Questions: High-rent districts The mean monthly rent for a one-bedroom apartment without a doorman in Manhattan is 2699. Assume the standard deviation is 513. A real estate firm samples 103 apartments. Use Excel. Part 1 of 5 (a) What is the probability that the sample mean rent is greater than 2769? Round the answer to at least four decimal places. The probability that the sample mean rent is greater than 2769 is 0.0831. Part 2 of 5 (b) What is the probability that the sample mean rent is between 2568 and 2668? Round the answer to at least four decimal places. The probability that the sample mean rent is between 2568 and 2668 is 0.2651. Part 3 of 5 (c) Find the 55th percentile of the sample mean. Round the answer to at least two decimal places. The 55th percentile of the sample mean rent is 2705.35. Part 4 of 5 (d) Would it be unusual if the sample mean were greater than 2795? Round the answer to at least four decimal places. (Choose one) because the probability that the sample mean is greater than 2795 is

High-rent districts The mean monthly rent for a one-bedroom apartment without a doorman in Manhattan is 2699. Assume the standard deviation is 513. A real estate firm samples 103 apartments. Use Excel.

Part 1 of 5
(a) What is the probability that the sample mean rent is greater than 2769? Round the answer to at least four decimal places.

The probability that the sample mean rent is greater than 2769 is 0.0831.

Part 2 of 5
(b) What is the probability that the sample mean rent is between 2568 and 2668? Round the answer to at least four decimal places. The probability that the sample mean rent is between 2568 and 2668 is 0.2651.

Part 3 of 5
(c) Find the 55th percentile of the sample mean. Round the answer to at least two decimal places.

The 55th percentile of the sample mean rent is 2705.35.

Part 4 of 5
(d) Would it be unusual if the sample mean were greater than 2795? Round the answer to at least four decimal places.
(Choose one) because the probability that the sample mean is greater than 2795 is
Transcript text: High-rent districts The mean monthly rent for a one-bedroom apartment without a doorman in Manhattan is $\$ 2699$. Assume the standard deviation is $\$ 513$. A real estate firm samples 103 apartments. Use Excel. Part 1 of 5 (a) What is the probability that the sample mean rent is greater than $\$ 2769$ Round the answer to at least four decimal places. The probability that the sample mean rent is greater than $\$ 2769$ is 0.0831 . Part 2 of 5 (b) What is the probability that the sample mean rent is between $\$ 2568$ and $\$ 2668$ Round the answer to at least four decimal places. The probability that the sample mean rent is between $\$ 2568$ and $\$ 2668$ is 0.2651 . Part 3 of 5 (c) Find the $55^{\text {th }}$ percentile of the sample mean. Round the answer to at least two decimal places. The $55^{\text {th }}$ percentile of the sample mean rent is $\$ 2705.35$. Part 4 of 5 (d) Would it be unusual if the sample mean were greater than $\$ 2795$ Round the answer to at least four decimal places. (Choose one) because the probability that the sample mean is greater than $\$ 2795$ is
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Solution

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Solution Steps

Step 1: Probability that the Sample Mean Rent is Greater than $2769

To find the probability that the sample mean rent is greater than \( \$2769 \), we calculate the Z-score for \( 2769 \):

\[ Z_{end} = \frac{2769 - 2699}{\frac{513}{\sqrt{103}}} \approx 1.3848 \]

Using the cumulative distribution function \( \Phi \), we find:

\[ P = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(\infty) - \Phi(1.3848) = 0.0831 \]

Thus, the probability that the sample mean rent is greater than \( \$2769 \) is \( 0.0831 \).

Step 2: Probability that the Sample Mean Rent is Between $2568 and $2668

Next, we calculate the probability that the sample mean rent is between \( \$2568 \) and \( \$2668 \). We find the Z-scores for both bounds:

\[ Z_{start} = \frac{2568 - 2699}{\frac{513}{\sqrt{103}}} \approx -2.5916 \] \[ Z_{end} = \frac{2668 - 2699}{\frac{513}{\sqrt{103}}} \approx -0.6133 \]

Using the cumulative distribution function \( \Phi \):

\[ P = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(-0.6133) - \Phi(-2.5916) = 0.2651 \]

Thus, the probability that the sample mean rent is between \( \$2568 \) and \( \$2668 \) is \( 0.2651 \).

Step 3: Finding the 55th Percentile of the Sample Mean

To find the 55th percentile of the sample mean, we first calculate the Z-score corresponding to the 55th percentile:

\[ Z_{55} \approx 0.12566134685507416 \]

Next, we calculate the standard error of the mean:

\[ SE = \frac{513}{\sqrt{103}} \approx 50.5 \]

Now, we can find the 55th percentile of the sample mean:

\[ P_{55} = 2699 + Z_{55} \cdot SE \approx 2705.35 \]

Thus, the 55th percentile of the sample mean rent is \( 2705.35 \).

Final Answer

The results are summarized as follows:

  • The probability that the sample mean rent is greater than \( \$2769 \) is \( 0.0831 \).
  • The probability that the sample mean rent is between \( \$2568 \) and \( \$2668 \) is \( 0.2651 \).
  • The 55th percentile of the sample mean rent is \( 2705.35 \).

\[ \boxed{P > 2769: 0.0831, \quad P(2568 < X < 2668): 0.2651, \quad 55^{th} \text{ percentile: } 2705.35} \]

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