Scores for a common standardized college aptitude test are normally distributed with a mean of 514 and a standard deviation of 109. Randomly selected men are given a Test Prepartion 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 592.7.

23.58% probability that his score is at least 592.7

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 592.7.

This is 1 subtracted by the pvalue of Z when X = 592.7. So

has a pvalue of 0.7642.

1 - 0.7642 = 0.2358

23.58% probability that his score is at least 592.7

7

step-by-step explanation:

step-by-step explanation:

milk=16 large

soda=280 kids

juice box=82 regular

63 soda +63 regular

124 large total