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

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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.

Tip

A quick memory aid: For the first $n$ natural numbers, mean is $\frac{n+1}{2}$ and variance is $\frac{n^2-1}{12}$ . Notice the denominator 12 — it’s the same as the variance of a continuous uniform distribution over $[0,1]$ , which is $1/12$ .

Common Exam Pitfall

Watch out

Do not confuse the variance of the first $n$ natural numbers with the variance of a sample from a larger population. Here, the set $\{1,2,\dots,n\}$ is the entire population, so we divide by $n$ , not $n-1$ . If a problem says “variance of the first $n$ natural numbers,” use $\frac{n^2-1}{12}$ .

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.

For the first $n$ natural numbers with $n$ odd, the mean is $\frac{n+1}{2}$ . The mean deviation about the mean simplifies to $\frac{n^2 - 1}{4n}$ , which is the average absolute distance of each number from the centre of the set.

The mean deviation about the mean is a measure of spread — it tells us, on average, how far each observation lies from the arithmetic mean. For the first $n$ natural numbers $1, 2, 3, \dots, n$ , the data is perfectly symmetric when $n$ is odd. The mean sits right at the middle number, and the deviations on either side mirror each other. This symmetry is the key to a clean calculation.

Let’s work through it.

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

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

Since $n$ is odd, $\frac{n+1}{2}$ is an integer — it is exactly the middle term of the sequence.

Set up the mean deviation formula. Mean deviation about the mean is

\text{MD} = \frac{1}{n} \sum_{i=1}^{n} |x_i - \bar{x}|.

Here $x_i = i$ , and $\bar{x} = \frac{n+1}{2}$ .

Exploit symmetry.

The numbers are $1, 2, \dots, \frac{n+1}{2}, \dots, n$ . The mean is at position $\frac{n+1}{2}$ . For any $k$ from $1$ to $\frac{n-1}{2}$ , the pair $( \frac{n+1}{2} - k, \frac{n+1}{2} + k )$ has the same absolute deviation $k$ . So the sum of absolute deviations is twice the sum of $k$ for $k = 1$ to $\frac{n-1}{2}$ , plus zero for the middle term itself.
Compute the sum.

\sum_{i=1}^{n} |i - \frac{n+1}{2}| = 2 \sum_{k=1}^{(n-1)/2} k.

The sum of the first $m$ natural numbers is $\frac{m(m+1)}{2}$ . Here $m = \frac{n-1}{2}$ , so

\sum_{k=1}^{(n-1)/2} k = \frac{\frac{n-1}{2} \cdot \frac{n+1}{2}}{2} = \frac{(n-1)(n+1)}{8}.

Therefore

\sum_{i=1}^{n} |i - \bar{x}| = 2 \cdot \frac{(n-1)(n+1)}{8} = \frac{n^2 - 1}{4}.

Divide by $n$ to get the mean deviation.

\text{MD} = \frac{1}{n} \cdot \frac{n^2 - 1}{4} = \frac{n^2 - 1}{4n}.

Tip

A quick check: for $n=3$ , the numbers are $1,2,3$ , mean is $2$ , deviations are $1,0,1$ , sum = $2$ , MD = $2/3$ . Our formula gives $\frac{9-1}{12} = \frac{8}{12} = \frac{2}{3}$ . Works.

Watch out

A common mistake is to forget that the mean itself is $\frac{n+1}{2}$ , not $\frac{n}{2}$ or something else. Also, when $n$ is odd, the middle term contributes zero deviation — don’t accidentally include it in the sum of positive deviations.

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