The variance of the first 50 even natural numbers is?

To find the variance of the first 50 even natural numbers, we first need to understand what variance is and how to calculate it. Variance measures how much the numbers in a set differ from the mean (average) of that set. Let's break down the steps to find the variance of the first 50 even natural numbers.

Identifying the Set of Numbers

The first 50 even natural numbers can be represented as:

• 2, 4, 6, 8, ..., 100

These numbers can also be expressed in a formulaic way: the n-th even number is given by \(2n\). Thus, for n = 1 to 50, the numbers are \(2 \times 1, 2 \times 2, \ldots, 2 \times 50\).

Calculating the Mean

The mean (average) of a set of numbers is calculated by dividing the sum of the numbers by the count of the numbers. First, we find the sum of the first 50 even numbers:

• Sum = 2 + 4 + 6 + ... + 100

This can be simplified using the formula for the sum of an arithmetic series:

• Sum = n/2 * (first term + last term)
• Here, n = 50, the first term is 2, and the last term is 100.

Plugging in the values:

Sum = \(50/2 \times (2 + 100) = 25 \times 102 = 2550\)

Now, we calculate the mean:

Mean = Sum / Count = \(2550 / 50 = 51\)

Finding the Variance

Variance is calculated using the formula:

Variance = \( \frac{\sum (x_i - \mu)^2}{N} \)

Where \(x_i\) represents each number in the set, \(\mu\) is the mean, and N is the number of values.

First, we need to calculate \( (x_i - \mu)^2 \) for each even number:

• For example, for the first number (2): \( (2 - 51)^2 = (-49)^2 = 2401 \)
• For the second number (4): \( (4 - 51)^2 = (-47)^2 = 2209 \)
• Continuing this for all 50 numbers will give us a series of squared differences.

Summing the Squared Differences

Instead of calculating each squared difference individually, we can use the formula for the variance of an arithmetic series:

Variance = \( \frac{(n^2 - 1) \cdot d^2}{12} \)

• Where n is the number of terms (50) and d is the common difference (2).

Plugging in the values:

Variance = \( \frac{(50^2 - 1) \cdot 2^2}{12} = \frac{(2500 - 1) \cdot 4}{12} = \frac{2499 \cdot 4}{12} = \frac{9996}{12} = 833\)

Final Result

The variance of the first 50 even natural numbers is 833. This value indicates how spread out the even numbers are from their mean, providing insight into their distribution.