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The question: A hotel has 230 rooms and uses an overbooking policy. The probability that a customer cancels or does not show up at the hotel is 0.12. Find the maximum number of rooms the hotel can book and still be 85% sure everyone who turns up will have room.

I have an issue regarding the steps to solving this type of question. I have workings which have given me two different answers, but I cannot work out which one is right, if either.

My working:

Method 1:

Take n=230 p=0.12 x=0,1,2,3...Z

Here Z would be the number of extra rooms over 230 the resort can book

Thus we do P(X=0) + P(X=1)+...+P(X=Z). When the sum of these probabilities is subtracted from 1 the answer should be greater than 0.85 but as close as possible. For this method I get 22. So the resort can book 252 customers.

Other method:

take n=Z p=0.88 x=0,1,2,3...230

If Z is the total number of rooms that can be booked for 85% confidence everyone will have a room, then we know that P(X=<230) = 1 - P(X>230). So working out P(X=<230) as P(X=0) + P(X=1)+...+P(X=230) then taking the sum of this from 1, we need a probability that is 0.15 or below. For this method I get 255. The problem with this method is the value must be inputted into excel and values of n must be guessed until you get a probability less than 0.15.

Would anyone be able to say which method is right? I neither would you be able to provide the correct ay of solving this problem.Trial and error in the second method seems unnecessary.

vexs
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  • Two thoughts: 1) have you considered using the normal approximation to the binomial distribution? 2) how exactly did you compute P(X=0) + P(X=1)+...+P(X=Z) in the first method? –  Mar 16 '16 at 23:49
  • I'm not sure how you do the normal approximation. In the first method I took n as the number of rooms and p as probability of them not turning up then the idea was to sum the binomial probabilities for increasing values of x until at a value of x the probability is less than 0.15 or if you do 1-cumulative then more than 0.85. Just dont really know the best method of solving this problem. – vexs Mar 17 '16 at 00:24
  • I meant, show how you computed P(X=n) for example. –  Mar 17 '16 at 00:51
  • If the you mean the calculation it was just a cumulative sum. So P(X=0) = [ 230!/((230-0)!(0!)) ] x〖(0.12〗^0 〖)(0.88〗^(230-0)) = 1.7 x 〖10〗^(-13). What I don't get is whether we are trying to find x given n, or whether we are trying to find n. – vexs Mar 17 '16 at 08:08

1 Answers1

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As mentioned it is probably a good idea to use the normal approximation to solve this problem and as you say you are not sure how to do that here is how:

The approximation to $X \sim B(n,p)$ is given by the normal distribution $N(np, np(1-p))$

The standard normal is

$$Z = \frac{X - \mu}{\sigma}$$

$$Z = \frac{X - np}{\sqrt{np(1-p)}}$$

$$Z = \frac{230 - 0.88n}{0.32\sqrt{n}}$$

Then you need the $15$% upper tail of $Z$ which is $\approx 1.04$. (You want to be $85$% sure $100-85=15$)

Now you solve this for $n$ to find the number of rooms you are looking for

$$1.04 = \frac{230 - 0.88n}{0.32\sqrt{n}}$$

JKnecht
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