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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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
$\dfrac{4}{3}$
›Reveal solutionSolution
For a binomial distribution, the mean is $np$ and the variance is $np(1-p)$ . Solving $np = \frac{4}{3}$ and $np(1-p) = \frac{10}{9}$ gives $n = 8$ , $p = \frac{1}{6}$ . Then $P(X \geq 6) = P(X=6)+P(X=7)+P(X=8) = \frac{741}{6^8}$ , so the correct option is (B).

Concept & Intuition

The binomial distribution models the number of successes in $n$ independent trials, each with success probability $p$ . Its mean is $np$ and variance is $np(1-p)$ . When given these two numbers, we can solve for $n$ and $p$ uniquely. Once we know the distribution, computing $P(X \geq 6)$ is just summing the probabilities for $X = 6, 7, 8$ (since $n$ turns out to be small). The trick is to notice that the variance is less than the mean, which forces $p < 1/2$ and gives a neat fraction.

Step-by-step solution

Set up equations from given mean and variance Mean: $np = \frac{4}{3}$ Variance: $np(1-p) = \frac{10}{9}$ Divide variance by mean:

$\frac{np(1-p)}{np} = 1-p = \frac{10/9}{4/3} = \frac{10}{9} \cdot \frac{3}{4} = \frac{30}{36} = \frac{5}{6}$
So $1-p = \frac{5}{6}$ , hence $p = \frac{1}{6}$ .

Find $n$ From $np = \frac{4}{3}$ and $p = \frac{1}{6}$ , we have

$n \cdot \frac{1}{6} = \frac{4}{3} \quad\Rightarrow\quad n = \frac{4}{3} \cdot 6 = 8.$
So the distribution is $X \sim \text{Binomial}(n=8, p=\frac{1}{6})$ .

Write the probability mass function

$P(X = k) = \binom{8}{k} \left(\frac{1}{6}\right)^k \left(\frac{5}{6}\right)^{8-k}.$

Compute $P(X \geq 6)$ This is $P(X=6) + P(X=7) + P(X=8)$ .

For $k=6$ : $\binom{8}{6} = 28$ , so

$P(6) = 28 \left(\frac{1}{6}\right)^6 \left(\frac{5}{6}\right)^2 = 28 \cdot \frac{25}{6^8}.$

For $k=7$ : $\binom{8}{7} = 8$ , so
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TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
$\dfrac{4}{3}$
›Reveal solutionSolution

Solving $np = \tfrac{4}{3}$ and $np(1-p) = \tfrac{10}{9}$ gives $n=8,\ p=\tfrac16$ . Then $P(X\ge 6) = \dfrac{741}{6^8}$ , option (C).

For $X \sim \text{Bin}(n,p)$ : mean $= np$ , variance $= np(1-p)$ .

Find $p$ : divide variance by mean,
$1-p = \frac{np(1-p)}{np} = \frac{10/9}{4/3} = \frac{10}{9}\cdot\frac{3}{4} = \frac{5}{6} \;\Rightarrow\; p = \frac{1}{6}.$
Find $n$ : $np = \dfrac{4}{3} \Rightarrow n\cdot\dfrac{1}{6} = \dfrac{4}{3} \Rightarrow n = 8.$ So $q = 1-p = \dfrac{5}{6}$ .

Compute $P(X\ge 6) = P(6)+P(7)+P(8)$ : …
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TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
$\sigma^2$
›Reveal solutionSolution

Mean $=11$ , so $M=\tfrac{56}{9}$ and $\sigma^2=\tfrac{464}{9}$ , giving $3(\sigma^2-M)=136$ .

Data: $1, 3, 5, 7, 11, 13, 17, 19, 23$ (nine values).

