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.

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.