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A spinner at a carnival has three equally sized slices. One slice is green, one slice is blue, and one slice is red. The amounts you can win (in dollars), based on which slice the dial lands on, are given below.

slice- green ; winning (in dollars) 1
slice- blue ; winning - 2
slice- red ; winnings- 3
Consider spinning the dial like sampling from a population. (The population mean of the winnings is and the population standard deviation is .)

(a)Suppose the dial is spun 2 times, and the winnings for each spin is recorded. The sequence of 2 spins that is observed is a sample of size 2 randomly selected from the population, with replacement. There are 9 possible samples. The winnings for several of the possible samples have been listed in the table below. Enter the winnings for the remaining possible samples. When you are done, select "Compute". In the "Sample mean, x" column, you will then see the sample mean of the winnings for each sample, along with the mean and standard deviation of all the column's values.
index 1- sample: green, green ; winnings- 1,1
index 2- sample: green, blue ; winnings- 1,2
index 3- sample: green, red ; winnings 1,3
index 4- sample: blue, green ; winnings [?], [?]
index 5- sample: blue, blue ; winnings [?], [?]
index 6- sample: blue, red ; winnings [?], [?]
index 7- sample: red, green ; winnings [3], [1]
index 8- sample: red, blue ; winnings [3], [2]
index 9- sample: red, red ; winnings [3]. [3]

5 answers

To find the winnings for the remaining possible samples, we need to determine the winnings based on the colors of the slices, knowing the winnings for each slice:

  • Green: $1
  • Blue: $2
  • Red: $3

Now we will fill in the winnings for the remaining samples:

  1. Sample: Blue, Green

    • Winnings: $2 (blue), $1 (green) → Total Winnings: (2, 1)

  2. Sample: Blue, Blue

    • Winnings: $2 (blue), $2 (blue) → Total Winnings: (2, 2)

  3. Sample: Blue, Red

    • Winnings: $2 (blue), $3 (red) → Total Winnings: (2, 3)

Now we can summarize the winnings for all the samples:

  1. Sample: Green, Green

    • Winnings: (1, 1)

  2. Sample: Green, Blue

    • Winnings: (1, 2)

  3. Sample: Green, Red

    • Winnings: (1, 3)

  4. Sample: Blue, Green

    • Winnings: (2, 1)

  5. Sample: Blue, Blue

    • Winnings: (2, 2)

  6. Sample: Blue, Red

    • Winnings: (2, 3)

  7. Sample: Red, Green

    • Winnings: (3, 1)

  8. Sample: Red, Blue

    • Winnings: (3, 2)

  9. Sample: Red, Red

    • Winnings: (3, 3)

Now, you have all the winnings for the possible samples. You can proceed to compute the sample mean for each sample and calculate the overall mean and standard deviation for these values.

(b)Use the table from part (a) to find u-x (the mean of the sampling distribution of the sample mean) and o-x (the standard deviation of the sampling distribution of the sample mean). Write your answers to two decimal places.

To find the mean (\( \mu_{\bar{x}} \)) of the sampling distribution of the sample mean and the standard deviation (\( \sigma_{\bar{x}} \)), we can follow these steps:

  1. Find the Mean of the Sample Means (\( \mu_{\bar{x}} \)): To calculate \( \mu_{\bar{x}} \), we need to find the sample means for each of the 9 possible samples, which we already have based on the winnings:
  • Sample 1: (1, 1) → Sample Mean \( x = \frac{1 + 1}{2} = 1.00 \)
  • Sample 2: (1, 2) → Sample Mean \( x = \frac{1 + 2}{2} = 1.50 \)
  • Sample 3: (1, 3) → Sample Mean \( x = \frac{1 + 3}{2} = 2.00 \)
  • Sample 4: (2, 1) → Sample Mean \( x = \frac{2 + 1}{2} = 1.50 \)
  • Sample 5: (2, 2) → Sample Mean \( x = \frac{2 + 2}{2} = 2.00 \)
  • Sample 6: (2, 3) → Sample Mean \( x = \frac{2 + 3}{2} = 2.50 \)
  • Sample 7: (3, 1) → Sample Mean \( x = \frac{3 + 1}{2} = 2.00 \)
  • Sample 8: (3, 2) → Sample Mean \( x = \frac{3 + 2}{2} = 2.50 \)
  • Sample 9: (3, 3) → Sample Mean \( x = \frac{3 + 3}{2} = 3.00 \)

Now, let's compute the average of these sample means:

\[ \mu_{\bar{x}} = \frac{(1.00 + 1.50 + 2.00 + 1.50 + 2.00 + 2.50 + 2.00 + 2.50 + 3.00)}{9} = \frac{18.00}{9} = 2.00 \]

So, \( \mu_{\bar{x}} = 2.00 \).

  1. Find the Standard Deviation of the Sample Means (\( \sigma_{\bar{x}} \)): To calculate \( \sigma_{\bar{x}} \), we first need to calculate the standard deviation of the sample means. The formula for the standard deviation of the sample means is:

\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \]

Where:

  • \( \sigma \) is the population standard deviation.
  • \( n \) is the sample size.

