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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.

(a) What is the probability that the sample mean rent is greater than 2713? Round the answer to at least four decimal places.

11 answers

To find the probability that the sample mean rent is greater than 2713, we will use the sampling distribution of the sample mean.

Given:

  • Population mean \(\mu = 2643\)
  • Population standard deviation \(\sigma = 502\)
  • Sample size \(n = 102\)

First, we need to find the standard error (SE) of the sample mean:

\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{502}{\sqrt{102}} \approx \frac{502}{10.0995} \approx 49.66 \]

Next, we can standardize the sample mean rent of 2713 into a z-score using the formula:

\[ z = \frac{\bar{x} - \mu}{SE} = \frac{2713 - 2643}{49.66} \approx \frac{70}{49.66} \approx 1.41 \]

Now, we will find the probability that the sample mean is greater than 2713, which corresponds to finding \(P(Z > 1.41)\).

Using a standard normal distribution table or calculator, we find the cumulative probability for \(z = 1.41\).

Using a TI-84 Plus calculator:

  1. Press 2nd then VARS to access the DISTR menu.
  2. Select normalcdf( for cumulative probabilities.
  3. Input 1.41, 1E99 where 1E99 represents a very large number (practically, this approximates infinity).

The command should look like this:

\[ P(Z > 1.41) = 1 - P(Z < 1.41) \approx 1 - normalcdf(-1E99, 1.41) \]

You can also use:

\[ normalcdf(1.41, 1E99) \]

Performing these calculations gives:

\[ P(Z > 1.41) \approx 0.0793 \]

Thus, the probability that the sample mean rent is greater than 2713 is approximately 0.0793 when rounded to four decimal places.

Your instructions for using the calculator are confusing!!

I apologize for any confusion! Let me clarify the steps to find the probability using the TI-84 Plus calculator in a clearer way.

Steps to Use the TI-84 Plus:

  1. Calculate the Standard Error (SE): \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{502}{\sqrt{102}} \approx \frac{502}{10.0995} \approx 49.66 \]

  2. Calculate the Z-score for the sample mean of 2713: \[ z = \frac{\bar{x} - \mu}{SE} = \frac{2713 - 2643}{49.66} \approx \frac{70}{49.66} \approx 1.41 \]

  3. Finding the Probability using the TI-84:

    • Turn on your TI-84 Plus calculator.

    • Press the 2nd button.

    • Press the VARS button to access the DISTR menu.

    • Scroll down to select 2:normalcdf( (this is for cumulative probabilities).

    • For the probability that \(Z > 1.41\), you need to calculate:

      \[ P(Z > 1.41) = 1 - P(Z < 1.41) \]

    • Type the following directly into the calculator after selecting normalcdf(:

      \[ normalcdf(1.41, 1E99) \]

      • Put 1.41 as the lower bound (the value you want to start from).
      • Put 1E99 as the upper bound (which represents infinity).

    • Your command will look like this:

      \[ normalcdf(1.41, 1E99) \]

  4. Press Enter to calculate the probability.

Result:

The output will give you the probability that the sample mean rent is greater than 2713.

The final result should be approximately 0.0793 (rounded to four decimal places).

If you follow these steps, it should be much clearer, and I hope this helps! Let me know if you have any other questions.

My results arent showing this!!

I apologize for any confusion in the calculations. Let's go through the steps methodically to ensure accuracy.

Given Data

  • Population Mean (μ): 2643
  • Population Standard Deviation (σ): 502
  • Sample Size (n): 102
  • Sample Mean (x̄): 2713

Step 1: Calculate the Standard Error (SE)

The standard error of the mean (SE) can be calculated as:

\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{502}{\sqrt{102}} \]

Calculating this directly:

  1. Calculate \(\sqrt{102}\):

    • On your calculator, type \(\sqrt{102}\) which gives approximately \(10.0995\).

  2. Now calculate the SE: \[ SE = \frac{502}{10.0995} \approx 49.66 \]

Step 2: Calculate the Z-score

Now, you will calculate the Z-score for the sample mean of 2713:

\[ z = \frac{x̄ - μ}{SE} = \frac{2713 - 2643}{49.66} \]

Calculating the Z-score:

  1. Calculate \(2713 - 2643\):

    • This gives \(70\).

  2. Then: \[ z = \frac{70}{49.66} \approx 1.41 \]

Step 3: Use the TI-84 to Find the Probability

You want to find \(P(Z > 1.41)\). This can be done in two steps or directly using the cumulative distribution function.

