The ratio of the variance of first n positive integral multiples of 4 to the variance of first n positive odd number is

Answer

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 sequences
1. The first \( n \) positive integral multiples of 4 are:
\[ 4, 8, 12, \ldots, 4n \]
This can be expressed as:
\[ 4 \times (1, 2, 3, \ldots, n) \]

2. The first \( n \) positive odd numbers are:
\[ 1, 3, 5, \ldots, (2n - 1) \]

### Step 2: Calculate the variance of the first \( n \) multiples of 4
The variance \( \sigma^2 \) of a set of numbers \( x_1, x_2, \ldots, x_n \) is given by:
\[ \sigma^2 = \frac{1}{n} \sum_{i=1}^{n} x_i^2 - \left( \frac{1}{n} \sum_{i=1}^{n} x_i \right)^2 \]

For the sequence \( 4, 8, 12, \ldots, 4n \):
- The mean \( \mu \) is:
\[ \mu = \frac{4 + 8 + 12 + \ldots + 4n}{n} = \frac{4(1 + 2 + 3 + \ldots + n)}{n} = \frac{4 \cdot \frac{n(n+1)}{2}}{n} = 2(n + 1) \]

- The sum of squares \( \sum_{i=1}^{n} (4i)^2 \):
\[ \sum_{i=1}^{n} (4i)^2 = 16 \sum_{i=1}^{n} i^2 = 16 \cdot \frac{n(n+1)(2n+1)}{6} \]

- Therefore, the variance of the first \( n \) multiples of 4 is:
\[ \sigma^2 = \frac{1}{n} \cdot 16 \cdot \frac{n(n+1)(2n+1)}{6} - (2(n + 1))^2 \]

### Step 3: Calculate the variance of the first \( n \) odd numbers
For the sequence \( 1, 3, 5, \ldots, (2n - 1) \):
- The mean \( \mu \) is:
\[ \mu = \frac{1 + 3 + 5 + \ldots + (2n - 1)}{n} = \frac{n^2}{n} = n \]

- The sum of squares \( \sum_{i=1}^{n} (2i - 1)^2 \):
\[ \sum_{i=1}^{n} (2i - 1)^2 = \sum_{i=1}^{n} (4i^2 - 4i + 1) = 4\sum_{i=1}^{n} i^2 - 4\sum_{i=1}^{n} i + n = 4 \cdot \frac{n(n+1)(2n+1)}{6} - 4 \cdot \frac{n(n+1)}{2} + n \]

- Therefore, the variance of the first \( n \) odd numbers is:
\[ \sigma^2 = \frac{1}{n} \left( 4 \cdot \frac{n(n+1)(2n+1)}{6} - 4 \cdot \frac{n(n+1)}{2} + n \right) - n^2 \]

### Step 4: Find the ratio of the variances
Now we can find the ratio of the variances:
\[ \text{Ratio} = \frac{\text{Variance of multiples of 4}}{\text{Variance of odd numbers}} = \frac{16 \cdot \text{Variance of } (1, 2, 3, \ldots, n)}{4 \cdot \text{Variance of } (1, 2, 3, \ldots, n)} \]
This simplifies to:
\[ \text{Ratio} = \frac{16}{4} = 4 \]

### Final Result
Thus, 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:
\[ \boxed{4} \]