The average amount of time that students use computers at a university computer center is 36 minutes with a standard deviation of 5 minutes. The times are known to be normally distributed. Around 10,000 uses are recorded each week in the computer center. The computer center administrative committee has decided that if more than 2000 uses of longer than 40 minutes at each sitting are recorded weekly, some new terminals must be purchased to meet usage needs. Should the computer center purchase the new computers? (Round your percent to two decimal places, and round the number of computer users to the nearest user.)

2119 students use the computer for more than 40 minutes. This number is higher than the threshold estabilished of 2000, so yes, the computer center should purchase the new computers.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

The first step to solve this question is finding the proportion of students which use the computer more than 40 minutes, which is 1 subtracted by the pvalue of Z when X = 40. So

has a pvalue of 0.7881.

1 - 0.7881 = 0.2119

So 21.19% of the students use the computer for longer than 40 minutes.

Out of 10000

0.2119*10000 = 2119

2119 students use the computer for more than 40 minutes. This number is higher than the threshold estabilished of 2000, so yes, the computer center should purchase the new computers.

there is no

1) 2/5

2) zero=0/1

3) zero=0/2

4) -4/5

5) undefined=3/0