The Problem A book store owner decides to sell children's talking books that will appeal to the middle 60% of his customers. The owner reads in a study that the mean price of children's talking books is \$10.52 with a standard deviation of \$1.08. Find the maximum and minimum prices of talking books the ownere should sell. Assume the variable is normally distributed.

We are looking for the x's at the bottom of the middle 60% (the minimum price) and at the top of the middle 60% (the maximum price). The middle 60% means 30% is above the mean (which is the 50th percentile) and 30% is below the mean.

Thus, the minimum price is at the 20th percentile because 50 - 30 = 20, and the maximum price is at the 80th percentile as 50 + 30 = 80. Both x values can be found on the TI83/84 calculator using the following function:

It is given that = 10.52, and = 1.08. To find the minimum price, press 2nd VARS[DISTR]. Press 3 to select 3:invNorm( , and then type .2,10.52,1.08) and press Enter. To find the maximum price, press 2nd VARS[DISTR]. Press 3 to select 3:invNorm( , and then type .8,10.52,1.08) and press Enter. See the results below.

The minimum x is about 9.61 and the maximum x is about 11.43. Thus, the store owner should pick books that will sell for between \$9.61 and \$11.43, if he wishes to cater to the middle 60% of customers.