  1. Finding \(P(Z < 1.41)\) using the calculator:

    • Press 2nd then VARS for the DISTR menu.
    • Select 2:normalcdf(.
    • For the cumulative probability, use: \[ normalcdf(-1E99, 1.41) \]
    • This gives you the probability that \(Z\) is less than or equal to \(1.41\).

  2. Finding \(P(Z > 1.41)\):

    • Alternatively, you can also compute it as: \[ 1 - normalcdf(-1E99, 1.41) \]

Complete Calculation

The Z-table (or using the calculator) for \(P(Z < 1.41)\) typically gives approximately 0.9207.

So to find \(P(Z > 1.41)\):

\[ P(Z > 1.41) \approx 1 - 0.9207 = 0.0793 \]

Result Confirmation

  • Make sure you use -1E99 and 1.41 — this represents the entire left side up to the Z-score you calculated.
  • Your result for \(P(Z > 1.41)\) should indeed be approximately 0.0793.

If you are still getting a different result, double-check the values you enter and ensure you are correctly using normalcdf. If issues persist, let me know what specific outputs you are receiving and we can troubleshoot further!

(b) What is the probability that the sample mean rent is between 2512 and 2600? Round the answer to at least four decimal places.

To find the probability that the sample mean rent is between 2512 and 2600, we can again use the normal distribution properties of the mean.

Given Data:

  • Population Mean (μ): 2643
  • Population Standard Deviation (σ): 502
  • Sample Size (n): 102

Step 1: Calculate the Standard Error (SE)

The standard error (SE) is calculated as:

\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{502}{\sqrt{102}} \approx 49.66 \]

Step 2: Calculate the Z-scores

Next, we will find the Z-scores for both sample means of 2512 and 2600.

  1. For 2512:

\[ z_1 = \frac{2512 - 2643}{SE} = \frac{2512 - 2643}{49.66} = \frac{-131}{49.66} \approx -2.64 \]

  1. For 2600:

\[ z_2 = \frac{2600 - 2643}{SE} = \frac{2600 - 2643}{49.66} = \frac{-43}{49.66} \approx -0.87 \]

Step 3: Find the Probabilities

Now we will use the TI-84 Plus calculator to find the probabilities corresponding to these Z-scores.

  1. Finding \(P(Z < z_1)\) for \(z_1 \approx -2.64\):

    • Press 2nd, then VARS for the DISTR menu.
    • Select 2:normalcdf(.
    • Input -1E99, -2.64.

    The command will be: \[ normalcdf(-1E99, -2.64) \]

  2. Finding \(P(Z < z_2)\) for \(z_2 \approx -0.87\):

    • Again, go to the DISTR menu.
    • Select 2:normalcdf(.
    • Input -1E99, -0.87.

    The command will be: \[ normalcdf(-1E99, -0.87) \]

Step 4: Calculate \(P(2512 < \bar{X} < 2600)\)

To find the probability that the sample mean is between 2512 and 2600, you will compute:

\[ P(2512 < \bar{X} < 2600) = P(Z < -0.87) - P(Z < -2.64) \]

Results:

  • Let's calculate these values:

  1. For \(z \approx -2.64\), calculate: \[ P(Z < -2.64) \approx 0.0041 \quad \text{(Using normalcdf)} \]

  2. For \(z \approx -0.87\), calculate: \[ P(Z < -0.87) \approx 0.1922 \quad \text{(Using normalcdf)} \]

Now, subtract these probabilities:

\[ P(2512 < \bar{X} < 2600) = 0.1922 - 0.0041 = 0.1881 \]

Conclusion

Thus, the probability that the sample mean rent is between 2512 and 2600 is approximately 0.1881 when rounded to four decimal places.

im not getting that result!

I apologize for any discrepancies in the calculations. Let's carefully walk through the problem step by step to ensure that we get the correct probability for the sample mean rent being between 2512 and 2600.

Given Information

  • Population Mean (μ): 2643
  • Population Standard Deviation (σ): 502
  • Sample Size (n): 102

Step 1: Calculate the Standard Error (SE)

The formula for the standard error (SE) of the mean is:

\[ SE = \frac{\sigma}{\sqrt{n}} \]

Calculating it:

\[ SE = \frac{502}{\sqrt{102}} \approx \frac{502}{10.0995} \approx 49.66 \]

Step 2: Calculate the Z-scores for 2512 and 2600

  1. Z-score for 2512:

\[ z_1 = \frac{2512 - \mu}{SE} = \frac{2512 - 2643}{49.66} = \frac{-131}{49.66} \approx -2.64 \]

  1. Z-score for 2600:

\[ z_2 = \frac{2600 - \mu}{SE} = \frac{2600 - 2643}{49.66} = \frac{-43}{49.66} \approx -0.87 \]

Step 3: Use the TI-84 to Find the Required Probabilities

We want to determine the probability that \(2512 < \bar{X} < 2600\), which corresponds to \(P(z_1 < Z < z_2)\).

