Suppose worker A averages 100 picks per hour, worker B averages 60 picks per hour, and worker C averages 40 picks per hour. [20 points] a. If the 3 workers work in a bucket brigade, in which order the workers should be arranged?
b. If the average order requires 100 picks, what would be the average rate of order completion of the team under bucket brigades?
c. Assume the 3 workers are assigned into zones with equal work in each. What is the average rate of order completion of the team when the workers are arranged in A à B à C order?

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step-by-step explanation:

irst, find the z score of a score of 80. this is given by the formula z = (x-u)/o, where x is the score, u is the mean, and o is the standard deviation. this gives

z = (80-78)/12 = 1/6 = .1667

now, find what percentile a z score of 1/6 correlates to in the normal distribution.

a z score of 1/6 = .1667 is equivalent to the 56.36th percentile of the distribution. so the probability a score is over this is 1-.5636 = .4364

the mean of a sample mean is equal to the original mean. however, the standard deviation of a sample mean is equal to the original standard deviation divided by the square root of n (sample size). so the new standard deviation is 12/sqrt(16) = 3.

the new z score is 80-78/3 = 2/3 = .667. this is equivalent to the 74.86th percentile. so the probability that the mean will be greater than this score is 1-.7486 = .2514.