The time it takes to give a man a shampoo and haircut is normally distributed with a mean of 22 minutes and standard deviation of 3 minutes. Customers are scheduled every 30 minutes. (a) What is the probability that a male customer will take longer than the allocated time? *(b) If three customers are scheduled sequentially on the half hour, what is the probability that all three will be finished withintheir allocated half-hours time?

http://davidmlane.com/hyperstat/z_table.html

set your mean and standard deviation on the graph, then calculate the area below 30, that is the probability of getting a customer done. Cube it for three customers sequentially.

To find the probabilities in this scenario, we will need to use the concept of z-scores and the standard normal distribution.

(a) To find the probability that a male customer will take longer than the allocated time (30 minutes), we need to calculate the z-score for a time longer than 30 minutes using the formula:

z = (x - μ) / σ

where x is the value we are interested in, μ is the mean, and σ is the standard deviation.

In this case, we want to find the probability of a time longer than 30 minutes, so x = 30 and μ = 22. The standard deviation is given as σ = 3. Substituting these values into the z-score formula:

z = (30 - 22) / 3 = 2.67

Next, we need to find the probability associated with this z-score. We can look up this value in the standard normal distribution table or use a calculator that can provide the cumulative probability.

Using the table, we find that the probability associated with a z-score of 2.67 (rounded to two decimal places) is approximately 0.9965.

Therefore, the probability that a male customer will take longer than the allocated time is approximately 0.9965 or 99.65%.

(b) To find the probability that all three customers will be finished within their allocated half-hour time, we need to find the probability that each individual customer will be finished within their allocated time and then multiply them together.

Using the same approach as in part (a), we can find the probability for one customer to be finished within the half-hour time:

z = (30 - 22) / 3 = 2.67

The probability associated with this z-score is approximately 0.9965.

Since the customers are scheduled sequentially, we assume that the times for each customer are independent. Therefore, we can simply multiply the probabilities together for each customer.

P(all three finished within allocated time) = P(customer 1) * P(customer 2) * P(customer 3)
= 0.9965 * 0.9965 * 0.9965
≈ 0.9895

Therefore, the probability that all three customers will be finished within their allocated half-hour time is approximately 0.9895 or 98.95%.