for \( \frac{1}{\lambda} \) is: \[ \frac{1}{\lambda} \approx 3.52. \] 4. Final Answer: - Round off to 2 decimal places: \[ \boxed{3.52} \]