Letxbe a random variable that represents checkout time (time spent in the actual checkout process) in minutes in the express lane of a large grocery. Based on a consumer survey, the mean of thexdistribution is about= 2.7minutes, with standard deviation= 0.6minute. Assume that the express lane always has customers waiting to be checked out and that the distribution ofxvalues is more or less symmetric and mound-shaped. What is the probability that thetotalcheckout time for the next 30 customers is less than 90 minutes? Let us solve this problem in steps.

(a) Letx_i(fori= 1, 2, 3, . . . , 30) represent the checkout time for each customer. For example,x₁is the checkout time for the first customer,x₂is the checkout time for the second customer, and so forth. Eachx_ihas mean= 2.7minutes and standard deviation= 0.6minute. Letw=x₁+x₂+ . . . +x₃₀.Explain why the problem is asking us to compute the probability thatwis less than 90.

wis the total waiting time for all customers, so we want to findP(w< 90).

wis the average waiting time for 30 customers, so we want to findP(w< 90).

wis the average of the waiting times for 30 customers, so we want to findP(w< 90).

wis the sum of the waiting times for 30 customers, so we want to findP(w< 90).

(b) Use a little algebra and explain whyw<90 is mathematically equivalent tow/30<3.

add 30

divide by 30

subtract 90

subtract 30

divide by 3

Sincewis the total of the 30xvalues, thenw/30 =x.Therefore, the statementx<3is equivalent to the statementw<90.From this we conclude that the probabilitiesP(x<3)andP(w<90)are equal.

(c) What does the central limit theorem say about the probability distribution ofx? Is it approximately normal? What are the mean and standard deviation of thexdistribution?

The probability distribution ofxis approximately normal with mean_x= 2.7 and standard deviation_x= 0.6.

The probability distribution ofxis not normal.

The probability distribution ofxis approximately normal with mean_x= 2.7 and standard deviation_x= 0.11.

The probability distribution ofxis approximately normal with mean_x= 2.7 and standard deviation_x= 0.02.

(d) Use the result of part (c) to computeP(x<3). (Round your answer to three decimal places.)

What does this result tell you aboutP(w<90)? (Round your answer to three decimal places.)

The probability that the total checkout time for the next 30 customers is less than 90 minutes is.