A hotel manager believes that 27% of the hotel rooms are booked. If the manager is correct, what is the probability that the proportion of rooms booked in a sample of 423 rooms would differ from the population proportion by less than 6%

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Answer:

The probability that the proportion of rooms booked in a sample of 423 rooms would differ from the population proportion by less than 6% is 0.9946.

Step-by-step explanation:

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

 [tex]\mu_{\hat p}=p[/tex]

The standard deviation of this sampling distribution of sample proportion is:

 [tex]\sigma_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]

The information provided here is:

p = 0.27

n = 423

As n = 423 > 30, the sampling distribution of sample proportion can be approximated by the Normal distribution.

The mean and standard deviation of the sampling distribution of sample proportion are:

[tex]\mu_{\hat p}=p=0.27\\\\\sigma_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}=\sqrt{\frac{0.27\times(1-0.27)}{423}}=0.0216[/tex]

Compute the probability that the proportion of rooms booked in a sample of 423 rooms would differ from the population proportion by less than 6% as follows:

[tex]P(|\hat p-p|<0.06)=P(p-0.06<\hat p<p+0.06)[/tex]

                           [tex]=P(0.27-0.06<\hat p<0.27+0.06)\\\\=P(0.21<\hat p<0.33)\\\\=P(\frac{0.21-0.27}{0.0216}<\frac{\hat p-\mu_{\hat p}}{\sigma_{\hat p}}<\frac{0.33-0.27}{0.0216})\\\\=P(-2.78<Z<2.78)\\\\=P(Z<2.78)-P(Z<-2.78)\\\\=0.99728-0.00272\\\\=0.99456\\\\\approx 0.9946[/tex]

*Use a z-table.

Thus, the probability that the proportion of rooms booked in a sample of 423 rooms would differ from the population proportion by less than 6% is 0.9946.