Respuesta :
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:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
In this problem, we have that:
A popular resort hotel has 300 rooms and is usually fully booked. This means that [tex]n = 300[/tex]
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 [tex]p = 0.93[/tex]
Approximating the binomial to the normal.
[tex]E(X) = np = 300*0.93 = 279[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{300*0.93*0.07} = 4.42[/tex]
The probability that at least 285 rooms will be occupied is 1 subtracted by the pvalue of Z when X = 285. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{285- 279}{4.42}[/tex]
[tex]Z = 1.36[/tex]
[tex]Z = 1.36[/tex] 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.
The reservation of the hotel rooms follow a normal distribution.
The probability that at least 285 rooms are occupied is 0.0874
The given parameters are:
[tex]\mathbf{n = 300}[/tex] --- number of rooms
[tex]\mathbf{p = 7\%}[/tex] --- proportion of rooms, reserved
The sample mean is calculated as:
[tex]\mathbf{\bar x = np}[/tex]
So, we have:
[tex]\mathbf{\bar x = 300 \times 7\%}[/tex]
[tex]\mathbf{\bar x = 21}[/tex]
The population mean, according to binomial distribution is:
[tex]\mathbf{\mu = n - \bar x}[/tex]
[tex]\mathbf{\mu = 300 - 21}[/tex]
[tex]\mathbf{\mu = 279}[/tex]
The standard deviation is:
[tex]\mathbf{\sigma_x=\sqrt{\bar x \times (1 - p)}}[/tex]
So, we have:
[tex]\mathbf{\sigma_x=\sqrt{21 \times (1 - 7\%)}}[/tex]
[tex]\mathbf{\sigma_x=\sqrt{21 \times (1 - 0.07)}}[/tex]
[tex]\mathbf{\sigma_x=\sqrt{19.53}}[/tex]
[tex]\mathbf{\sigma_x=4.42}[/tex]
Calculate the z-score
[tex]\mathbf{z = \frac{x - \mu}{\sigma_x}}[/tex]
So, we have:
[tex]\mathbf{z = \frac{285 - 279}{4.42}}[/tex]
[tex]\mathbf{z = \frac{6}{4.42}}[/tex]
[tex]\mathbf{z = 1.357}[/tex]
The probability that at least 285 rooms are occupied is:
[tex]\mathbf{P(x > 285) = P(z > 1.357)}[/tex]
From z-score of probabilities, we have:
[tex]\mathbf{P(x > 285) = 0.0873906}[/tex]
Approximate
[tex]\mathbf{P(x > 285) = 0.0874}[/tex]
Hence, the probability that at least 285 rooms are occupied is 0.0874
Read more about probabilities at:
https://brainly.com/question/11234923
