In: Statistics and Probability
You just received a grant to study the motivation effects of pay level on employee performance. Suppose you examine job performance scores for 630 employees, where a high score indicates good work performance. The results of your study determine the average performance score is 86, with a standard deviation of 11 points. Assuming a normal distribution, determine the percent and number of employees who:
a. What is the job performance score at the 29th percentile (6 pts)?
b. What is the job performance score at the 86th percentile (6 pts)?
Given that,
mean = = 86
standard deviation = = 11
n = 630
= 86
= / n = 11/ 630 = 0.4383
a. The z-distribution of the 29% is,
P(Z < z) = 29%
= P(Z < z ) = 0.29
= P(Z < -0.553 ) = 0.29
z = -0.553
Using z-score formula
= z * +
= -0.553 * 0.4383 + 86
= 85.76
Answer :- 85.76
b. The z-distribution of the 86% is,
P(Z < z) = 86%
= P(Z < z ) = 0.86
= P(Z < 1.080 ) = 0.86
z = 1.080
Using z-score formula
= z * +
= 1.080 * 0.4383 + 86
= 86.47
Answer :- 86.47