# Normal Distribution

• Nov 1st 2012, 06:55 PM
samanthah8625
Normal Distribution
A sell-out crowd of 42,200 is expected at Cleveland's Jacob field. 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 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?
• Nov 1st 2012, 07:05 PM
chiro
Re: Normal Distribution
Hey samanthah8625.

This is actually a proportion/Bernoulli estimation problem but since you have so many people, you can use a Normal approximation.

Hint: Do you know how to calculate the standard deviation of the mean? (Given that the variance is p*(1-p) with n observations)?
• Nov 1st 2012, 07:43 PM
samanthah8625
Re: Normal Distribution
Yeah i know the formula X=(np(1-p))^(1/2) Z+ np which would be X= 99.7 Z + 16036 but the problem I'm having is what the probability would be. P( Z < ?)
• Nov 1st 2012, 09:29 PM
chiro
Re: Normal Distribution
Well if you want to have 20% not exceeding supply you need to find the distribution of the sample mean X and show that P(X < a) = 0.8 where you need to find a.

You are given the distribution of X as normal with sample mean p = 0.38 and variance p*(1-p)/N where N is the number of people in the stadium so you have complete information to get the distribution and solve for the probability (it will be a quantile calculation).

Once you have this, then you have an upper limit of the proportion of hot-dogs that you need since a will be a proportion (it should be given this problem).

If you assume that one person will buy a hot-dog then the limit proportion will b a*N where N again is the number of people.