Service time for a customer coming through a checkout counter in a retail store

Question

Service time for a customer coming through a checkout counter in a retail store is a random variable with the mean of 4.0 minutes and standard deviation of 0.5 minutes. Suppose that the distribution of service time is fairly close to a normal distribution. Suppose there are two counters in a store, ni = 1 1 customers in the first line and n2-27 customers in the second line. Find the probability that the difference between the mean service time for the shorter line X1 and the mean service time for the longer one X2 is more than 0.6 minutes. Assume that the service times for each customer can be regarded as independent random variables. Round your answer to two decimal places (e.g. 98.76)

Explanation / Answer

here mean differnce in tme =0

and std error of mean difference =sqrt(0.52/11+0.52/27)=0.1788

therefore probability that difference is more than 0.6 =P(Xbar>0.6)=P(Z>(0.6-0)/0.1788)=P(Z>3.36)

=0.00 ( please try 00.04 if this comes wrong)

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