Calculate the mean deviation about the mean of the set of first n natural numbers when n is an even 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. …

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For the first $n$ natural numbers with $n$ even, the mean deviation about the mean is $\dfrac{n}{4}$ .

Setup

The first $n$ natural numbers are $1,2,\dots,n$ . Their mean is

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

When $n$ is even, $\bar{x}=\dfrac{n+1}{2}$ is a half-integer sitting exactly between the two middle numbers $\tfrac{n}{2}$ and $\tfrac{n}{2}+1$ .

The deviations

\text{M.D.}=\frac{1}{n}\sum_{i=1}^{n}\left\lvert i-\frac{n+1}{2}\right\rvert.

The numbers split symmetrically about $\bar{x}$ . For the lower half $i=1,2,\dots,\tfrac{n}{2}$ the deviations $\dfrac{n+1}{2}-i$ are

\frac{n-1}{2},\ \frac{n-3}{2},\ \dots,\ \frac{3}{2},\ \frac{1}{2},

i.e. the half-integers $\tfrac12,\tfrac32,\dots,\tfrac{n-1}{2}$ . The upper half gives the same set by symmetry.

Summing

Sum over one half:

$\sum_{i=1}^{n/2}\left(\frac{n+1}{2}-i\right)=\frac{1}{2}\big(1+3+5+\cdots+(n-1)\big)=\frac{1}{2}\left(\frac{n}{2}\right)^{2}=\frac{n^{2}}{8},$ …

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Q.Calculate the mean deviation about the mean of the set of first nnn natural numbers when nnn is an even number.

Setup

The deviations

Summing

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

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