To solve this problem, we need to find the probability that a randomly selected customer will have to wait between 26 minutes and 32 minutes, given that the waiting time is normally distributed with a mean of 30 minutes and a standard deviation of 2 minutes.

Step 1: Standardize the values

We first convert the raw scores (26 minutes and 32 minutes) to z-scores using the formula:

$z = \frac{X - \mu}{\sigma}$

Where:

• $X$ is the value of interest (either 26 or 32 minutes),
• $\mu$ is the mean (30 minutes),
• $\sigma$ is the standard deviation (2 minutes).

For $X = 26$ minutes: $z_1 = \frac{26 - 30}{2} = \frac{-4}{2} = -2$

For $X = 32$ minutes: $z_2 = \frac{32 - 30}{2} = \frac{2}{2} = 1$

Step 2: Find the probability corresponding to the z-scores

Using the standard normal distribution table (or a calculator that provides cumulative probabilities for the standard normal distribution):

• The cumulative probability for $z_1 = -2$ is approximately 0.0228.
• The cumulative probability for $z_2 = 1$ is approximately 0.8413.

Step 3: Calculate the probability between 26 and 32 minutes

To find the probability that the waiting time is between 26 and 32 minutes, subtract the cumulative probability at $z_1$ from the cumulative probability at $z_2$:

$P(26 < X < 32) = P(z_2) - P(z_1) = 0.8413 - 0.0228 = 0.8185$

Thus, the probability that a randomly selected customer will have to wait between 26 minutes and 32 minutes is approximately 0.819 (to the nearest thousandth).

Would you like more details or have any questions about this solution?

Here are five related questions:

1. How would the probability change if the standard deviation were 3 minutes instead of 2 minutes?
2. What is the probability that a customer will wait more than 34 minutes?
3. What is the probability that a customer will wait less than 28 minutes?
4. If a customer has to wait 35 minutes, is this an unusual event based on the distribution?
5. How would you calculate the waiting time for the top 5% of customers?

Tip: Understanding z-scores is crucial as they allow you to compare individual data points to a distribution, making it easier to find probabilities and make inferences.