A sell-out crowd of 42,200 is expected at Cleveland\'s Jacobs Field for next Tue

Question

A sell-out crowd of 42,200 is expected at Cleveland's Jacobs Field for next Tuesday's game with the Baltimore Orioles, the last before a long road trip. The ballpark's concession manager is trying to decide how much food to have on hand. Looking at records from games played earlier in the season, she knows that, on the average, 38% of all those in attendance will buy a hot dog. How large an order should she place if she wants to have no more than a 20% chance of demand exceeding supply?

Explanation / Answer

If 38% of people order a hot dog, then we could expect - if 42200 people show up - that 16036 hotdogs will be ordered for that game. Then, we want to know for what value of X is P(X>=16036)=.2, where X is the number of hotdogs that are expected to be ordered. Since X is a binomial random variable (think of each of the 42200 attendee's as a Bernouili trial - where the either buy a hotdog or they dont, with prob p = 0.38 that they do), then the standard deviation of X is npq or sqrt[(42200)(.38)(.62)]= 99.71. Then we want P(X>16036) = P[(X-16036)/99.71>z]=.2 From the z-chart we see that a z-score of .84 is approxiamately .8 (.7995). Then the probability of z being bigger than .84 is .2. Thus we have (X-16036)/99.71=0.84 => X=16036+(0.84)(99.71)=16 119.7564. Then the largest order she should place is 16120 hotdogs.

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