There are 300 hotel rooms at a hotel. 15% chance of cancellation for each room. What is the max # of rooms you can book with a 95% chance that you won't be overbooked.
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There are 300 hotel rooms at a hotel. 15% chance of cancellation for each room. What is the max # of rooms you can book with a 95% chance that you won't be overbooked.
That do not make no sense to me.
Repeat question, from the way I understand it, the answer is impossible.
p(a reservation will be kept) = .85 and those who don't keep their reservation simply don't show up, thus taking up a room without paying for it
Thus probability of getting a hotel is .85 and you want to have $\displaystyle n$ such as probability at least one is .95 or more. If you have $\displaystyle n$ what is the probability that you do not get a room? Well, each one is .15 and there are $\displaystyle n$ thus the probability is OF GETTING AT LEAST ONE ROOM, isQuote:
Originally Posted by magicpunt
$\displaystyle 1-(.15)^n$ and you want that $\displaystyle P(n)\geq .95$. Thus,
$\displaystyle 1-(.15)^n\geq .95$
Thus,
$\displaystyle -(.15)^n\geq -.05$
Thus,
$\displaystyle (.15)^n\leq .05$
Make chart, for values,
$\displaystyle
\left\{
\begin{array}{cc}
n&(.15)^n\\
1&.1500\\
2&.0225\\
3&.0034
$
Note at $\displaystyle n=2$ we have already it less than .05
However, I am sure if this is what you mean.
I appreciate the help but let me rephrase.
You are the hotel manager. You have 300 hotel rooms in your hotel. There is an 85% chance that those who book a room will actually show up and a 15% chance that they will not (thus the room is "booked" but not used).
You want to book as many rooms as possible so that there is a 95% chance that all of those who book with you will have a room (a 95% confidence that you won't be overbooked).
Let P(N|M) be the probability that there are exactly N shows out of M bookings.Quote:
Originally Posted by magicpunt
Then the question is:
What is the largest M such that
P=P(0|M)+P(1|M) + P(2|M) + .. + P(300|M)>=0.95 ?
(that is: what is the maximum number of bookings M so that there is less
than a 5% chance of more than 300 shows?).
RonL
If the manager books $\displaystyle n$ rooms, the number of rooms taken will have a binomial distribution with probability of success $\displaystyle p = .85$ and number of trials $\displaystyle n.$ You want to know what is the highest value of $\displaystyle n$ that leaves a .95 probability that the number of rooms taken is no more than 300. Since calculating that probability from the binomial distribution is messy and probably not intended by your instructor, you should use the normal approximation to the binomial distribution.Quote:
Originally Posted by magicpunt
Treat the number of rooms taken as a normal variable $\displaystyle S_n$ with mean $\displaystyle pn$ and standard deviation $\displaystyle \sqrt{p(1-p)/n} .$ Find the maximum $\displaystyle n$ with the probability $\displaystyle P(S_n \le 300) \ge .95 .$ Can you take it from here?
Continuity correction!Quote:
Originally Posted by JakeD
$\displaystyle P(S_n \le 300.5) \ge .95 .$
RonL
Quote:
Originally Posted by JakeD
Standard deviation of a binomial RV with probability $\displaystyle p$ of
success on single trial and $\displaystyle n$ trials is:
$\displaystyle \sigma=\sqrt{n\ p(1-p)}$
RonL
Thank you, Captain! Confusing the standard deviation of $\displaystyle S_n /n,$ which is $\displaystyle \sqrt{p(1-p) /n},$ with the standard deviation of $\displaystyle S_n,$ which is $\displaystyle \sqrt{p(1-p)n}$ as you say, is a common mistake for me and it seems it's not the last time I'll make it. When working this kind of problem, I'll check whether the results are reasonable, and then have to go back and make that correction.Quote:
Originally Posted by CaptainBlack
As for the continuity correction, there ain't no stinking continuity correction in the Central Limit Theorem! ;)
Then how would you use the normal approximation to evaluate:Quote:
Originally Posted by JakeD
B(721, 0.232, 10342) - the probability of 721 success in 10342 trials
each with a probability of a favourable outcome of 0.232? :)
RonL
