Two movie theaters compete for the business of 1, 000 customers. Assume that each customer chooses between the movie theaters completely at random. If they find the chosen theater is full, then they will return home, instead of going to the other theater. Let N denote the number of seats in each theater.
- For one of the movie theater A, using a binomial model, find an expression for N that will guarantee that the probability of turning away a customer (because of a full house) is less than 1%.
My Attempt Let X be a random variable representing the number of people that choose a certain movie theater. So, $X\sim\text{Bin}(1000,\frac{1}{2})$.
The probability that we turn away a customer is $\mathbb{P}(X>N)$ and we know that this is the same as $\mathbb{P}(X=N+1)+\mathbb{P}(X=N+2)+...$. So we have $$\sum^{1000}_{X=N+1}\binom{1000}{X}\left(\frac{1}{2}\right)^X\left(1-\frac{1}{2}\right)^{1000-X}<0.01$$.
My question is, how would I go about finding the value of N for this. I tried guessing but the expression is too much for my calculator and I get an overflow error. (I could cheat and use a normal approximation to get my N but I want to know how to solve it without that). Thank you
Assuming customers are equally likely to choose either theater, and choices are independent, then we want the least $N$ such that $\Pr[X > N] < 0.01$, where $X \sim \operatorname{Binomial}(n = 1000, p = 1/2)$. Because the mean is $np = 500$ and the standard deviation is $\sqrt{np(1-p)} = 5\sqrt{10} \approx 15.8$, in order for the right-tailed probability to be less than $0.01$, we'd have to be somewhere between $2$ and $3$ standard deviations above the mean. This estimate uses the $68$-$95$-$99.7$ rule. The actual $z$-score should be $$z_{0.01}^* = \Phi^{-1}(0.99) \approx 2.32635.$$ So using a continuity-corrected approximation, $$\Pr[X > N] = \Pr\left[\frac{X - \mu}{\sigma} > \frac{N - 500 + 1/2}{5 \sqrt{10}}\right] \approx \Pr\left[Z > \frac{N - 499.5}{5 \sqrt{10}}\right],$$ hence we require $$\frac{N - 499.5}{5 \sqrt{10}} \ge z_{0.01}^* = 2.232635,$$ or $N \ge 537$. This happens to be the exact bound, mainly because $n = 1000$ is so large and $p = 1/2$ is optimal for the approximation.
How then do we calculate the exact probability? It requires computing $$\Pr[N > 537] = 2^{-1000} \sum_{x=538}^{1000} \binom{1000}{x}.$$ This is highly problematic without a computer, since the exact value is
$$\frac{23656541122627242261775011991151552914411782707957602657343671920897378833066433409453904020724128142312409463116182645799035563749820974026705212163870470815028374811940015472552709729473060463950492383604650435801919238682815367875510082809740313834734023280158062894359172404052828968053317369907}{2678771517965668302371062622650004526403512029263834018609375970925877627812340306232995947039239645318986682293882867062967863214230785108996144393674643700983641943706057746355268651265592785469488545538261618745895485316849691889791385986519265728642799119421635541915107457913156096709301417017344}.$$
And you have to show that adding $\binom{1000}{537}2^{-1000}$ makes this value greater than $0.01$. I see no way to do this analytically, by hand, without a computer. Perhaps some calculators have a built-in function to compute the binomial CDF, but this is essentially the same thing as using a computer.