Question

The times that a cashier spends processing individual customer’s order are independent random variables with mean 3 minutes and standard deviation 2 minutes.  What is the approximate probability that it will take less than 4 hours to process the orders of 100 people? If a sample of 80 customers is selected, what is the probability that the sample mean (¯ Y ) of their processing times is within 1.5 standard deviations of the sample mean (σ_{¯ Y} ) from the true mean?

Answer 1

1)

for 100 people ; expected mean =3*100=300 minute

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

therefore probability that it will take less than 4 hours to process the orders of 100 people =P(X<240 minutes)

=P(Z<(240-300)/20)=P(Z<-3)=0.0013

2)

probability that the sample mean (¯ Y ) of their processing times is within 1.5 standard deviations of the sample mean (σ_{¯ Y} ) from the true mean =P(-1.5<Z<1.5)=0.9332-0.0668 =0.8664