Acompany 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.
number of days absent probability
0 0.60
1 0.20
2 0.12
3 0.04
4 0.04
5 0.00
what is the variance of the number of days absent? select one:
a. 1.1616
b. 1.41
c. 5.00
d. 55.52

a. 1.1616

Explanation:

For computing the variance, first we have to determine the mean which is shown below:

Mean = Number of Days × Absent Probability

= 0 × 0.60 + 1 × 0.20 + 2 × 0.12 + 3 × 0.04 + 4 × 0.04

= 0 + 0.20 + 0.24 + 0.12 + 0.16

= 0.72

Now the variance equal to

= (Number of Days Absent - Mean) ^2  × Probability

So,

=  (0 - 0.72)^2 × 0.60 +(1 - 0.72)^2  × 0.20 +(2 - 0.72)^2  × 0.12 +(3 - 0.72)^2  × 0.04 + (4 - 0.72)^2  × 0.04

= 1.1616

explanation:

kool beans

its ii) i did this test

explanation: