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.
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