To find the variance of the first 20 positive integral multiples of 4, we can follow these steps:
### Step 1: Identify the first 20 positive integral multiples of 4.
The first 20 positive integral multiples of 4 are:
\[ 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80 \]
### Step 2: Define the formula for variance.
The variance \( V \) of a set of numbers is given by the formula:
\[
V = \frac{\sigma x_i^2}{n} - \left(\frac{\sigma x_i}{n}\right)^2
\]
where:
- \( \sigma x_i^2 \) is the sum of the squares of the numbers,
- \( n \) is the total number of terms,
- \( \sigma x_i \) is the sum of the numbers.
### Step 3: Calculate \( n \).
Here, \( n = 20 \) (the number of terms).
### Step 4: Calculate \( \sigma x_i \).
The sum of the first 20 multiples of 4 can be calculated as:
\[
\sigma x_i = 4 + 8 + 12 + \ldots + 80 = 4(1 + 2 + 3 + \ldots + 20)
\]
Using the formula for the sum of the first \( n \) natural numbers:
\[
1 + 2 + 3 + \ldots + n = \frac{n(n + 1)}{2}
\]
For \( n = 20 \):
\[
1 + 2 + 3 + \ldots + 20 = \frac{20 \times 21}{2} = 210
\]
Thus,
\[
\sigma x_i = 4 \times 210 = 840
\]
### Step 5: Calculate \( \sigma x_i^2 \).
Now, we need to calculate the sum of the squares of the first 20 multiples of 4:
\[
\sigma x_i^2 = 4^2(1^2 + 2^2 + 3^2 + \ldots + 20^2) = 16(1^2 + 2^2 + 3^2 + \ldots + 20^2)
\]
Using the formula for the sum of the squares of the first \( n \) natural numbers:
\[
1^2 + 2^2 + 3^2 + \ldots + n = \frac{n(n + 1)(2n + 1)}{6}
\]
For \( n = 20 \):
\[
1^2 + 2^2 + 3^2 + \ldots + 20^2 = \frac{20 \times 21 \times 41}{6} = 2870
\]
Thus,
\[
\sigma x_i^2 = 16 \times 2870 = 45920
\]
### Step 6: Substitute into the variance formula.
Now we can substitute into the variance formula:
\[
V = \frac{\sigma x_i^2}{n} - \left(\frac{\sigma x_i}{n}\right)^2
\]
\[
V = \frac{45920}{20} - \left(\frac{840}{20}\right)^2
\]
Calculating each part:
\[
\frac{45920}{20} = 2296
\]
\[
\frac{840}{20} = 42 \quad \text{and} \quad 42^2 = 1764
\]
Thus,
\[
V = 2296 - 1764 = 532
\]
### Final Answer:
The variance of the first 20 positive integral multiples of 4 is \( \boxed{532} \).
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