Question

To qualify for a police academy, applicants are given a test of physical fitness. The scores are normally distributed with a standard deviation of 9 and a mean of 81. Find the minimum score for the top 30% of the applicants who took the test.

a) 87
b) 96
c) 93
d) 90

Answer 1

To find the minimum score for the top 30% of applicants, use a calculator to find the 70th percentile based on the given mean and standard deviation. Enter the inverse normal distribution function with the area as 0.7 and the given mean of 81 and standard deviation of 9.

Step-by-step explanation:

The student asked how to find the minimum score for the top 30% of applicants to a police academy, given that the test scores are normally distributed with a mean of 81 and a standard deviation of 9. To solve this, one would use a z-score table or a calculator to find the z-score that corresponds to the top 30% (or the 70th percentile since it's the minimum score for the top 30%). This z-score typically relates to an area of 0.7 on the standard normal distribution curve. Using a calculator like the TI-83, 83+, 84, or 84+:

1. Enter the distribution function, often found under a 'DISTR' menu.
2. Choose the inverse normal distribution function, often labeled 'invNorm'.
3. Enter the area to the left of the z-score as 0.7 to find the 70th percentile.
4. Input the mean and standard deviation of the score distribution (mean = 81, standard deviation = 9).
5. The calculator will give you the minimum score for the top 30%.

Unfortunately, the information provided about a 90th percentile of 69.4 does not fit the context as it contradicts the given mean of 81 and would not be required to answer this specific question.