# CLT so we have enough burgers!

An event has 32 people. Caterers need to make burgers for these people. They expect that a person at this event might need 0, 1 or 2 burgers with probabilities 0.2, 0.55, and 0.25 respectively. We assume that the amount of burgers each person eats is independent. How many burgers should the caterers make so that they are 99% sure that there isn't a lack of burger supply.

I have computed the following but not sure where to go from here:

$$\bar{x} = E(X) = 0(0.2) + 1(0.55) + 2(0.25) = 1.05$$ $$s^2 = Var(X) = E(X^2) - [E(X)]^2 = 0^2(0.2)+1^2(0.55)+2^2(0.25)-1.05^2= 0.4475$$ $$Z = 2.576 \:\:\:\ \text{(Z-value for desired confidence level of 0.99)}$$

I know I have to somehow use the CLT where $$\bar{x} \sim N(\mu, \frac{\sigma^2}{n})$$ for i.i.d random variables with mean $$\mu$$ and variance $$\sigma^2$$ for sufficiently large $$n$$.

• This appears to be a problem with the total, not the mean. It seems to me you need to use a normal approximation with $N(n\mu,n\sigma^2)$. – Gregg H Apr 27 at 22:15
• @GreggH So, $N(33.6, 14.32)$? I believe my confusion lies in what to consider $n$? The people or the burgers? And how I actually get to calculate the required amount of burgers for the 32 people. I've tried sample size determination and CIs, but those don't seem to work. – User_13 Apr 27 at 22:28
• In this context, n is the number of random experiments, which would be the number of people. (The random variable is the number of burgers.) I might suggest another approach to get to the desired count. You can ask what would be the upper bound on the mean of the number burgers eaten by 32 randomly chosen people. This would use the CLT. Then, with your final answer, you simply multiply the upper bound for the average by 32 (the number of people at the party). – Gregg H Apr 27 at 22:35
• Correct, but note that you know $n$ here. (I'd suggest posting your solution as an answer to your initial question.) – Gregg H Apr 27 at 22:41
• It is a conservative estimate for the average number of burgers a group of that size will have. Recall, the CLT describes the sampling distribution...in this case, the average number of burgers eaten, not the number of burgers eaten by one person. With a bound for that average, we can "work back the math" to get the total number of burgers that a group that size would have had to have eaten. – Gregg H Apr 27 at 22:46

$$32\left(1.05 + \dfrac{2.576(\sqrt{0.4475})}{\sqrt{32}}\right) = 43.348... \rightarrow 44$$