Question

An office will increase salary to its top 8% employees on the basis of a performance score the office created for each employee. The performance score is approximately normal with mean 82.5 and standard deviation 9.25. How high must an employee score in order to qualify for increase in the salary?

Answer 1

Answer:

An employee's score in order to qualify for increase in the salary must be higher than 95.45.

Step-by-step explanation:

Let X represent the performance score of employees.

It is provided that X follows a normal distribution with parameters μ = 82.5 and σ - 9.25.

It is provided that the office will increase salary of its top 8% employees on the basis of a performance score the office created for each employee.

That is, the probability to qualify for increase in the salary is,

P (X > x) = 0.08

⇒ P (X < x) = 0.92

⇒ P (Z < z) = 0.92

The corresponding z-value is,

z = 1.40

Compute the value of x as follows:

$z=\frac{x-\mu}{\sigma}\\\\1.40=\frac{x-82.5}{9.25}\\\\x=82.5+(1.40\times 9.25)\\\\x=95.45$

Thus, an employee's score in order to qualify for increase in the salary must be higher than 95.45.