Chapter 18, Problem 51E

(a)

To determine

To estimate the probability that he will earn at least $\$500$ in tips.

Answer to Problem 51E

  $0.0003$ .

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 $\$5.40$ . The waiter usually waits on about $40$ parties over a weekend of work. Thus, let us check the conditions for the following as:

Random condition: It is satisfied as we assume that the tips from $40$ parties can be considered a representative sample of all tips.

Independent condition: It is reasonable to think that the tips are mutually independent.

$10\%$ condition: The sample size of $40$ parties represents less than $10\%$ of all tips.

Large enough sample condition: A sample of $40$ parties is large enough thus, it is satisfied.

Thus, we have the mean and standard deviation for the mean sample as:

  $\begin{array}{l}{\mu }_{\overline{y}}=\$9.60\\ {\sigma }_{\overline{y}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{5.40}{\sqrt{40}}\\ =\$0.8538\end{array}$

Thus, the z -score will be as:

  $\begin{array}{l}{z}_{\frac{\$500}{40\text{Parties}}}=\frac{12.50-9.60}{0.8538}\\ =3.397\end{array}$

Thus, the probability that he will earn at least $\$500$ in tips is as:

  $\begin{array}{l}P(z\ge 3.397)=normalcdf(3.397,E99,0,1)\\ =0.0003\end{array}$

Thus, the probability that he will earn at least $\$500$ in tips in a weekend is approximately $0.0003$ .

(b)

To determine

To find out how much does he earn on the best $10\%$ of such weekends.

Answer to Problem 51E

The waiter can expect to have a mea tip of about $\$10.69$ per party which corresponds to about $\$427.77\text{ for 40}$ parties.

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 $\$5.40$ . The waiter usually waits on about $40$ parties over a weekend of work. Thus, we have that,

  $invNorm(0.90,0,1)=1.282$

Thus, put the value of above for z -score as:

  $\begin{array}{l}1.282=\frac{y-9.60}{0.8538}\\ \Rightarrow 1.282\times 0.8538=y-9.60\\ \Rightarrow 1.094=y-9.60\\ \Rightarrow y=10.69\end{array}$

Thus, the waiter can expect to have a mean tip of about $\$10.69$ per party which corresponds to about $\$427.77\text{ for 40}$ parties.