Question

QUESTION 2 The times that a cashier spends serving individual customers are independent random variables with mean 2.5 minutes and standard deviation 2 minutes. What is approximate probability that (a) the average time to serve a sample of 64 customers is within 0.5 minutes of the true mean above? (b) it will take more than 4 hours to process the orders of 100 customers?

Answer 1

a)

here std error =std deviation/sqrt(n)=2/sqrt(64)=0.25

hence from normal approximation ; P(sample mean is within 0.5 minute of true mean):

P(2<Xbar<3)=P((2-2.5)/0.25<Z<(3-2.5)/0.25)=P(-2<Z<2)=0.9772-0.0228=0.9544

b)

expected time to process 100 customers =100*2.5=250 minute

and std deviation=2*sqrt(100)=20

hence P(it take more then 4 hours (240 minutes))=P(X>240)=P(Z>(240-250)/20)=P(Z>-0.5)=0.6915