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