SOLUTION: a company has 23,5000 employees. the average length of service for the employees is 9.3 years with a standard deviation of 4.2 years. assuming that the length of service at this co
Algebra ->
Probability-and-statistics
-> SOLUTION: a company has 23,5000 employees. the average length of service for the employees is 9.3 years with a standard deviation of 4.2 years. assuming that the length of service at this co
Log On
Question 1100921: a company has 23,5000 employees. the average length of service for the employees is 9.3 years with a standard deviation of 4.2 years. assuming that the length of service at this company is normally distributed, find the lengths of service (to the nearest one-tenth of a year) for the middle 80% of the employees. Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! The middle 80% is the 10th and 90th percentiles
10th percentile is z=-1.28; 90th is +1.28
z=(x-mean)/sd
1.28=(x-9.3)/4.2
x-9.3=5.38
x=14.68
-1.28=(x-9.3)/4.2
x=3.92
(3.9, 14.68) units are years
