Let x be 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 the x distribution is about = 2.7 minutes, with standard deviation = 0.6 minute. Assume that the express lane always has customers waiting to be checked out and that the distribution of x values is more or less symmetric and mound-shaped. What is the probability that the total checkout time for the next 30 customers is less than 90 minutes? Let us solve this problem in steps.

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 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.The probability distribution ofxis not normal.

(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 .