The variance of the first 20 positive integral multiples of 4 is equal to

Options

A. 532
B. 133
C. 266
D. 600

Answer

To find the variance of the first 20 positive integral multiples of 4, we will 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: Calculate the mean
The mean (\(\mu\)) of these numbers can be calculated using the formula:
\[ \mu = \frac{\text{Sum of the numbers}}{n} \]
Where \(n\) is the number of terms (which is 20).

The sum of the first 20 multiples of 4 can be calculated as:
\[ \text{Sum} = 4(1 + 2 + 3 + \ldots + 20) = 4 \cdot \frac{20 \cdot 21}{2} = 4 \cdot 210 = 840 \]
Thus, the mean is:
\[ \mu = \frac{840}{20} = 42 \]

### Step 3: Calculate the variance
Variance (\(\sigma^2\)) is given by the formula:
\[ \sigma^2 = \frac{\sum (x_i^2)}{n} - \mu^2 \]
Where \(x_i\) are the individual data points.

#### Step 3.1: Calculate \(\sum (x_i^2)\)
We need to calculate the sum of the squares of the first 20 multiples of 4:
\[ \sum (x_i^2) = 4^2(1^2 + 2^2 + 3^2 + \ldots + 20^2) = 16 \cdot \frac{20 \cdot 21 \cdot 41}{6} \]
Calculating \(1^2 + 2^2 + \ldots + 20^2\):
\[ \sum_{i=1}^{20} i^2 = \frac{20 \cdot 21 \cdot 41}{6} = 2870 \]
Thus,
\[ \sum (x_i^2) = 16 \cdot 2870 = 45920 \]

#### Step 3.2: Substitute into the variance formula
Now we can substitute back into the variance formula:
\[ \sigma^2 = \frac{45920}{20} - 42^2 \]
Calculating:
\[ \sigma^2 = 2296 - 1764 = 532 \]

### Conclusion
The variance of the first 20 positive integral multiples of 4 is:
\[ \boxed{532} \]