Almost all companies utilize some type of year-end performance review for their employees. Human Resources (HR) at the University of Texas 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 15. a. What is the lowest rating you should give to an "exemplary" employee if you follow the University of Texas HR guidelines?
b. What is the lowest rating you should give to a "competent" employee if you follow the University of Texas HR guidelines?

Answer and Step-by-step explanation:

Solution:

Given:

µ = 50

∂ = 15

Lowest rating given to exemplary.

The top 10 %of the employees are related exemplary:

P(x > L) = 10 % =0.10

The average probability table in the appendix contains the only probability of the form.

P (0 < x< L ) = p ( x > 0 ) – P ( x > L )

                   = 0.5 -0.10

                    = 0.40

z- Score in the normal probability table in the appendix corresponding to a probability of 0.40 or the closest probability.

Z = 1.28

The standardized score is the value x decreased by the mean and then divided by the standard deviation.

Z = x - µ / σ

L = x = µ +z0

 = 50 + 1.28(15)

= 69.2

Lowest rating giving to competent:

The lower boundary of competent is boundary of bottom 30%

P(x < L ) = 30% = 0.30

The normal probability table contains only probability of the form p( 0≤z≤z0)

P( 0< x < -L) = P(L < x < 0)

                    = p(x < 0) – p(x < L)

                    =0.5 – 0.30

                     = 0.20

Z score in normal probability corresponding to the probability of 0.20

  -z = 0.52

Z= - 0.52

Value x decreased by the mean and  divide by standard deviation:

Z = x - µ / ∂

L = x = µ + zo

   = 50 – 0.52(15)

  = 42.2

phosphate, calcium ion, chloride ion, neon atom

step-by-step explanation:

i believe the correct answer is b.