Finding the z-Value that Corresponds to a Given Area

Question

Suppose that preschoolers spend a mean of 25 hours per week in daycare with a standard deviation of 5 hours per week. A newspaper journalist wants to point out that preschoolers are staying in day cares too long, from his perspective. If he only looks at the extreme end of the distribution, that is, the top 3%, what's the minimum number of hours per week those preschoolers spend in daycare? Assume that the numbers of hours per week that preschoolers spend in daycare are normally distributed.

Answer 1

If the mean is 25 and the standard deviation is 5, then 99% of the cases will fall between 10 and 40 hours per week. If the journalist looks at the extreme end of the distribution, the top 3% (or 2.5%, actually), that would be about 35 hours per week or more. Over 40 hours per week would be less than 0.5% of the cases.

Answer 2

The Z score for the standard normal distribution that associates with 97% of the distribution to the left and 3% to the right is 1.881.

When setting 1.881 = (x - 25)/5 you solve for x and find that you get

1.881 (5) = x - 25

9.405 = x - 25

x = 9.405 + 25 = 34.41 hours spent in daycare.