Chapter 18, Problem 49E

(a)

To determine

To explain why you cannot find out the probability that a given party will tip him at least $\$20$ .

Explanation of Solution

It is given in the question, a waiter believes the distribution of his tip has a model that is slightly skewed to the right with a mean of $\$9.60$ and a standard deviation of $\$5.40$ . Thus, since the distribution of tips is skewed to the right we cannot use the Normal model to find out the probability that a given party will tip at least $\$20$ .

(b)

To determine

To explain can you estimate the probability that the next four parties will tip an average of at least $\$15$ .

Answer to Problem 49E

No, we cannot estimate.

Explanation of Solution

It is given in the question, a waiter believes the distribution of his tip has a model that is slightly skewed to the right with a mean of $\$9.60$ and a standard deviation of $\$5.40$ . Thus, we cannot estimate the probability that the next four parties will tip an average of at least $\$15$ because a sample of four parties is probably not a large enough sample for the Central Limit Theorem to allow us to use the Normal model to estimate the distribution of averages.

(c)

To determine

To explain is it likely that his ten parties today will tip an average of at least $\$15$ .

Answer to Problem 49E

It is not likely that his ten parties today will tip an average of at least $\$15$ .

Explanation of Solution

It is given in the question, a waiter believes the distribution of his tip has a model that is slightly skewed to the right with a mean of $\$9.60$ and a standard deviation of $\$5.40$ . Thus, a sample of ten parties may not be large enough to allow the use of a normal model to describe the distribution of averages. It would be risky to attempt to estimate the probability that his next ten parties to an average of $\$15$ . However, since the distribution of tips has mean of $\$9.60$ and a standard deviation of $\$5.40$ , we still know that the mean and the standard deviation of the sampling distribution model is as:

  $\begin{array}{l}{\mu }_{\overline{y}}=\$9.60\\ {\sigma }_{\overline{y}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{5.40}{\sqrt{10}}\\ =\$1.71\end{array}$

Thus, we do not know the exact shape of the distribution but we can still access the likelihood of specified means. A mean tip of $\$15$ is over three standard deviations above the expected mean tip for ten parties. That is not very likely to happen.