This can be calculated as:

\[ P(Z < z_2) - P(Z < z_1) \]

  1. Finding \(P(Z < z_1)\) (for \(z_1 \approx -2.64\)):

    • Press 2nd, then VARS to access the DISTR menu.
    • Select 2:normalcdf(.
    • Enter the values: \[ normalcdf(-1E99, -2.64) \]

  2. Finding \(P(Z < z_2)\) (for \(z_2 \approx -0.87\)):

    • Again, go to the DISTR menu.
    • Select 2:normalcdf(.
    • Enter the values: \[ normalcdf(-1E99, -0.87) \]

Step 4: Evaluate the Results

  1. For \(P(Z < -2.64)\):

    • Calculate: \[ P(Z < -2.64) \approx 0.0041 \]

  2. For \(P(Z < -0.87)\):

    • Calculate: \[ P(Z < -0.87) \approx 0.1922 \]

Final Calculation

So:

\[ P(2512 < \bar{X} < 2600) = P(Z < -0.87) - P(Z < -2.64) \approx 0.1922 - 0.0041 = 0.1881 \]

The previously mentioned answer of 0.1881 seems correct. However, if your TI-84 is outputting something different, please check the inputs for any possible errors.

Clarification of Inputs

  1. Make sure when you're using normalcdf that you're inputting negative infinity (-1E99) and the Z-scores correctly.
  2. The commands should be precisely:
    • normalcdf(-1E99, -2.64) to find \(P(Z < -2.64)\)
    • normalcdf(-1E99, -0.87) to find \(P(Z < -0.87)\)

If There Are Still Issues

If you still face discrepancies, please let me know the exact values you're calculating or the commands you're using within the TI-84 Plus, and I can help troubleshoot further!

Im using normal cdf
lower -1E99
upper: -2.64
u:0
o:1

and not getting 0.0041

Thank you for your patience, and I apologize for the confusion with the calculator settings. Let's clarify how to properly use the TI-84 Plus for this calculation.

Important Points for Using normalcdf

When using the normalcdf function on the TI-84 Plus, you must ensure you are using the correct parameters:

  1. The first two parameters should represent the lower and upper bounds of your Z-scores.
  2. Use 0 for the mean (μ) and 1 for the standard deviation (σ) if you are working with Z-scores, since you are essentially working with a standard normal distribution.

Correct Steps to Calculate \(P(2512 < \bar{X} < 2600)\)

Let's go through the calculations again carefully:

  1. Calculate the Z-scores (already completed):

    • \(z_1 = -2.64\) for the sample mean of 2512.
    • \(z_2 = -0.87\) for the sample mean of 2600.

  2. Using the Calculator: You will need to perform two separate calculations to find \(P(Z < -0.87)\) and \(P(Z < -2.64)\):

For \(P(Z < -2.64)\):

  • Use: \[ normalcdf(-1E99, -2.64, 0, 1) \] This represents the cumulative probability of Z going from negative infinity up to \(-2.64\).

For \(P(Z < -0.87)\):

  • Use: \[ normalcdf(-1E99, -0.87, 0, 1) \] This represents the cumulative probability of Z going from negative infinity up to \(-0.87\).

Step 3: Probability Calculation

Now, you want to calculate:

  1. First calculate \(P(Z < -2.64)\):

    Enter:

    normalcdf(-1E99, -2.64, 0, 1)

    You should get approximately 0.0041 for this calculation.

  2. Then calculate \(P(Z < -0.87)\):

    Enter:

    normalcdf(-1E99, -0.87, 0, 1)

    You should get approximately 0.1922 for this calculation.

Step 4: Difference

Now, subtract the two results to get: \[ P(2512 < \bar{X} < 2600) = P(Z < -0.87) - P(Z < -2.64) \]

Using the earlier approximate values (if you confirm them):

  • \(P(Z < -2.64) \approx 0.0041\)
  • \(P(Z < -0.87) \approx 0.1922\)

Thus: \[ P(2512 < \bar{X} < 2600) \approx 0.1922 - 0.0041 = 0.1881 \]

Conclusion

Please ensure you are entering the values into the calculator exactly as stated, particularly with the mean and standard deviation of the standard normal distribution (0 and 1, respectively). If you follow these corrected instructions and still encounter discrepancies, please let me know what your specific outputs are, and I can help troubleshoot further!