To find the ratio of the variance of the first \( n \) positive integral multiples of 4 to the variance of the first \( n \) positive odd numbers, we can follow these steps:
### Step 1: Identify the first \( n \) positive integral multiples of 4
The first \( n \) positive integral multiples of 4 are:
\[
4, 8, 12, \ldots, 4n
\]
This can be expressed as:
\[
x_i = 4i \quad \text{for } i = 1, 2, \ldots, n
\]
### Step 2: Calculate the variance of the multiples of 4
The variance \( \sigma_x^2 \) of a dataset is given by:
\[
\sigma^2 = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2
\]
Where \( \bar{x} \) is the mean of the dataset.
First, we calculate the mean \( \bar{x} \):
\[
\bar{x} = \frac{1}{n} \sum_{i=1}^{n} 4i = \frac{4}{n} \cdot \frac{n(n+1)}{2} = 2(n+1)
\]
Now, we calculate the variance:
\[
\sigma_x^2 = \frac{1}{n} \sum_{i=1}^{n} (4i - 2(n+1))^2
\]
This can be simplified using the property of variance. Since we can factor out the constant:
\[
\sigma_x^2 = 4^2 \cdot \sigma^2 \quad \text{(where } \sigma^2 \text{ is the variance of the first } n \text{ natural numbers)}
\]
### Step 3: Variance of the first \( n \) natural numbers
The variance of the first \( n \) natural numbers \( 1, 2, 3, \ldots, n \) is:
\[
\sigma^2 = \frac{(n^2 - 1)}{12}
\]
Thus, the variance of the multiples of 4 becomes:
\[
\sigma_x^2 = 16 \cdot \frac{(n^2 - 1)}{12} = \frac{4(n^2 - 1)}{3}
\]
### Step 4: Identify the first \( n \) positive odd numbers
The first \( n \) positive odd numbers are:
\[
1, 3, 5, \ldots, (2n-1)
\]
### Step 5: Calculate the variance of the odd numbers
The mean \( \bar{y} \) of the first \( n \) odd numbers is:
\[
\bar{y} = \frac{1}{n} \sum_{i=1}^{n} (2i - 1) = \frac{1}{n} \cdot n^2 = n
\]
Now, we calculate the variance:
\[
\sigma_y^2 = \frac{1}{n} \sum_{i=1}^{n} ((2i - 1) - n)^2
\]
Using the same property of variance:
\[
\sigma_y^2 = 2^2 \cdot \sigma^2 = 4 \cdot \frac{(n^2 - 1)}{12} = \frac{n^2 - 1}{3}
\]
### Step 6: Calculate the ratio of the variances
Now we can find the ratio of the variances:
\[
\text{Ratio} = \frac{\sigma_x^2}{\sigma_y^2} = \frac{\frac{4(n^2 - 1)}{3}}{\frac{(n^2 - 1)}{3}} = \frac{4(n^2 - 1)}{(n^2 - 1)} = 4
\]
### Final Answer
The ratio of the variance of the first \( n \) positive integral multiples of 4 to the variance of the first \( n \) positive odd numbers is:
\[
\boxed{4}
\]
