A restaurant has 15 tables. Given that 70% of booked guests show up, how many reservations can be made to limit the risk of beeing overbooked?

Question

I have this problem which I need some help with.

"A restaurant has 15 tables, and it is known that 70% of guests who make reservations actually show up. To compensate for this, the restaurant has a policy of taking more than 15 reservations, thus running the risk of become overbooked. How many reservations can they take to limit this risk to at most 5%?"

I'm trying to solve the problem in the following way, please explain where my reasoing goes wrong.

If we consider each (booked) person that can possibly show up to be a Bernoulli trial, which we denote $Y_i$, we can note that $P(Y_i = 1) = 0.7 \forall i=1,2,...,N$, where $N$ is the number of trials in our "experiment". If we now consider $X$ to be the number of succesive trials (i.e. the number of guests that actually shows up) $X$ is clearly a binomial distribution with parameters $n=N$ and $p=0.7$. We now want to find the number of trials, $N$ (i.e. number of reservations).

So, we know that $E[X]=$"The expected number of guests showing up" should satisfy $E[X] \leqq 15 + 15*0.05$ (I think it is here I'm doing something wrong) and because we know that (for binomial distribution) $E[X] = Np$ we can simply solve for N. This gives the number of reservations to be less than 22.5. The book says the answer should be 17. What am I doing wrong?

Thanks!

Answer 1

You want $P(X \le 15) \ge .95,$ where $X \sim \mathsf{Binom}(n, .7).$ In R statistical software pnorm is a binomial CDF. (Ignore brackets [ ].)

n = 15:20;  p.ok = pbinom(15, n, .7)
cbind(n, p.ok)
      n      p.ok
[1,] 15 1.0000000
[2,] 16 0.9966767
[3,] 17 0.9807249
[4,] 18 0.9400478
[5,] 19 0.8668290
[6,] 20 0.7624922

Seems 17 is safe and 18 isn't.

If you are unable or unwilling to use software, a normal approximation with continuity correction would suffice. If $n = 17$ and $p = .7,$ then $$P(X \le 15.5) = P\left(\frac{X - np}{\sqrt{np(1-p)}} \le \frac{15.5-17(.7)}{\sqrt{17(.7)(.3)}}\right) \approx P(Z \le 1.91) = 0.9719.$$ So 17 is safe, and then you could try 18 (which gives about 0.93, and so isn't safe).

More generally, using a normal approximation, you need $n$ such that $\frac{15.5-.7n}{\sqrt{.21n}} > 1.645.$