At a local restaurant, the amount of time that customers have to wait for their food is normally distributed with a mean of 30 minutes and a standard deviation of 2 minutes. What is the probability that a randomly selected customer will have to wait between 26 minutes and 32 minutes, to the nearest thousandth?


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.

Learn how to calculate the probability of waiting time in a normal distribution problem with a mean of 30 minutes and a standard deviation of 2 minutes. Understand how to use z-scores and the standard normal distribution table to find probabilities. Suitable for high school students learning about probability and normal distribution.

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