In the article of Journal ManagementSolutionn170 mu 12 and

In the article of Journal Management

Solution

n=170, mu = 1.2 and sigma = 1.1 for the time lost

n=170, mu = 1.2 and sigma = 1.1 for the time paid lost

Unpaid time lost is mu =1 and s = 1.6

We have to calculate the probability that average for paid time lost for 100 workers exceed 1.5 days1

For our sample with 100 workers std error = 1.1/rt 100 = 0.11

Thus X the average time paid lost for 100 workers is N(1.2,0.11)

Reqd prob = P(X>=1.5)

= P(Z>=0.3/0.11)

= P(z>2.72)

= 0.0033