Scores for a common standardized college aptitude test are normally distributed with a mean of 506 and a standard deviation of 114. 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 582.5.

25.14% probability that his score is at least 582.5.

Step-by-step explanation:

When the distribution is normal, we use 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 question, we have that:

If 1 of the men is randomly selected, find the probability that his score is at least 582.5.

This is 1 subtracted by the pvalue of Z when X = 582.5. So

has a pvalue of 0.7486

1 - 0.7486 = 0.2514

25.14% probability that his score is at least 582.5.

1.5%

step-by-step explanation:

46 dollars is the answer.

step-by-step explanation:

6*6 = 36,         36+10=46