Question

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.

(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).