Question

14. The average waiting time to be seated for dinner at a popular restaurant is 23.5 minutes, with a standard deviation of 3.6 minutes. Assume the variable is normally distributed. When a patron arrives at the restaurant for dinner, find the probability that the patron will have to wait the following time

Answer 1

Answer:

Incomplete question, but you use the normal distribution to solve it, with $\mu = 23.5$ and $\sigma = 3.6$

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

The average waiting time to be seated for dinner at a popular restaurant is 23.5 minutes, with a standard deviation of 3.6 minutes.

This means that $\mu = 23.5, \sigma = 3.6$

Thus

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

$Z = \frac{X - 23.5}{3.6}$

Find the probability that the patron will have to wait:

More than X minutes:

1 subtracted by the p-value of Z

Between A and B minutes:

p-value of Z when X = A subtracted by the p-value of Z when Z = B.

Less than X minutes.

p-value of Z.