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.
(a) What is the probability that the sample mean rent is greater than 2627? Round the answer to at least four decimal places.
Answers
GPT-4o mini
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).
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).