Question

The length of time required to complete a college achievement test is found to be normally distributed with mean 92 minutes and standard deviation 9 minutes. When should the test be terminated if we wish to allow sufficient time for 92% of the students to complete the test? (Round your answer to two decimal places.)

Answer 1

To accommodate 92% of the students completing a college achievement test normally distributed with a mean of 92 minutes and a standard deviation of 9 minutes, the test should be terminated at approximately 104.70 minutes.

Step-by-step explanation:

To determine when the college achievement test should be terminated to allow 92% of the students to complete it, we need to find the test duration that corresponds to the 92nd percentile of the normally distributed test completion times. The mean test completion time is given as 92 minutes, and the standard deviation is 9 minutes.

First, we look up the z-score that corresponds to the 92nd percentile in a standard normal distribution table or use a calculator that provides this functionality. The z-score for the 92nd percentile is approximately 1.41.

Next, we use the z-score formula:

z = (X - μ) / σ

Where z is the z-score, X is the value from the normal distribution (the test duration we are solving for), μ is the mean, and σ is the standard deviation.

Plugging in the values we have:

1.41 = (X - 92) / 9

Now we solve for X:

X = 1.41 * 9 + 92

X = 12.69 + 92

X = 104.69

Therefore, the test should be terminated at 104.69 minutes, which can be rounded to 104.70 minutes, to allow enough time for 92% of the students to complete it.