Using the normal distribution, it is found that 0.0329 = 3.29% of the population are considered to be potential leaders.
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
In this problem:
The proportion of the population considered to be potential leaders is 1 subtracted by the p-value of Z when X = 780, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{780 - 550}{125}[/tex]
[tex]Z = 1.84[/tex]
[tex]Z = 1.84[/tex] has a p-value of 0.9671.
1 - 0.9671 = 0.0329
0.0329 = 3.29% of the population are considered to be potential leaders.
To learn more about the normal distribution, you can take a look at https://brainly.com/question/24663213