Mathematics, 02.07.2021 15:50 KitKat22Rose9

A hotel manager believes that 23% of the hotel rooms are booked. If the manager is right, 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%

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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 $\mu$ and standard deviation $\sigma$ , the z-score 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 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 $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

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 $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

A hotel manager believes that 23% of the hotel rooms are booked.

This means that p = 0.23

Sample of 610 rooms

This means that n = 610

Mean and standard deviation:

$\mu = p = 0.23$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.23*0.77}{610}} = 0.017$

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

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

By the Central Limit Theorem

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

$Z = \frac{0.26 - 0.23}{0.017}$

Z = 1.76

Z = 1.76 has a p-value of 0.9608

X = 0.2

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

$Z = \frac{0.2 - 0.23}{0.017}$

Z = -1.76

Z = -1.76 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%

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