Questions: Suppose that two electronic components in the guidance system for a missile operate independently and that each has a length of life governed by the exponential distribution with mean 5 (with measurements in hundreds of hours). (a) Find the probability density function for the average length of life of the two components. f(u)= , v ≥ , , elsewhere (b) Find the mean and variance of this average, using the answer in part (o). Check your answer by computing the mean and variance, using this theorem. mean variance

Solution Steps

The mean \( \mu \) of an exponential distribution is given by the formula:

\[ \mu = \frac{1}{\lambda} \]

For our case, the rate parameter \( \lambda \) is \( 0.2 \). Thus, we calculate:

\[ \mu = \frac{1}{0.2} = 5.0 \]

The variance \( \sigma^2 \) of an exponential distribution is calculated using the formula:

\[ \sigma^2 = \frac{1}{\lambda^2} \]

Substituting \( \lambda = 0.2 \):

\[ \sigma^2 = \frac{1}{0.2^2} = \frac{1}{0.04} = 25.0 \]

The mean of the average length of life of the two independent components is the same as the mean of each individual component, which we found to be:

\[ \text{Mean of the average} = 5.0 \]

The variance of the average length of life of the two independent components is given by the variance of each component divided by the number of components (which is 2):

\[ \text{Variance of the average} = \frac{\sigma^2}{2} = \frac{25.0}{2} = 12.5 \]

Final Answer