4. Now imagine that you roll two dice, one red and one green. What are the possible sample means that you could have? Complete the following table, entering the sample mean number of pips for each of the 36 possible outcomes. Some have been done for you. MAT 120 - Introduction to Statistics ACTIVITY #10: Central Limit Theorem. In this activity, we investigate the sampling distribution of the sample mean, X | 1 | 2 · 3 4 5 6 1 1. Imagine that you roll a single, fair, six-sided die once. Define the random variable X, to be the number of dots (pips) on the upward face. Complete the probability distribution table for X. 2 |1.5 3 3 2.5 4 5 P(x) 6 1 5. Using the 36 sample means found in # 4, complete the probability distribution table for X, the mean number of pips on the faces when 2 dice are rolled. 3 4 5 P(x) 6 1 1.5 2 2.5 3 3.5 2. Sketch the probability histogram for X below. Let the x-axis be the OUTCOME and let the y-axis be the PROBABILITY. 4 4.5 5 5.5 6

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**MAT 120 – Introduction to Statistics**

**ACTIVITY #10: Central Limit Theorem**

In this activity, we investigate the sampling distribution of the sample mean, \(\overline{X}\).

1. Imagine that you roll a single, fair, six-sided die once. Define the random variable \(X\) to be the number of dots (pips) on the upward face. Complete the probability distribution table for \(X\).

   | \(x\) | \(P(x)\) |
   |---|---|
   | 1 |   |
   | 2 |   |
   | 3 |   |
   | 4 |   |
   | 5 |   |
   | 6 |   |

2. Sketch the probability histogram for \(X\) below. Let the x-axis be the OUTCOME and let the y-axis be the PROBABILITY.

   (Graph paper is shown without a plotted histogram)

3. Determine the mean and standard deviation for \(X\). As a reminder, you determine the mean of a discrete probability distribution using \(\mu = \sum xp(x)\) and the standard deviation is found most simply using the formula \(\sigma = \sqrt{\sum x^2P(x) - \mu^2}\) (see page 322).

4. Now imagine that you roll two dice, one red and one green. What are the possible sample means that you could have? Complete the following table, entering the sample mean number of pips for each of the 36 possible outcomes. Some have been done for you.

   | \(\overline{X}\) | 1  |   2  | 3  | 4  | 5  | 6  |
   |---|---|---|---|---|---|---|
   | 1 |   |   |   |   |   |   |
   | 2 | 1.5|   |   |   |   |   |
   | 3 |   | 2.5|   |   |   |   |
   | 4 |   |   |   |   |   |   |
   | 5 |   |   |   |   |   |   |
   | 6 |   |   |   |   |   | 6 |

5. Using the 36 sample means found in #4,
Transcribed Image Text:**MAT 120 – Introduction to Statistics** **ACTIVITY #10: Central Limit Theorem** In this activity, we investigate the sampling distribution of the sample mean, \(\overline{X}\). 1. Imagine that you roll a single, fair, six-sided die once. Define the random variable \(X\) to be the number of dots (pips) on the upward face. Complete the probability distribution table for \(X\). | \(x\) | \(P(x)\) | |---|---| | 1 | | | 2 | | | 3 | | | 4 | | | 5 | | | 6 | | 2. Sketch the probability histogram for \(X\) below. Let the x-axis be the OUTCOME and let the y-axis be the PROBABILITY. (Graph paper is shown without a plotted histogram) 3. Determine the mean and standard deviation for \(X\). As a reminder, you determine the mean of a discrete probability distribution using \(\mu = \sum xp(x)\) and the standard deviation is found most simply using the formula \(\sigma = \sqrt{\sum x^2P(x) - \mu^2}\) (see page 322). 4. Now imagine that you roll two dice, one red and one green. What are the possible sample means that you could have? Complete the following table, entering the sample mean number of pips for each of the 36 possible outcomes. Some have been done for you. | \(\overline{X}\) | 1 | 2 | 3 | 4 | 5 | 6 | |---|---|---|---|---|---|---| | 1 | | | | | | | | 2 | 1.5| | | | | | | 3 | | 2.5| | | | | | 4 | | | | | | | | 5 | | | | | | | | 6 | | | | | | 6 | 5. Using the 36 sample means found in #4,
Certainly! Here is the transcription and explanation of the content for an educational website:

---

**Transcription:**
   
7. Compute the mean and standard deviation of the distribution in #5.

8. Verify that your results in #3 and #7 agree with the information in the text on page 401, "The mean of the sampling distribution of the sample mean is equal to the mean of the underlying population, and the standard deviation of the sampling distribution of the sample mean is \( \frac{\sigma}{\sqrt{n}} \), regardless of the size of the sample."

9. If we were to repeat the process in #1 - 6 for n=3 dice, there would be 216 possible outcomes (why?); we would find the mean of the \( \overline{X} \) distribution equal to 3.5 and the standard deviation equal to 1.71/\( \sqrt{3} \). The graph of the distribution is shown below.

*Image description:*
A histogram labeled "Histogram of means, n=3 dice" is presented. It displays a symmetric distribution of sample means, with the x-axis labeled from 1 to 6 in increments of 0.33. The y-axis ranges from 0 to 30. The distribution has a peak around 3.33, tapering off evenly on both sides.

Explain how the dice example illustrates the Central Limit Theorem, cited on page 401 of the text:

*Separate text box:*

**The Central Limit Theorem**
   
Regardless of the shape of the underlying population, the sampling distribution of \( \overline{x} \) becomes approximately normal as the sample size, \( n \), increases.

---

**Explanation of the Histogram:**

The histogram shows the distribution of the means when rolling three dice. The data illustrates the Central Limit Theorem (CLT), which states that regardless of the population distribution, the distribution of the sample means will approach a normal distribution with increased sample size. Here, even with a small sample size of three dice, the sample means form an approximately normal distribution centered around the population mean (3.5), demonstrating the CLT in practice.
Transcribed Image Text:Certainly! Here is the transcription and explanation of the content for an educational website: --- **Transcription:** 7. Compute the mean and standard deviation of the distribution in #5. 8. Verify that your results in #3 and #7 agree with the information in the text on page 401, "The mean of the sampling distribution of the sample mean is equal to the mean of the underlying population, and the standard deviation of the sampling distribution of the sample mean is \( \frac{\sigma}{\sqrt{n}} \), regardless of the size of the sample." 9. If we were to repeat the process in #1 - 6 for n=3 dice, there would be 216 possible outcomes (why?); we would find the mean of the \( \overline{X} \) distribution equal to 3.5 and the standard deviation equal to 1.71/\( \sqrt{3} \). The graph of the distribution is shown below. *Image description:* A histogram labeled "Histogram of means, n=3 dice" is presented. It displays a symmetric distribution of sample means, with the x-axis labeled from 1 to 6 in increments of 0.33. The y-axis ranges from 0 to 30. The distribution has a peak around 3.33, tapering off evenly on both sides. Explain how the dice example illustrates the Central Limit Theorem, cited on page 401 of the text: *Separate text box:* **The Central Limit Theorem** Regardless of the shape of the underlying population, the sampling distribution of \( \overline{x} \) becomes approximately normal as the sample size, \( n \), increases. --- **Explanation of the Histogram:** The histogram shows the distribution of the means when rolling three dice. The data illustrates the Central Limit Theorem (CLT), which states that regardless of the population distribution, the distribution of the sample means will approach a normal distribution with increased sample size. Here, even with a small sample size of three dice, the sample means form an approximately normal distribution centered around the population mean (3.5), demonstrating the CLT in practice.
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