The mean monthly rent for a one-bedroom apartment without a doorman in Manhattan is 2557. Assume the standard deviation is 486. A real estate firm samples apartments. Use the TI-84 Plus calculator.

Question

(a) What is the probability that the sample mean rent is greater than 2627? Round the answer to at least four decimal places.

Answer 1

To find the probability that the sample mean rent is greater than 2627, we first need to use the Central Limit Theorem.

The sample mean \(\bar{x}\) is normally distributed if the sample size is sufficiently large (n > 30 is a common rule of thumb). Let's assume we have a sufficiently large sample size \(n\).

Step 1: Determine the parameters for the sample mean

1. Population mean (\(\mu\)): \[ \mu = 2557 \]

2. Population standard deviation (\(\sigma\)): \[ \sigma = 486 \]

3. Standard deviation of the sample mean (\(\sigma_{\bar{x}}\)): \[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \]

Since we don’t have \(n\) explicitly, we'll denote it for now.

Step 2: Calculate the z-score

The z-score is calculated as follows:

\[ z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} \]

We want to find the probability that \(\bar{x} > 2627\). This can be converted to a z-score:

\[ z = \frac{2627 - 2557}{\sigma_{\bar{x}}} = \frac{70}{\sigma_{\bar{x}}} \]

Step 3: Find the probability

The probability we seek is:

\[ P(\bar{x} > 2627) = P(z > \frac{70}{\sigma_{\bar{x}}}) \]

To find this probability, we can use the TI-84 Plus calculator to find the area to the right of the z-score.

Example Calculation

Let's assume a sample size \(n = 30\) (you can change this as necessary based on your context).

\[ \sigma_{\bar{x}} = \frac{486}{\sqrt{30}} \approx 88.79 \]

Then we calculate z:

\[ z = \frac{2627 - 2557}{88.79} \approx \frac{70}{88.79} \approx 0.79 \]

Using TI-84 to find the probability

To find \(P(z > 0.79)\):

1. Press the 2nd button, then VARS to access the distribution menu.
2. Select normalcdf( (for cumulative probability) which is typically for left-tail, so we need the complementary probability.
3. We will find 1 - normalcdf(-1E99, 0.79).

This function call will provide the area to the left of z = 0.79, therefore you need to subtract this from 1.

Calculating \(P(z > 0.79)\):

P(z > 0.79) = 1 - normalcdf(-1E99, 0.79)

The approximate value we get is:

\[ \approx 1 - 0.7852 = 0.2148 \]

Final Answer:

Thus, \(P(\bar{x} > 2627) \approx 0.2148\).

So, the probability that the sample mean rent is greater than 2627 is approximately 0.2148 (rounded to four decimal places).