An aptitude test is designed to measure leadership abilities of the test subjects. Suppose that the scores on the test are normally distributed with a mean of 580 and a standard deviation of 120. The individuals who exceed 750 on this test are considered to be potential leaders. What proportion of the population are considered to be potential leaders

0.0778 = 7.78% of the population are considered to be potential leaders

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean and standard deviation , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean of 580 and a standard deviation of 120.

This means that

What proportion of the population are considered to be potential leaders?

Proportion of those who exceed 750, that is, 1 subtracted by the vpalue of Z when X = 750.

has a pvalue of 0.9222

1 - 0.9222 = 0.0778

0.0778 = 7.78% of the population are considered to be potential leaders

idk

step-by-step explanation: