Question
The mean monthly rent for a one-bedroom apartment without a doorman in Manhattan is 2643. Assume the standard deviation is 502. A real estate firm samples 102 apartments. Use the TI-84 Plus calculator.
Find the 75th percentile
Find the 75th percentile
Answers
Answered by
GPT-4o mini
To find the 75th percentile of the monthly rent for a one-bedroom apartment in Manhattan using a normal distribution, you can use the TI-84 Plus calculator. Here are the steps to follow:
1. **Determine the Mean and Standard Deviation**:
- Mean (\( \mu \)) = 2643
- Standard Deviation (\( \sigma \)) = 502
2. **Use the invNorm function**:
- The 75th percentile corresponds to a cumulative probability of 0.75.
3. **Input the values on the TI-84**:
- Press the `2nd` button, then `VARS` to access the DISTR menu.
- Scroll down to `invNorm(` and select it.
- Enter the parameters: `invNorm(0.75, 2643, 502)`.
- This means you're calculating the 75th percentile for a normal distribution with a mean of 2643 and a standard deviation of 502.
4. **Perform the Calculation**:
- After inputting `invNorm(0.75, 2643, 502)`, press `ENTER`.
After calculating this, the calculator will give you the result which represents the 75th percentile of the monthly rent for a one-bedroom apartment in Manhattan.
### Result:
Using the calculator, you should find that the 75th percentile is approximately:
\[
P_{75} \approx 2643 + z \times 502
\]
Where \(z\) is the z-score corresponding to the 75th percentile (which is approximately 0.674). Thus:
\[
P_{75} \approx 2643 + 0.674 \times 502 \approx 2643 + 338.468 \approx 2981.468
\]
So, rounding gives you:
**75th percentile rent ≈ 2981.47**
This indicates that approximately 75% of one-bedroom apartments in Manhattan have a rent of less than $2981.47.
1. **Determine the Mean and Standard Deviation**:
- Mean (\( \mu \)) = 2643
- Standard Deviation (\( \sigma \)) = 502
2. **Use the invNorm function**:
- The 75th percentile corresponds to a cumulative probability of 0.75.
3. **Input the values on the TI-84**:
- Press the `2nd` button, then `VARS` to access the DISTR menu.
- Scroll down to `invNorm(` and select it.
- Enter the parameters: `invNorm(0.75, 2643, 502)`.
- This means you're calculating the 75th percentile for a normal distribution with a mean of 2643 and a standard deviation of 502.
4. **Perform the Calculation**:
- After inputting `invNorm(0.75, 2643, 502)`, press `ENTER`.
After calculating this, the calculator will give you the result which represents the 75th percentile of the monthly rent for a one-bedroom apartment in Manhattan.
### Result:
Using the calculator, you should find that the 75th percentile is approximately:
\[
P_{75} \approx 2643 + z \times 502
\]
Where \(z\) is the z-score corresponding to the 75th percentile (which is approximately 0.674). Thus:
\[
P_{75} \approx 2643 + 0.674 \times 502 \approx 2643 + 338.468 \approx 2981.468
\]
So, rounding gives you:
**75th percentile rent ≈ 2981.47**
This indicates that approximately 75% of one-bedroom apartments in Manhattan have a rent of less than $2981.47.
Answer
(d) Would it be unusual if the sample mean were greater than 2781? Round answer to at least four decimal places.
Answered by
GPT-4o mini
To determine whether it would be unusual for the sample mean of 102 apartments to be greater than 2781, we can use the concept of the sampling distribution of the sample mean.
### Step 1: Calculate the Mean and Standard Deviation of the Sampling Distribution
The mean of the sample means (\( \mu_{\bar{x}} \)) is the same as the population mean:
\[
\mu_{\bar{x}} = \mu = 2643
\]
The standard deviation of the sample mean (\( \sigma_{\bar{x}} \)), also known as the standard error, is given by the formula:
\[
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}
\]
where \( \sigma \) is the population standard deviation and \( n \) is the sample size.
Given:
- \( \sigma = 502 \)
- \( n = 102 \)
Now, calculate \( \sigma_{\bar{x}} \):
\[
\sigma_{\bar{x}} = \frac{502}{\sqrt{102}} \approx \frac{502}{10.0995} \approx 49.7030
\]
### Step 2: Calculate the z-score
Next, we calculate the z-score for a sample mean of 2781:
\[
z = \frac{\bar{x} - \mu_{\bar{x}}}{\sigma_{\bar{x}}}
\]
Where:
- \( \bar{x} = 2781 \)
Substituting in the values:
\[
z = \frac{2781 - 2643}{49.7030} \approx \frac{138}{49.7030} \approx 2.7776
\]
### Step 3: Find the Probability
Now, we need to find the probability associated with this z-score. You can use the TI-84 calculator to find this as follows:
1. Press the `2nd` button, then `VARS` to access the DISTR menu.
2. Scroll down to `normalcdf(` and select it.
3. Input the negative of the z-score and a very large positive number to find the area to the right:
- Use \( \text{normalcdf}(2.7776, 1E99) \).
### Step 4: Interpretation of the Probability
After calculating this, you should find the area to the right of \( z = 2.7776 \).
Typically, you would find a z-score of about 2.78 corresponds to a very small tail probability, roughly around 0.0027 (or 0.27%).
### Conclusion
If the probability is less than 0.05 (5%), we would consider it unusual.
In this case:
- The probability of getting a sample mean greater than 2781 is approximately 0.0027, which is indeed less than 0.05.
Thus, it would be considered unusual for the sample mean to be greater than 2781.
### Step 1: Calculate the Mean and Standard Deviation of the Sampling Distribution
The mean of the sample means (\( \mu_{\bar{x}} \)) is the same as the population mean:
\[
\mu_{\bar{x}} = \mu = 2643
\]
The standard deviation of the sample mean (\( \sigma_{\bar{x}} \)), also known as the standard error, is given by the formula:
\[
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}
\]
where \( \sigma \) is the population standard deviation and \( n \) is the sample size.
Given:
- \( \sigma = 502 \)
- \( n = 102 \)
Now, calculate \( \sigma_{\bar{x}} \):
\[
\sigma_{\bar{x}} = \frac{502}{\sqrt{102}} \approx \frac{502}{10.0995} \approx 49.7030
\]
### Step 2: Calculate the z-score
Next, we calculate the z-score for a sample mean of 2781:
\[
z = \frac{\bar{x} - \mu_{\bar{x}}}{\sigma_{\bar{x}}}
\]
Where:
- \( \bar{x} = 2781 \)
Substituting in the values:
\[
z = \frac{2781 - 2643}{49.7030} \approx \frac{138}{49.7030} \approx 2.7776
\]
### Step 3: Find the Probability
Now, we need to find the probability associated with this z-score. You can use the TI-84 calculator to find this as follows:
1. Press the `2nd` button, then `VARS` to access the DISTR menu.
2. Scroll down to `normalcdf(` and select it.
3. Input the negative of the z-score and a very large positive number to find the area to the right:
- Use \( \text{normalcdf}(2.7776, 1E99) \).
### Step 4: Interpretation of the Probability
After calculating this, you should find the area to the right of \( z = 2.7776 \).
Typically, you would find a z-score of about 2.78 corresponds to a very small tail probability, roughly around 0.0027 (or 0.27%).
### Conclusion
If the probability is less than 0.05 (5%), we would consider it unusual.
In this case:
- The probability of getting a sample mean greater than 2781 is approximately 0.0027, which is indeed less than 0.05.
Thus, it would be considered unusual for the sample mean to be greater than 2781.