My youngest brother used to own a cafe in Calgary and I visited him frequently for free meals. One day, Indiana Jones rushed in and wanted to order two dozen roasted chickens. But there was a catch. He insisted that all the chickens should be heavier than 5.5 lbs, or else my brother would face very serious consequences.
My brother had to decide if he should take Indiana's order. He would like to have the business, but he did not want to risk Indiana's wrath. He turned to me for advice as I had been enjoying frequent free meals at his joint.
What I knew at that point was that the weight of all his roasted chickens was normally distributed, with a mean weight of 5.2 lbs and a standard deviation of 0.35 lb. And he had 100 chickens roasting on the rack, which should be ready in less than 5 minutes.
Indiana was getting very impatient, and how should I advise my brother? Take the order or not?
m = 5.2, s = 0.35, n = 100, x = 5.5
z = (x - m)/(s/ sqrt n)
z = (5.5 - ...
In a brief, but concise, step-wise manner, this solution illustrates how to compute the z-statistic in order to make a decision on whether to accept the order or not. This solution is neat and very easy to follow.