Scores for a common standardized college aptitude test are normally distributed with a mean of 493 and a standard deviation of 95. Randomly selected men are given a Test Preparation Course before taking this test. Assume, for sake of argument, that the test has no effect. If 1 of the men is randomly selected, find the probability that his score is at least 532.1.

34.01% probability that his score is at least 532.1.

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:

If 1 of the men is randomly selected, find the probability that his score is at least 532.1.

This is 1 subtracted by the pvalue of Z when X = 532.1. So

has a pvalue of 0.6591

1 - 0.6591 = 0.3409

34.01% probability that his score is at least 532.1.

$4034.40

the perimeter of the field is the sum of the lengths of the four sides:

47 yd + 158 yd + 47 yd + 158 yd = 410 yd

the cost of 3 ft (1 yd) of fence is 3×$3.28 = $9.84.

so the cost of jake's wood perimeter fance is

(410 yd) × ($9.84 /yd) = $4034.40