Questions: The rent for a one-bedroom apartment in Southern California has a mean of 2,200 per month and a standard deviation of 250 per month. A sample of 50 one-bedroom apartments is randomly selected. a. What is the probability that the sample mean is at least 2100 per month? Round answer to four decimal places.

Solution Steps

The rent for a one-bedroom apartment in Southern California has a mean (\( \mu \)) of \( 2200 \) dollars per month and a standard deviation (\( \sigma \)) of \( 250 \) dollars per month. A sample of \( n = 50 \) one-bedroom apartments is randomly selected.

To find the probability that the sample mean is at least \( 2100 \) dollars, we first calculate the standard error of the mean (\( \sigma_{\bar{x}} \)): \[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{250}{\sqrt{50}} \approx 35.3553 \]

Next, we calculate the Z-score for the sample mean of \( 2100 \): \[ Z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} = \frac{2100 - 2200}{35.3553} \approx -2.8284 \]

Using the Z-score, we find the probability that the sample mean is at least \( 2100 \): \[ P(\bar{x} \geq 2100) = P(Z \geq -2.8284) = 1 - P(Z < -2.8284) \] From the standard normal distribution, we have: \[ P(Z < -2.8284) \approx 0.0023 \] Thus, \[ P(\bar{x} \geq 2100) = 1 - 0.0023 = 0.9977 \]

Final Answer