Respuesta :
Answer:
0.9216 = 92.16% probability that the proportion of rooms booked in a sample of 610 rooms would differ from the population proportion by less than 3%
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 believes that 23% of the hotel rooms are booked.
This means that [tex]p = 0.23[/tex]
Sample of 610 rooms
This means that [tex]n = 610[/tex]
Mean and standard deviation:
[tex]\mu = p = 0.23[/tex]
[tex]s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.23*0.77}{610}} = 0.017[/tex]
What is the probability that the proportion of rooms booked in a sample of 610 rooms would differ from the population proportion by less than 3%?
p-value of Z when X = 0.23 + 0.03 = 0.26 subtracted by the p-value of Z when X = 0.23 - 0.03 = 0.2. So
X = 0.26
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.26 - 0.23}{0.017}[/tex]
[tex]Z = 1.76[/tex]
[tex]Z = 1.76[/tex] has a p-value of 0.9608
X = 0.2
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.2 - 0.23}{0.017}[/tex]
[tex]Z = -1.76[/tex]
[tex]Z = -1.76[/tex] has a p-value of 0.0392
0.9608 - 0.0392 = 0.9216
0.9216 = 92.16% probability that the proportion of rooms booked in a sample of 610 rooms would differ from the population proportion by less than 3%
