Chapter 18, Problem 50E

(a)

To determine

To explain why you cannot find out the probability that the next Sunday customer will spend at least $\$40$ .

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 $\$20$ . Thus, since the distribution of Sunday purchases is skewed we cannot use the Normal model to find out the probability that the next Sunday customer will spend at least $\$40$ .

(b)

To determine

To explain can you estimate the probability that the next ten Sunday customers will spend an average of at least $\$40$ .

Answer to Problem 50E

No, we cannot estimate.

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 $\$20$ . Thus, we cannot estimate the probability that the next ten Sunday customers will spend an average of at least $\$40$ because a sample of ten customers may not be large enough for the CLT to allow the use of a Normal model for the sampling distribution model. If the distribution of Sunday purchases is only slightly skewed, the sample may be large enough, but if the distribution is heavily skewed, it would be very risky to attempt to determine the probability.

(c)

To determine

To explain is it likely that the next $50$ Sunday customers will spend an average of at least $\$40$ .

Answer to Problem 50E

It is likely that the next $50$ Sunday customers will spend an average of at least $\$40$ .

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 $\$20$ . Thus, for the mean Sunday customers we have,

  $\begin{array}{l}{\mu }_{\overline{y}}=\$32\\ {\sigma }_{\overline{y}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{20}{\sqrt{50}}\\ =\$2.83\end{array}$

And let us then find the z -score as:

  $\begin{array}{l}z=\frac{4.-32}{2.83}\\ =2.827\end{array}$

Thus, the probability that the next $50$ Sunday customers will spend an average of at least $\$40$ will be as:

  $\begin{array}{l}P(z\ge 2.827)=normalcdf(2.827,E99,0,1)\\ =0.0023\\ \overline{y}\end{array}$

Thus, the probability that the next $50$ Sunday customers will spend an average of at least $\$40$ is $0.0023$ . So, it is likely that the next $50$ Sunday customers will spend an average of at least $\$40$ .