Suppose that, on average, in a Whole Foods Market in Dallas, 3.4 customers want to check out every minute. Based on this figure, store management wants to staff checkout lines such that less than 1% o

a) To ensure that less than 1% of the time demand for checkout cannot be met, store management would need to staff for the maximum number of customers that can be handled in one minute, which would be the 99th percentile of demand. To find this number, we need to use the normal distribution and convert the percentile into a z-score. Using a standard normal table or a calculator, we find that the z-score for the 99th percentile is \approx imately 2.33. We can then use the formula z = (x - ?) / ?, where ? is the mean and ? is the standard deviation, to solve for x. Plugging in ? = 3.4 and ? = sqrt(3.4) = 1.84 (since demand follows a Poisson distribution), we get: 2.33 = (x - 3.4) / 1.84 x = 7.71 Therefore, store management would need to staff for at least 8 customers every minute (rounding up for practical reasons) to ensure that demand for checkout can be met 99% of the time. b) The number of customers who want to check out in a two-minute period follows a Poisson distribution with parameter ? = 3.4 x 2 = 6.8 (since the average per minute is 3.4). To find the probability that there are 12 or more customers who want to check out in a two-minute period, we can use the Poisson formula: P(X ? 12) = 1 - P(X ? 11) = 1 - ? (e^-6.8) (6.8^k) / k! where ? goes from k = 0 to 11. Using a calculator or a software program, we find that P(X ? 12) ? 0.0125, or about 1.25%. Therefore, the store would expect to have 12 or more customers who want to check out in any two-minute period only about 1.25% of the time.