6. (10 pts) Sick-leave time used by employees of a firm in the course of one mon

Question

6. (10 pts) Sick-leave time used by employees of a firm in the course of one month has approximately a normal distribution, with a mean of 200 hours and a variance of 400 hours. (a) Find the probability that total sick leave for the next month will be less than 168 hours (b) In planning schedules for next month, how much time should be budgeted for sick leave if that amount is to be exceeded with a probability of only 0.10? (c) Assume months are independent with regards to sick-leave time. What is the probability that at most 3 months out of the next 12 months have a total sick leave that exceeds 215 hours?

Explanation / Answer

Ans:

Given that

mean=200

standard deviation=sqrt(400)=20

a)

z=(168-200)/20=-1.6

P(z<-1.6)=0.0548

b)

P(Z>=z)=0.1

P(Z<=z)=1-0.1=0.9

z=normsinv(0.9)=1.282

x=200+1.282*20=225.63 hrs.

c)First find P(x>215)

z=(215-200)/20=0.75

P(z>0.75)=0.2266

Let y be the number of months having sick leave hrs exceeding 215 hrs.

Then,y has binomial distribution with n=12,p=0.2266

P(y<=3)=binomcdf(12,0.2266,3)=0.7197

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