Answer:
0.0778 = 7.78% of the population are considered to be potential leaders
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:
![Z = (X - \mu)/(\sigma)](https://img.qammunity.org/2022/formulas/mathematics/college/bnaa16b36eg8ubb4w75g6u0qutzsb68wqa.png)
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
![\mu = 580, \sigma = 120](https://img.qammunity.org/2022/formulas/mathematics/college/5li3x1se3pz0qusnlppcizznnxjyubh3g2.png)
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.
![Z = (X - \mu)/(\sigma)](https://img.qammunity.org/2022/formulas/mathematics/college/bnaa16b36eg8ubb4w75g6u0qutzsb68wqa.png)
![Z = (750 - 580)/(120)](https://img.qammunity.org/2022/formulas/mathematics/college/gxiwfavzffztiuwnx1yl3hvppqmkbbzyyf.png)
![Z = 1.42](https://img.qammunity.org/2022/formulas/mathematics/college/9axfc8bjn8pvu6q3nu9kxo5mwp82fjyvba.png)
has a pvalue of 0.9222
1 - 0.9222 = 0.0778
0.0778 = 7.78% of the population are considered to be potential leaders