The problem you provided asks to find the probability that a component will last between 14 and 20 months, given a probability density function (PDF):

$f(x) = \frac{1}{4} e^{-x/4}, \text{ for } x \in [0, \infty).$

This is the PDF of an exponential distribution with rate parameter $\lambda = \frac{1}{4}$. For an exponential distribution, the cumulative distribution function (CDF) can be used to calculate probabilities.

The CDF for an exponential distribution with PDF $f(x) = \lambda e^{-\lambda x}$ is:

$F(x) = 1 - e^{-\lambda x}.$

Here, $\lambda = \frac{1}{4}$, so the CDF is:

$F(x) = 1 - e^{-x/4}.$

To find the probability that the component lasts between 14 and 20 months, we need to calculate $P(14 \leq X \leq 20)$, which is:

$P(14 \leq X \leq 20) = F(20) - F(14).$

Substituting into the CDF:

$P(14 \leq X \leq 20) = \left(1 - e^{-20/4}\right) - \left(1 - e^{-14/4}\right).$

Simplifying the exponents:

$P(14 \leq X \leq 20) = \left(1 - e^{-5}\right) - \left(1 - e^{-3.5}\right).$

This becomes:

$P(14 \leq X \leq 20) = e^{-3.5} - e^{-5}.$

Now you can calculate the numerical values for $e^{-3.5}$ and $e^{-5}$:

$e^{-3.5} \approx 0.0302, \quad e^{-5} \approx 0.0067.$

Thus:

$P(14 \leq X \leq 20) \approx 0.0302 - 0.0067 = 0.0235.$

So the probability that the component will last between 14 and 20 months is approximately 0.0235, or 2.35%.

Would you like further details on how this was calculated?

Here are some related questions you might explore:

1. How do you interpret the result in terms of the component's reliability?
2. How does the rate parameter $\lambda = \frac{1}{4}$ influence the distribution?
3. What would the probability be for a different time range, say between 10 and 30 months?
4. How would you calculate the mean and variance of this exponential distribution?
5. How does the memoryless property of the exponential distribution apply in this scenario?

Tip: The exponential distribution is often used to model waiting times or the lifespan of products that have a constant failure rate.