To solve this problem, let's go through each part step-by-step. The density function provided is:

$f(z) = \frac{1}{10} e^{-z/10}, \quad 0 < z < \infty$

This is the probability density function (pdf) of an exponential distribution with a mean of 10 minutes, as we can recognize from the form $\lambda e^{-\lambda z}$ where $\lambda = \frac{1}{10}$.

Part (a)

What is the probability that there are no calls within a 20-minute time interval?

For an exponential distribution, the probability that there are no calls within a given time $t$ is $P(Z > t)$, which can be calculated as:

$P(Z > t) = 1 - F(t)$

where $F(t)$ is the cumulative distribution function (CDF) for the exponential distribution. The CDF for an exponential distribution is:

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

Thus:

$P(Z > 20) = e^{-20/10} = e^{-2}$

Calculating this:

$P(Z > 20) \approx 0.1353$

Part (b)

What is the probability that the first call comes within 10 minutes of opening?

Here, we want $P(Z \leq 10)$, which is simply the CDF at $z = 10$:

$P(Z \leq 10) = F(10) = 1 - e^{-10/10} = 1 - e^{-1}$

Calculating this:

$P(Z \leq 10) \approx 1 - 0.3679 = 0.6321$

Part (c)

What is the CDF?

As derived earlier, the cumulative distribution function $F(z)$ for the exponential distribution with $\lambda = \frac{1}{10}$ is:

$F(z) = 1 - e^{-z/10}$

This function gives the probability that a call will occur within $z$ minutes.

Let me know if you need further clarification on any part of this solution.

Related Questions

1. What is the expected value of the time between calls in this scenario?
2. How would you calculate the variance of the time between calls?
3. How would you determine the probability of getting at least two calls within 15 minutes?
4. If the rate parameter changed, how would that affect the probability calculations?
5. How could this model be adapted if calls occurred at different rates throughout the day?

Tip

For exponential distributions, remember that the mean is the reciprocal of the rate parameter ($\lambda$), and it's memoryless—meaning the probability of waiting an additional time is independent of how much time has already passed.