Question

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%

Answer 1

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:

 $\mu_{\hat p}=p$

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

 $\sigma_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}$

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:

$\mu_{\hat p}=p=0.27\\\\\sigma_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}=\sqrt{\frac{0.27\times(1-0.27)}{423}}=0.0216$

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:

$P(|\hat p-p|$

                           $=P(0.27-0.06$

*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.