Explore Westonci.ca, the premier Q&A site that helps you find precise answers to your questions, no matter the topic. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
Answer:
0.0735 = 7.35% probability that the proportion of rooms booked in a sample of 556 rooms would be less than 10%.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]
A hotel manager calculates that 12% of the hotel rooms are booked.
This means that [tex]p = 0.12[/tex]
Sample of 556 rooms
This means that [tex]n = 556[/tex]
Mean and standard deviation:
[tex]\mu = p = 0.12[/tex]
[tex]s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.12*0.88}{556}} = 0.0138[/tex]
What is the probability that the proportion of rooms booked in a sample of 556 rooms would be less than 10%?
This is the p-value of Z when X = 0.1. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.1 - 0.12}{0.0138}[/tex]
[tex]Z = -1.45[/tex]
[tex]Z = -1.45[/tex] has a p-value of 0.0735
0.0735 = 7.35% probability that the proportion of rooms booked in a sample of 556 rooms would be less than 10%.
Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.