Find the standard deviation of the first n natural numbers.

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Concept understanding — Mean Variance Natural Numbers

Mean and Variance of Natural Numbers

Let’s start with something you already know: the mean (average) and variance (spread) of a set of numbers. If I give you the first five natural numbers — 1, 2, 3, 4, 5 — you can compute their mean and variance easily. But what if I ask: What is the mean of all natural numbers? That’s infinite, so it doesn’t make sense directly. Instead, we ask: What is the mean of the first $n$ natural numbers? And then we see how it behaves as $n$ grows.

That’s the core idea: we study the mean and variance of the first $n$ natural numbers as a function of $n$ , and often look at what happens when $n$ becomes very large.

Intuition First

Imagine you line up the numbers $1, 2, 3, \dots, n$ on a number line. Their average is somewhere in the middle — roughly $n/2$ . More precisely, the mean of the first $n$ natural numbers is $\frac{n+1}{2}$ . For $n=5$ , that’s $3$ , which matches your intuition.

Now, variance measures how spread out the numbers are around that mean. For small $n$ , the spread is small; for large $n$ , the spread grows. The variance of the first $n$ natural numbers turns out to be $\frac{n^2 - 1}{12}$ . For $n=5$ , that’s $\frac{25-1}{12} = 2$ , which is a moderate spread.

Note

These formulas assume we are using population variance (dividing by $n$ , not $n-1$ ). In exam contexts, always check which variance definition is expected — but for natural numbers, population variance is standard.

Precise Statement

Let $X$ be a random variable that takes values $1, 2, 3, \dots, n$ with equal probability $1/n$ . Then:

\text{Mean: } \mu_n = \frac{n+1}{2}

These are exact formulas for any positive integer $n$ .

Derivation (Why These Formulas?)

Mean

The sum of the first $n$ natural numbers is $1 + 2 + \dots + n = \frac{n(n+1)}{2}$ .

Since there are $n$ numbers, the mean is:

\mu_n = \frac{1}{n} \cdot \frac{n(n+1)}{2} = \frac{n+1}{2}

Variance

Variance is the average of squared deviations from the mean:

\sigma_n^2 = \frac{1}{n} \sum_{k=1}^n (k - \mu_n)^2

A cleaner way uses the identity: $\sigma^2 = E[X^2] - (E[X])^2$ .

First, $E[X^2] = \frac{1}{n} \sum_{k=1}^n k^2$ . The sum of squares formula is $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$ . So:

E[X^2] = \frac{1}{n} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6}

Now, $(E[X])^2 = \left( \frac{n+1}{2} \right)^2 = \frac{(n+1)^2}{4}$ .

Therefore:

\sigma_n^2 = \frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4}

Factor $(n+1)$ :

\sigma_n^2 = (n+1) \left[ \frac{2n+1}{6} - \frac{n+1}{4} \right]

Compute the bracket: common denominator 12:

\frac{2(2n+1) - 3(n+1)}{12} = \frac{4n+2 - 3n - 3}{12} = \frac{n - 1}{12}

Thus:

\sigma_n^2 = (n+1) \cdot \frac{n-1}{12} = \frac{n^2 - 1}{12}

What This Tells You

The mean grows linearly with $n$ — roughly half of $n$ .
The variance grows quadratically — roughly $n^2/12$ for large $n$ .
For large $n$ , the standard deviation $\sigma_n \approx \frac{n}{\sqrt{12}} \approx 0.2887n$ , meaning the spread is about 29% of the range. …

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The standard deviation of the first $n$ natural numbers is $\sqrt{\frac{n^2 - 1}{12}}$ . This comes from the fact that variance is the average of squares minus the square of the average, and for natural numbers both the mean and the mean of squares have neat closed forms.

The first $n$ natural numbers are $1, 2, 3, \dots, n$ . When we talk about standard deviation, we're measuring how spread out these numbers are around their mean. For equally spaced data like this, the spread depends only on $n$ and the spacing (which is 1). The formula turns out to be surprisingly clean.

The key insight: standard deviation is the square root of variance. And variance can be computed as:

\text{Variance} = \frac{1}{n}\sum_{i=1}^n x_i^2 - \left(\frac{1}{n}\sum_{i=1}^n x_i\right)^2

This is often easier than summing squared deviations directly. So we just need two things: the mean of the numbers, and the mean of their squares.

Find the mean of the first $n$ natural numbers. The sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$ . So the mean is:

\bar{x} = \frac{1}{n} \cdot \frac{n(n+1)}{2} = \frac{n+1}{2}

Find the mean of the squares. The sum of squares of the first $n$ natural numbers is $\frac{n(n+1)(2n+1)}{6}$ . So the mean of the squares is:

\overline{x^2} = \frac{1}{n} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6}

Compute the variance. Using the formula:

\sigma^2 = \overline{x^2} - (\bar{x})^2 = \frac{(n+1)(2n+1)}{6} - \left(\frac{n+1}{2}\right)^2

Put everything over a common denominator of 12:

\sigma^2 = \frac{2(n+1)(2n+1)}{12} - \frac{3(n+1)^2}{12}

Factor $(n+1)$ out of the numerator:

$\sigma^2 = \frac{n+1}{12} \left[ 2(2n+1) - 3(n+1) \right]$ …

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