Question

The average cost of a one-bedroom apartment in a town is $850 per month. What is the probability of randomly selecting a sample of 50 one-bedroom apartments in this town and getting a sample mean of less than $830 if the population standard deviation is $100 ? Appendix A Statistical Tables (Round the values of z to 2 decimal places, e.g. 0.75. Round your answer to 4 decimal places, e.g. 0.7845.)

Answer 1

To find the probability of getting a sample mean of less than $830, calculate the z-score and use a z-table to find the probability.

Step-by-step explanation:

To find the probability of randomly selecting a sample of 50 one-bedroom apartments and getting a sample mean of less than $830, we need to calculate the z-score and use a z-table to find the corresponding probability.

The formula for the z-score is:

z = (sample mean - population mean) / (population standard deviation / sqrt(sample size))

Plugging in the values, we have:

z = ($830 - $850) / ($100 / sqrt(50)) = -2.1213

Using the z-table, we can find that the probability of getting a z-score of -2.1213 or less is approximately 0.0164.

Answer 2

The probability of randomly selecting a sample of 50 one-bedroom apartments and obtaining a sample mean of less than $830, given a population standard deviation of $100, is approximately 0.1151.

Step-by-step explanation:

To calculate this probability, we use the z-score formula:

`\[ z = \frac{{\bar{X} - \mu}}{{(\sigma)/(√(n))}} \]`

where
`\(\bar{X}\)` is the sample mean,
`\(\mu\)` is the population mean,
`\(\sigma\)`is the population standard deviation, and n is the sample size.

In this case:

`\[ z = \frac{{830 - 850}}{{(100)/(√(50))}} \]`

Calculating the value of z, we find the z-score. Next, we consult the z-table to find the probability associated with this z-score. The probability of obtaining a sample mean less than $830 is given by the area under the standard normal distribution curve to the left of this z-score.

Therefore, the final probability is approximately 0.1151, meaning there's a 11.51% chance of randomly selecting a sample of 50 one-bedroom apartments and getting a sample mean of less than $830.