Question

A popular resort hotel has 300 rooms and is usually fully booked. 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? Use the binomial distribution to find the exact valu

Answer 1

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:

$E(X) = np$

The standard deviation of the binomial distribution is:

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

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

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

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 $\mu = E(X)$, $\sigma = \sqrt{V(X)}$.

In this problem, we have that:

A popular resort hotel has 300 rooms and is usually fully booked. This means that $n = 300$

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 $p = 0.93$

Approximating the binomial to the normal.

$E(X) = np = 300*0.93 = 279$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{300*0.93*0.07} = 4.42$

The probability that at least 285 rooms will be occupied is 1 subtracted by the pvalue of Z when X = 285. So

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

$Z = \frac{285- 279}{4.42}$

$Z = 1.36$

$Z = 1.36$ 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.