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.
Continuous Probability Distributions
Probability distributions are of two types, which are continuous probability distributions and discrete probability distributions. A continuous probability distribution contains an infinite number of values. For example, if time is infinite: you could count from 0 to a trillion seconds, billion seconds, so on indefinitely. A discrete probability distribution consists of only a countable set of possible values.
Normal Distribution
Suppose we had to design a bathroom weighing scale, how would we decide what should be the range of the weighing machine? Would we take the highest recorded human weight in history and use that as the upper limit for our weighing scale? This may not be a great idea as the sensitivity of the scale would get reduced if the range is too large. At the same time, if we keep the upper limit too low, it may not be usable for a large percentage of the population!
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.

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