To calculate the variance of the data set {4, 16, 21, 32, 11, 12}, we follow these steps:

Step 1: Find the Mean (Average)

The mean is calculated by summing all the numbers and dividing by the number of data points.

$\text{Mean} = \frac{4 + 16 + 21 + 32 + 11 + 12}{6} = \frac{96}{6} = 16$

Step 2: Calculate the Squared Differences from the Mean

Now, we subtract the mean from each data point and square the result.

• For $4$: $(4 - 16)^2 = (-12)^2 = 144$
• For $16$: $(16 - 16)^2 = 0^2 = 0$
• For $21$: $(21 - 16)^2 = 5^2 = 25$
• For $32$: $(32 - 16)^2 = 16^2 = 256$
• For $11$: $(11 - 16)^2 = (-5)^2 = 25$
• For $12$: $(12 - 16)^2 = (-4)^2 = 16$

Step 3: Find the Average of the Squared Differences

Sum the squared differences:

$144 + 0 + 25 + 256 + 25 + 16 = 466$

Now, divide by the number of data points (for population variance) or by $n - 1$ (for sample variance).

Population Variance

If this is the entire population, divide by $n = 6$:

$\text{Variance} = \frac{466}{6} \approx 77.67$

Sample Variance

If this is a sample from a larger population, divide by $n - 1 = 5$:

$\text{Variance} = \frac{466}{5} = 93.2$

Final Answer:

• Population Variance: $77.67$
• Sample Variance: $93.2$

Would you like to know more details or have any other questions?

Here are some related questions:

1. What is the relationship between variance and standard deviation?
2. How do you calculate standard deviation from variance?
3. What is the significance of dividing by $n - 1$ in sample variance?
4. What are the implications of a higher variance in a data set?
5. How does variance change if one data point is significantly larger than the others?

Tip: The square root of variance gives the standard deviation, a useful measure of data dispersion.