Find the mean and variance of the first n natural numbers.
Their sum is ∑k = n(n+1)/2
So, the mean is easy, right?
The sum of their squares is ∑k^2 = n(n+1)(2n+1)/6
Now you can get the variance using the usual formula.
To find the mean and variance of the first n natural numbers, let's break it down step by step:
Step 1: Determine the numbers
The first n natural numbers are 1, 2, 3, ..., n.
Step 2: Calculate the mean
The mean of a set of numbers is the sum of all the numbers divided by the count of numbers.
The sum of the first n natural numbers can be found using the formula for the sum of an arithmetic series:
Sum = (n/2) * (first number + last number)
= (n/2) * (1 + n)
Therefore, the mean is given by:
Mean = Sum / Count
= [(n/2) * (1 + n)] / n
= (1 + n) / 2
So, the mean of the first n natural numbers is (1 + n) / 2.
Step 3: Calculate the variance
The variance of a set of numbers measures the variability or spread of the numbers around the mean. To calculate the variance, we need to first find the squared deviations of each number from the mean, then compute the average of those squared deviations.
The squared deviation of a number x from the mean is (x - mean)^2.
For the first n natural numbers, we can represent them mathematically as:
1, 2, 3, ..., n
To find the squared deviation of each number from the mean, we substitute the values into the formula:
(1 - mean)^2 + (2 - mean)^2 + (3 - mean)^2 + ... + (n - mean)^2
Expanding and simplifying, we get:
[(1 - mean)^2 + (2 - mean)^2 + (3 - mean)^2 + ... + (n - mean)^2] / n
Substituting the mean value, [(1 + n) / 2], we get:
[(1 - (1 + n) / 2)^2 + (2 - (1 + n) / 2)^2 + (3 - (1 + n) / 2)^2 + ... + (n - (1 + n) / 2)^2] / n
Simplifying the formula, we have:
[(1 - 1 - n / 2)^2 + (2 - 1 - n / 2)^2 + (3 - 1 - n / 2)^2 + ... + (n - 1 - n / 2)^2] / n
[(1 - n / 2)^2 + (2 - n / 2)^2 + (3 - n / 2)^2 + ... + (n - n / 2)^2] / n
Further simplifying, we get:
[(1 - n / 2)^2 + (2 - n / 2)^2 + (3 - n / 2)^2 + ... + (n - n / 2)^2] / n
[(1 - n / 2)^2 + (2 - n / 2)^2 + (3 - n / 2)^2 + ... + (n / 2)^2] / n
Calculating the sum of squared deviations, we get:
[(1 - n / 2)^2 + (2 - n / 2)^2 + (3 - n / 2)^2 + ... + (n / 2)^2] = [n(2n^2 + 6n + 3)] / 12
Finally, dividing by n, we get the variance of the first n natural numbers:
Variance = [n(2n^2 + 6n + 3)] / (12 * n)
= (2n^2 + 6n + 3) / 12
So, the variance of the first n natural numbers is (2n^2 + 6n + 3) / 12.