Let's go through each part of this problem related to the exponential distribution with a mean of 10 minutes.

Part (a)

An exponential distribution with a mean of 10 minutes implies that:

• The distribution is right-skewed (a characteristic of exponential distributions).
• The mean $\mu = 10$ minutes.
• The standard deviation $\sigma = \mu = 10$ minutes, since, in an exponential distribution, the mean and standard deviation are equal.

So, the correct answer here is likely Option E: "The distribution is right-skewed with a mean of 10 minutes and a standard deviation of 10 minutes."

Part (b)

To compute the probabilities, we use the exponential probability density function (PDF) and cumulative distribution function (CDF):

Given:

• Mean $\mu = 10$ minutes, so the rate $\lambda = \frac{1}{\mu} = 0.1$.

For an exponential distribution, the probability that $T \leq t$ is given by: $P(T \leq t) = 1 - e^{-\lambda t}$

(i) Probability that the time passing is less than 5 minutes:

$P(T < 5) = 1 - e^{-0.1 \times 5}$

(ii) Probability that the time passing is between 9.5 and 18 minutes:

$P(9.5 \leq T \leq 18) = P(T \leq 18) - P(T \leq 9.5)$ Calculate each term separately:

• $P(T \leq 18) = 1 - e^{-0.1 \times 18}$
• $P(T \leq 9.5) = 1 - e^{-0.1 \times 9.5}$

(iii) Probability that the time passing is more than 27 minutes:

$P(T > 27) = e^{-0.1 \times 27}$

Part (c)

This is a conditional probability question. Given that 14 minutes have passed, we want the probability that at least another 10 minutes will pass. For exponential distributions, the memoryless property tells us: $P(T > 24 \mid T > 14) = P(T > 10) = e^{-0.1 \times 10}$

Part (d)

To find the time $t$ such that 93% of the time the check-in interval is at most $t$: $P(T \leq t) = 0.93$ Solving for $t$: $1 - e^{-0.1 t} = 0.93$ $e^{-0.1 t} = 0.07$ $-0.1 t = \ln(0.07)$ $t = -\frac{\ln(0.07)}{0.1}$

I'll proceed with these calculations now.Here are the results for each part of the calculations:

• (b)(i) Probability that the time passing is less than 5 minutes: $\approx 0.3935$.

• (b)(ii) Probability that the time passing is between 9.5 and 18 minutes: $\approx 0.2214$.

• (b)(iii) Probability that the time passing is more than 27 minutes: $\approx 0.0672$.

• (c) Probability that at least another 10 minutes will pass after 14 minutes have already passed: $\approx 0.3679$.

• (d) 93% of the time, the number of minutes passing between successive customer check-ins is at most approximately 26.59 minutes.

Would you like further explanations on any part, or details on how these values were derived?

Here are 5 related questions to deepen understanding:

1. How would the results change if the mean time between check-ins were different?
2. Why does the memoryless property apply to exponential distributions specifically?
3. What is the impact of the rate parameter $\lambda$ on the shape of the distribution?
4. How do we interpret probabilities in the context of an exponential distribution?
5. Could we use a different distribution if the check-in times followed a different pattern?

Tip: For exponential distributions, the mean and standard deviation are always equal, making it unique among distributions.