Question

Question 4:

The times that a cashier spends processing each person’s transaction are independent and identically distributed random variables with a mean of µ and a variance of σ² . Thus, if X_i is the processing time for each transaction, E(X _i) = µ and Var(X_i) = σ² .

Let Y be the total processing time for 100 orders: Y = X₁ + X₂ + · · · + X₁₀₀

(a) What is the approximate probability distribution of Y , the total processing time of 100 orders? Hint: Y = 100X, where X = 1 100 P100 i=1 Xi is the sample mean.

(b) Suppose for Z ∼ N(0, 1), a standard normal random variable:

P(a < Z < b) = 100(1 − α)%

Using your distribution from part (a), show that an approximate 100(1 − α)% confidence interval for the unknown population mean µ is:

(Y − 10bσ)/100 < µ < (Y − 10aσ)/100

(c) Now suppose that the population mean processing time is known to be µ = 1.5 minutes, and the population standard deviation processing time is known to be σ = 1 minute. What is the probability that it takes less than 120 minutes to process the 100 orders? If you use R, please provide the commands used to determine the probability. Could you show all steps in the hand written working for this question please.

Answer 1

We are given that for the random variable

and

And the random variable Y is given as:

So we have

and

So the estimated probability distribution of Y is:

(b) Given that Z is a standard normal variate that is

Now we have

The above is with probability

(c) The mean processing time is given as and . Then we have

The required probability is given as: