Question

The average waiting time to be seated for dinner at a popular restaurant is 23.8 minutes with a standard deviation of 6 minutes. What is the probability that 20 people will wait less than 21 minutes? Leave your answer exactly as it is taken from the table.

The probability that 20 customers will wait less than 21 minutes to be seated is

Answer 1

Using the normal distribution, it is found that there is a 0.3192 = 31.92% probability that 20 customers will wait an average of less than 21 minutes to be seated.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

• The z-score measures how many standard deviations the measure is above or below the mean.

• Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

• By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

The parameters for this problem are given as follows:

$\mu = 23.8, \sigma = 6, n = 20, s = \frac{6}{\sqrt{20}} = 1.3416$

The probability that 20 customers will wait an average of less than 21 minutes to be seated is the p-value of Z when X = 21, hence:

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

By the Central Limit Theorem:

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

$Z = \frac{21 - 23.8}{6}$

Z = -0.47

Z = -0.47 has a p-value of 0.3192.

0.3192 = 31.92% probability that 20 customers will wait an average of less than 21 minutes to be seated.

More can be learned about the normal distribution at https://brainly.com/question/4079902

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