A company is studying the number of monthly absences among its 125 employees. The following probability distribution shows the likelihood that people were absent 0, 1, 2, 3, 4, or 5 days last month.

-What is the variance of the number of days absent?
A)1.99
B)1.41
C)5.00
D)55.52

Question

A company is studying the number of monthly absences among its 125 employees. The following probability distribution shows the likelihood that people were absent 0, 1, 2, 3, 4, or 5 days last month.

-What is the variance of the number of days absent?
A)1.99 
B)1.41 
C)5.00 
D)55.52

Quizplus · Accepted Answer

The variance is the average of the squared differences from the mean. The mean is 5. The squared differences from the mean are (0-5)^2 = 25, (1-5)^2 = 16, (2-5)^2 = 9, (3-5)^2 = 4, (4-5)^2 = 1. The average of these squared differences is (25+16+9+4+1)/5 = 1.99.

A Company Is Studying the Number of Monthly Absences Among