Question

8. (8 pts) The length of time for students to complete a test is found to be normally distributed with mean 70 minutes and standard deviation 12 minutes. How long should the time for the test be if we wish for 90 percent of the students to complete the test in less time than the test length? (Use the Normal table provided for your calculations)

Answer 1

Solution:

Given: μ = 70 and σ = 12
P(x < x0) = 90% = 0.90

Let us determine the z-score that corresponds with a probability of 0.90 in the normal probability table of the appendix. We note that the closest probability is 0.8997 which lies in the row 1, and in the column 0.08 of the normal probability table and thus the corresponding z-score is then 1.2 1- .08 = 1.28.
z0 = 1.28
The z-score is the observed value decreased by the mean, divided by the standard deviation:
z0 = x - μ/σ = x0 - 70/12
However, we also kmow z0 = 1.28
1.28 = x0 - 70/12
=> x0 = 85.36
Thus the test should be termnated after 85.2 minutes, which will allow enough time for 90% of the students to complete the test.