Answer:
8.69% probability that at least 285 rooms will be occupied.
Step-by-step explanation:
For each booked hotel room, there are only two possible outcomes. Either there is a cancelation, or there is not. So we use concepts of the binomial probability distribution to solve this question.
However, we are working with a big sample. So i am going to aproximate this binomial distribution to the normal.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
![E(X) = np](https://tex.z-dn.net/?f=E%28X%29%20%3D%20np)
The standard deviation of the binomial distribution is:
![\sqrt{V(X)} = \sqrt{np(1-p)}](https://tex.z-dn.net/?f=%5Csqrt%7BV%28X%29%7D%20%3D%20%5Csqrt%7Bnp%281-p%29%7D)
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that
,
.
In this problem, we have that:
A popular resort hotel has 300 rooms and is usually fully booked. This means that ![n = 300](https://tex.z-dn.net/?f=n%20%3D%20300)
About 7% of the time a reservation is canceled before the 6:00 p.m. deadline with no pen-alty. What is the probability that at least 285 rooms will be occupied?
Here a success is a reservation not being canceled. There is a 7% probability that a reservation is canceled, and a 100 - 7 = 93% probability that a reservation is not canceled, that is, a room is occupied. So we use ![p = 0.93](https://tex.z-dn.net/?f=p%20%3D%200.93)
Approximating the binomial to the normal.
![E(X) = np = 300*0.93 = 279](https://tex.z-dn.net/?f=E%28X%29%20%3D%20np%20%3D%20300%2A0.93%20%3D%20279)
![\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{300*0.93*0.07} = 4.42](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Csqrt%7BV%28X%29%7D%20%3D%20%5Csqrt%7Bnp%281-p%29%7D%20%3D%20%5Csqrt%7B300%2A0.93%2A0.07%7D%20%3D%204.42)
The probability that at least 285 rooms will be occupied is 1 subtracted by the pvalue of Z when X = 285. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{285- 279}{4.42}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B285-%20279%7D%7B4.42%7D)
![Z = 1.36](https://tex.z-dn.net/?f=Z%20%3D%201.36)
has a pvalue of 0.9131.
So there is a 1-0.9131 = 0.0869 = 8.69% probability that at least 285 rooms will be occupied.