Questions: High-rent district: The mean monthly rent for a one-bedroom apartment without a doorman in Manhattan 2,674. Assume the standard deviation is 508. (a) What is the probability that the sample mean rent is greater than 2,744? Round the answer to at least four decimal places. The probability that the sample mean rent is greater than 2,744 is 0.2939

Solution Steps

We are given the following parameters for the monthly rent of a one-bedroom apartment in Manhattan:

• Mean monthly rent (\( \mu \)): \( 2674 \)
• Standard deviation (\( \sigma \)): \( 508 \)
• We want to find the probability that the sample mean rent is greater than \( 2744 \).

To find the probability, we first calculate the Z-score for the value \( 2744 \) using the formula:

\[ Z = \frac{X - \mu}{\sigma} \]

Substituting the values:

\[ Z = \frac{2744 - 2674}{508} = \frac{70}{508} \approx 0.1378 \]

Next, we need to find the probability that the sample mean is greater than \( 2744 \). This can be expressed as:

\[ P(X > 2744) = 1 - P(X \leq 2744) = 1 - \Phi(Z) \]

Where \( \Phi(Z) \) is the cumulative distribution function (CDF) of the standard normal distribution. From the output, we have:

\[ P(X \leq 2744) = \Phi(0.1378) \approx 0.4452 \]

Thus, the probability that the sample mean rent is greater than \( 2744 \) is:

\[ P(X > 2744) = 1 - 0.4452 = 0.5548 \]

Final Answer