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The question: A sell-out crowd of $42,200$ is expected at Cleveland's Jacobs Field for next Tuesday's game against 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 that a $20$% chance of demand exceeding supply?

I know I need to use a normal distribution for this but I am having trouble figuring out what the specific parts are, for example, what is n?

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38% is the proportion of people expected to buy hot dogs, $p$, and 42,200 is the total number of people expected at the game, $n$. You should approximate the distribution of people buying hot dogs, which is binomial, with a normal distribution. – caburke Feb 1 at 4:49
@caburke how do I use these to find out the answer? I think I am really stuck on how to use these numbers correctly even though I have thought about this problem for several days now. – user59633 Feb 6 at 1:06

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So you can start with normal approximation to the binomial with $\mu=np$ and $\sigma^2=np(1-p)$, where $p=0.38$ and $n=42,200$. Let $X$ then denote the order that needs to be placed. You just need to find $R_{hotdogs}$ where $P(X<R_{hotdogs})=1-0.2$.

$R_{hotdogs}$ is the number of hotdogs that she needs to put orders for if she wants to have no more that a $20\%$ chance of demand exceeding supply. I am hoping you can calculate the quantile here.

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I am still confused. How do you set this problem up? How do you get $R_{hotdogs}$? Do you set up your probability and let that equal $0.2$? – user59633 Feb 3 at 15:27
@user59633 Please review your understanding of normal distribution and their quantile calculations and you would know. – jay-sun Feb 3 at 17:48
But how do you use the 20% in the normal distribution? – user59633 Feb 5 at 0:10
@user59633 I was hoping you can find it out. You have to find the value at which the probability for this distribution is $80\%$, i.e., $0.8$. If you can't, then I think you seriously need to brush up you probability concepts again. – jay-sun Feb 5 at 1:33
Do I find $P[0 < \frac{X-16036}{\sqrt{9942.32}} <0.2]$? That is do I find $P[0 < \frac{X-np}{\sqrt{np(1-p)}} <0.2]$? – user59633 Feb 5 at 21:44

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