Almost all companies utilize some type of year-end performance review for their employees. Human Resource (HR) at a university's Health Science Center provides guidelines for supervisors rating their subordinates. For example, raters are advised to examine their ratings for a tendency to be either too lenient or too harsh. According to HR, ``if you have this tendency, consider using a normal distribution -- 10% of employees (rated) exemplary, 20% distinguished, 40% competent, 20% marginal, and 10% unacceptable.'' Suppose you are rating an employee's performance on a scale of 1 (lowest) to 100 (highest). Also, assume the ratings follow a normal distribution with a mean of 50 and a standard deviation of 20.
What is the lowest rating you should give to an ``exemplary'' employee if you follow the university's HR guidelines?
A. 76
B. 85
C. 90
D. 25

Given the normal distribution " 10% of employees (rated) exemplary, 20% distinguished, 40% competent, 20% marginal, and 10% unacceptable'', we can see that exemplary employees are top 10% rated employees.

We have the formula for normal distribution:

z=(X-M)÷σ

where z is the minimum z-score for top 10% employee, X is the minimum score for top 10% employee, M is the mean of the score distribution, σ is the standard deviation of the score distribution.

The z-score we are looking for is the value "a" that separates the highest 10% from the lowest 90% i.e. P(z≤a)=0.90

If we look at z-table, corresponding value for a is 1.28155

We can now put the values in the formula:

1.28155=[tex]\frac{X-50}{20}[/tex]

So X=(1.28155×20)+50=75.631

Therefore minimum score for exemplary employee is 76.