To calculate the population variance $\sigma^2$ and population standard deviation $\sigma$ for the given data set: $10, 13, 17, 19, 21$, we follow these steps:

Step 1: Find the mean ($\mu$) of the data.

The mean is calculated by summing the numbers and dividing by the total count: $\mu = \frac{10 + 13 + 17 + 19 + 21}{5} = \frac{80}{5} = 16$

Step 2: Calculate the squared deviations from the mean.

Now, we subtract the mean from each data point, square the result, and then find the average of these squared differences: $(10 - 16)^2 = (-6)^2 = 36$ $(13 - 16)^2 = (-3)^2 = 9$ $(17 - 16)^2 = (1)^2 = 1$ $(19 - 16)^2 = (3)^2 = 9$ $(21 - 16)^2 = (5)^2 = 25$

Step 3: Calculate the population variance $\sigma^2$.

The population variance is the average of these squared deviations: $\sigma^2 = \frac{36 + 9 + 1 + 9 + 25}{5} = \frac{80}{5} = 16$

Step 4: Calculate the population standard deviation $\sigma$.

The population standard deviation is the square root of the variance: $\sigma = \sqrt{16} = 4$

Final Answer:

• The population variance ($\sigma^2$) is $16$.
• The population standard deviation ($\sigma$) is $4$.

You can fill in the answers for the question as follows:

• For $\sigma^2 =$ $16$,
• For $\sigma =$ $4$.

Would you like a more detailed explanation or have any questions?

Here are 5 relative questions you can explore:

1. How does population variance differ from sample variance?
2. What is the formula for standard deviation when dealing with a sample instead of a population?
3. How would the variance change if an outlier were added to the data set?
4. Can the standard deviation ever be negative?
5. How does variance relate to the range of a data set?

Tip: Always check whether you're asked for population or sample variance/standard deviation, as the formulas differ slightly!