Chapter 18, Problem 52E

(a)

To determine

To estimate the probability that the store revenue were at least $\$10000$ .

Answer to Problem 52E

  $0.482$ .

Explanation of Solution

It is given in the question that a grocery store receipts show that Sunday customer purchases have a skewed distribution with a mean of $\$32$ and a standard deviation of $\$20$ . Suppose the store has $312$ customers this Sunday.

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

  $\begin{array}{l}{\mu }_{\overline{y}}=\$32\\ {\sigma }_{\overline{y}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{20}{\sqrt{312}}\\ =\$1.1323\end{array}$

Thus, the z -score will be as:

  $\begin{array}{l}{z}_{\frac{\$10000}{312\text{ customers}}}=\frac{32.05-32}{1.1323}\\ =0.044\end{array}$

Thus, the probability that the store revenue were at least $\$10000$ is as:

  $\begin{array}{l}P(z\ge 0.044)=normalcdf(0.044,E99,0,1)\\ =0.482\end{array}$

Thus, the probability that the store revenue were at least $\$10000$ is at least $0.482$ .

(b)

To determine

To find out how much does the store take in on the worst $10\%$ of such days, if on a typical Sunday the store serves $312$ customers.

Answer to Problem 52E

The mean Sunday purchase of such days is approximately $\$30.55$ so $312$ customers are expected to spend about $\$9531.27$ .

Explanation of Solution

It is given in the question that a grocery store receipts show that Sunday customer purchases have a skewed distribution with a mean of $\$32$ and a standard deviation of $\$20$ . Suppose the store has $312$ customers this Sunday.Thus, we have that,

  $invNorm(0.10,0,1)=-1.282$

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

  $\begin{array}{l}-1.282=\frac{y-32}{1.1323}\\ \Rightarrow -1.282\times 1.1323=y-32\\ \Rightarrow -1.452=y-32\\ \Rightarrow y=\$30.55\end{array}$

Thus, the mean Sunday purchase of such days is approximately $\$30.55$ so $312$ customers are expected to spend about $\$9531.27$ .