Question

Workers’ unscheduled absence survey. Each year CCH, Inc., a firm that provides human resources and employment law information, conducts a survey on absenteeism in the workplace. The latest CCH Unscheduled Absence Surveyfound that of all unscheduled work absences, 34% are due to “personal illness,” 22% for “family issues,” 18% for “personal needs,” 13% for “entitlement mentality,” and 13% due to “stress.” Consider a randomly selected employee who has an unscheduled work absence.

a. List the sample points for this experiment.

b. Assign reasonable probabilities to the sample points.

c. What is the probability that the absence is due to something other than “personal illness”?

Short answer

1. The sample points are PI, F, P and E.
2. The probability are 0.390, 0.252, 0.206 and 0.149.
3. The probability is 0.609

Step-by-step solution

Important formula

The formula for probability is $\mathbf{P}\mathbf{=}\frac{\mathrm{favourable}\mathrm{outcomes}}{\mathrm{total}\mathrm{out}\mathrm{comes}}$.

The sample points.

a)

The sample points are

Here PI=personal illness

F=family issue

P=personal needs

E=entitlement mentality.

Thus, the sample points are PI, F, P, E.

Step 3:The possible probabilities.

b)

$\begin{array}{c}\mathrm{P}\left(\mathrm{PI}\right)=\frac{0.34}{0.87}\\ =0.391\end{array}$

$\begin{array}{c}P\left(F\right)=\frac{0.22}{0.87}\\ =0.253\end{array}$

$\begin{array}{c}P\left(P\right)=\frac{0.18}{0.87}\\ =0.207\end{array}$

$\begin{array}{c}P\left(E\right)=\frac{0.13}{0.87}\\ =0.149\end{array}$

Hence, the probability is 0.391, 0.253, 0.207, and 0.149.

The probability that the absence is due to something other than “personal illness”.

c)

$\begin{array}{c}\mathrm{P}\left(\mathrm{F}\text{and}\mathrm{P}\text{and}\mathrm{E}\right)=\frac{0.22+0.18+0.13}{0.87}\\ =0.609\end{array}$

Therefore, the probability is 0.609