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.
Picture

col1# of days absent 0 1 2 3 4 5
col2Probability 0.60 0.20 0.12 0.04 0.04 0
What is the variance of the number of days absent?

a. 1.1616
b. 55.52
c. 5.00
d. 1.41

Answer 1

Answer:

a. 1.1616

Step-by-step explanation:

Variance is given by:

$\sum x_i^2p_i -\mu^2$

Where "xi" is each individual number of absences, and "pi" are their respective probability. The mean number of absences is:

$\mu=0*0.6+1*0.2+2*0.12+3*0.04+4*0.04+5*0\\\mu= 0.72\ absences$

Applying the given data, the variance is:

$V=0^2*0.6+1^2*0.2+2^2*0.12+3^2*0.04+4^2*0.04+5^2*0 - 0.72^2\\V= 1.1616\ absences$

The variance of the number of days absent is 1.1616 days