Q.Calculate the mean deviation about the mean of the set of first $n$ natural numbers when $n$ is an odd number.

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.

Precise Statement

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

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:

Variance

Variance is the average of squared deviations from the mean:

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:

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

Therefore:

Factor $(n+1)$:

Compute the bracket: common denominator 12:

Thus:

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.

Common Exam Pitfall

Quick Check

For $n=1$: mean = $1$, variance = $0$ (only one number, no spread). Formula gives $\frac{1^2-1}{12}=0$ — correct.

For $n=2$: numbers $1,2$, mean = $1.5$, variance = $\frac{(1-1.5)^2 + (2-1.5)^2}{2} = \frac{0.25+0.25}{2}=0.25$. Formula gives $\frac{4-1}{12}=0.25$ — correct.

You now have the complete picture: from intuition to derivation to exam-ready formulas.

Mean and Variance of the First n Natural Numbers is a classic result taught in the NCERT Class 11 Mathematics chapter on Statistics, matching searches like "mean and variance of natural numbers formula" or "statistics important questions class 11 maths". Because it combines the sum-of-squares formula with statistics, it's a frequently asked derivation-and-apply question in both CBSE boards and JEE Main.