Question

Suppose that the time between two absences of a supermarket employee in a company follows an exponential distribution. Based on the past data on number of supermarket employees absent in a particular month, it has been observed that on an average 5 supermarket employees tend to be absent in any given month. What is the probability that the time between two absences for a particular supermarket employee is:

1. Less than 2 weeks

2. Between 2 weeks and 4 weeks

3. More than a month

Answer 1

Time between two absences follows an exponential distribution.

One month has 4 weeks.

Given: On an average 5 supermarket employees tend to be absent in any given month.

4/5 =0.8 week per absence or 0.8/4 =0.2 months occur between successive absences on an average.

So, the mean, =0.2

m = =1/0.2 =5

Cumulative probability is: P(T < t) = 1 – e^-mt

1)

The probability that the time between two absences for a particular supermarket employee is less than 2 weeks:

2 weeks is 1/2 month. So,

P(T < 2 weeks) =P(T < 1/2 month)

=1 - e^-5(1/2) =1 - e^-2.5 =0.9179

2)

The probability that the time between two absences for a particular supermarket employee is between 2 weeks and 4 weeks is:

P(2 weeks < T < 4 weeks) =P(1/2 month < T < 1 month)

=P(T < 1 month) - P(T < 1/2 month) =(1 - e^-5(1)) - (1 - e^-5(1/2)) =0.9933 - 0.9179 =0.0754

3)

The probability that the time between two absences for a particular supermarket employee is more than a month is:

P(T > 1 month) =1 - P(T < 1 month) =1 - 0.9933 =0.0067