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Let's analyze the given table and solve the problem.

### The data provided:
- **List A**: 1, 2, 3, 4, 5
- **List B**: 2, 3, 3, 4, 5

The question asks us to compare the **means** and **standard deviations** of the two lists. We will calculate both values for each list and check the correct answer among the given options.

#### Step 1: Calculate the mean of each list
The mean (\(\mu\)) is calculated as:
\[
\mu = \frac{\sum x_i}{n}
\]
where \(x_i\) is each value and \(n\) is the total number of values.

- **List A**: 
  \[
  \mu_A = \frac{1 + 2 + 3 + 4 + 5}{5} = \frac{15}{5} = 3
  \]
- **List B**: 
  \[
  \mu_B = \frac{2 + 3 + 3 + 4 + 5}{5} = \frac{17}{5} = 3.4
  \]

#### Step 2: Calculate the standard deviation of each list
The formula for standard deviation (\(\sigma\)) is:
\[
\sigma = \sqrt{\frac{\sum (x_i - \mu)^2}{n}}
\]

- **List A**:
  \[
  \sigma_A = \sqrt{\frac{(1 - 3)^2 + (2 - 3)^2 + (3 - 3)^2 + (4 - 3)^2 + (5 - 3)^2}{5}}
  \]
  \[
  = \sqrt{\frac{4 + 1 + 0 + 1 + 4}{5}} = \sqrt{\frac{10}{5}} = \sqrt{2} \approx 1.414
  \]

- **List B**:
  \[
  \sigma_B = \sqrt{\frac{(2 - 3.4)^2 + (3 - 3.4)^2 + (3 - 3.4)^2 + (4 - 3.4)^2 + (5 - 3.4)^2}{5}}
  \]
  \[
  = \sqrt{\frac{1.96 + 0.16 + 0.16 + 0.36 + 2.56}{5}} = \sqrt{\frac{5.2}{5}} = \sqrt{1.04} \approx 1.02
  \]

#### Step 3: Compare the means and standard deviations
- **Means**: 
  - Mean of List A = 3
  - Mean of List B = 3.4
  - The means are different.

- **Standard Deviations**:
  - Standard deviation of List A ≈ 1.414
  - Standard deviation of List B ≈ 1.02
  - The standard deviations are different.

### Conclusion:
The correct answer is **C) The means are different, and the standard deviations are different.**

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Would you like more details on any of the steps? Here are five related questions for further exploration:

1. How would you calculate the variance of the two lists?
2. Why is standard deviation a useful measure of dispersion?
3. Can the standard deviation ever be negative?
4. What happens to the standard deviation if all values in a list are the same?
5. How does changing one value in a list affect the mean and standard deviation?

**Tip**: Standard deviation gives insight into how spread out values are from the mean; the larger the standard deviation, the more dispersed the data.

The table shows two lists of numbers. Which of the following is a true statement comparing list A and list B?

Analyze the comparison of two lists of numbers based on their means and standard deviations. This problem involves calculating the mean and standard deviation of List A and List B and determining whether they are the same or different. Suitable for high school students in Grades 9-12.

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