Step to find population standard deviation (\( \sigma \)):

First, let's find the population mean (\( \mu \)), and then we'll use it to find the population standard deviation.

The winnings are: 1 (green), 2 (blue), 3 (red).

\[ \mu = \frac{(1 + 2 + 3)}{3} = \frac{6}{3} = 2.00 \]

Now, calculate the variance (\( \sigma^2 \)):

\[ \sigma^2 = \frac{(1 - 2)^2 + (2 - 2)^2 + (3 - 2)^2}{3} = \frac{(1)^2 + (0)^2 + (1)^2}{3} = \frac{1 + 0 + 1}{3} = \frac{2}{3} \approx 0.67 \]

Then, standard deviation (\( \sigma \)):

\[ \sigma = \sqrt{\frac{2}{3}} \approx 0.8165 \text{ (rounded to 4 decimal places)} \]

Substituting into the formula for \( \sigma_{\bar{x}} \):

Now, substituting into the standard error formula:

\[ \sigma_{\bar{x}} = \frac{0.8165}{\sqrt{2}} \approx \frac{0.8165}{1.4142} \approx 0.5773 \]

Thus, rounding to two decimal places, we have:

  • \( \mu_{\bar{x}} = 2.00 \)
  • \( \sigma_{\bar{x}} \approx 0.58 \)

Final Answers:

  • \( \mu_{\bar{x}} = 2.00 \)
  • \( \sigma_{\bar{x}} = 0.58 \)
(c)Graph the frequency histograms for the population distribution and the sampling distribution of the sample mean.

Creating histograms for both the population distribution and the sampling distribution of the sample mean can help illustrate the differences between them. Since I don't have the capability to create visual content directly, I will guide you on how to plot these histograms using the winnings from the spinner and sample means obtained.

Population Distribution Histogram

  1. Data Points: We have three winnings corresponding to the three slices:

    • Green: $1
    • Blue: $2
    • Red: $3

  2. Frequencies: Since each slice is equally likely, the distribution is:

    • Winnings of $1 (1 occurrence)
    • Winnings of $2 (1 occurrence)
    • Winnings of $3 (1 occurrence)

  3. Histogram Setup:

    • X-axis: Winnings ($1, $2, $3)
    • Y-axis: Frequency (each has a frequency of 1)

Histogram Data:

  • Value 1: Frequency 1
  • Value 2: Frequency 1
  • Value 3: Frequency 1

Sampling Distribution Histogram

  1. Sample Means: We derived the sample means based on the results from the previous entries. Here are the sample means we previously calculated:

    • Sample 1: 1.00
    • Sample 2: 1.50
    • Sample 3: 2.00
    • Sample 4: 1.50
    • Sample 5: 2.00
    • Sample 6: 2.50
    • Sample 7: 2.00
    • Sample 8: 2.50
    • Sample 9: 3.00

  2. Frequencies:

    • 1.00 (1 occurrence)
    • 1.50 (2 occurrences)
    • 2.00 (3 occurrences)
    • 2.50 (2 occurrences)
    • 3.00 (1 occurrence)

  3. Histogram Setup:

    • X-axis: Sample Means (1.00, 1.50, 2.00, 2.50, 3.00)
    • Y-axis: Frequency

Histogram Data:

  • Value 1.00: Frequency 1
  • Value 1.50: Frequency 2
  • Value 2.00: Frequency 3
  • Value 2.50: Frequency 2
  • Value 3.00: Frequency 1

Plotting the Histograms

To plot these histograms, you can use a software tool like Excel, Google Sheets, Python (with Matplotlib), or any graphing software.

Example Steps in Python with Matplotlib

import matplotlib.pyplot as plt

# Population distribution
population_winnings = [1, 2, 3]
population_frequencies = [1, 1, 1]

plt.figure(figsize=(10, 5))

# Create the population distribution histogram
plt.subplot(1, 2, 1)
plt.bar(population_winnings, population_frequencies, color='green', alpha=0.7)
plt.title('Population Distribution')
plt.xlabel('Winnings ($)')
plt.ylabel('Frequency')
plt.xticks(population_winnings)

# Sampling distribution of the sample mean
sample_means = [1.00, 1.50, 2.00, 1.50, 2.00, 2.50, 2.00, 2.50, 3.00]
sample_mean_frequencies = [1, 2, 3, 2, 1]  # for 1.00, 1.50, 2.00, 2.50, 3.00

plt.subplot(1, 2, 2)
plt.bar([1.00, 1.50, 2.00, 2.50, 3.00], sample_mean_frequencies, color='blue', alpha=0.7)
plt.title('Sampling Distribution of the Sample Mean')
plt.xlabel('Sample Mean ($)')
plt.ylabel('Frequency')
plt.xticks([1.00, 1.50, 2.00, 2.50, 3.00])

plt.tight_layout()
plt.show()

Conclusion

Use the above information to create the histograms. The left histogram will represent the population distribution with winnings of the spinner, while the right histogram will represent the sampling distribution of the sample mean from the samples you've calculated. The sampling distribution will typically show more clustering towards the mean compared to the population distribution, illustrating the Central Limit Theorem even with a small sample size in this case.