Mean: $\bar{x}=\dfrac{1+3+5+7+11+13+17+19+23}{9}=\dfrac{99}{9}=11.$

Mean deviation from the mean $M$ : the absolute deviations $|x_i-11|$ are

$10, 8, 6, 4, 0, 2, 6, 8, 12$ , whose sum is $56$ , so
$M=\frac{56}{9}.$
Variance $\sigma^2$ : the squared deviations $(x_i-11)^2$ are …
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TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
$\begin{array}{c|cccccc} X = x_i & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline P(X = x_i) & 0.1 & k & 0.2 & 2k & 3k & k \\ \end{array}$
›Reveal solutionSolution

Solve $\sum P = 1$ for $k=0.1$ , then Var $= E(X^2) - [E(X)]^2 = 2.8 - 0.64 = 2.16$ (option C).

Step 1 — Find $k$ . Total probability is $1$ :
$0.1 + k + 0.2 + 2k + 3k + k = 1 \implies 0.3 + 7k = 1 \implies k = 0.1$
So the probabilities are:

$x_i$ $-2$ $-1$ $0$ $1$ $2$ $3$
$P$ $0.1$ $0.1$ $0.2$ $0.2$ $0.3$ $0.1$

Step 2 — Mean.
$E(X) = \sum x_i P_i = (-0.2) + (-0.1) + 0 + 0.2 + 0.6 + 0.3 = 0.8$
Step 3 — $E(X^2)$ . …
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TG EAPCET 2022Set eng-2022-07-19-FN1 markMCQ
$Q.The mean deviation from the mean for the observations 1, 3, 5, 7, 11, 13, 17, 19, 23 is (A) 6 (B) 11\dfrac{4}{9} (C) 11 (D) 6\dfrac{2}{9}$
›Reveal solutionSolution
To find the mean deviation from the mean, we first calculate the mean of the observations, then find the absolute difference of each observation from this mean, and finally average these absolute differences. The mean deviation from the mean for the given observations is $\boxed{6\frac{2}{9}}$ .

The mean deviation is a measure of dispersion that tells us, on average, how much the observations in a dataset differ from a central value (in this case, the mean). We use absolute values for the deviations because we are interested in the magnitude of the difference, regardless of whether an observation is above or below the mean. If we didn't take absolute values, the sum of deviations from the mean would always be zero, which wouldn't give us any useful information about spread.

Here's how to calculate it step-by-step:

Calculate the mean ( $\bar{x}$ ) of the observations. The given observations are $1, 3, 5, 7, 11, 13, 17, 19, 23$ . There are $n = 9$ observations. The sum of observations is: $1 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 = 99$ The mean is:

$\bar{x} = \frac{\sum x_i}{n} = \frac{99}{9} = 11$

Calculate the absolute deviation ( $|x_i - \bar{x}|$ ) for each observation. This means finding the absolute difference between each observation and the mean we just calculated.

$|1 - 11| = |-10| = 10$

$|3 - 11| = |-8| = 8$

$|5 - 11| = |-6| = 6$

$|7 - 11| = |-4| = 4$

$|11 - 11| = |0| = 0$

$|13 - 11| = |2| = 2$

$|17 - 11| = |6| = 6$

$|19 - 11| = |8| = 8$

$|23 - 11| = |12| = 12$ …
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TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
$\overline{x}$
›Reveal solutionSolution
The mean of the absolute deviations from the mean is, by definition, the mean deviation (also called the mean absolute deviation). The correct option is (D).

Concept and Intuition

When we talk about "spread" or "dispersion" in a data set, there are several common measures: variance, standard deviation, and mean deviation. Each captures how far the observations are from the center (usually the mean), but they differ in how they treat the distances.

Variance averages the squared deviations: $\frac{1}{n}\sum (x_i - \overline{x})^2$ .

Standard deviation is the square root of variance.

Mean deviation (or mean absolute deviation) averages the absolute deviations: $\frac{1}{n}\sum |x_i - \overline{x}|$ .

The question literally asks: "What is the mean of the absolute deviations from $\overline{x}$ ?" That is exactly the definition of the mean deviation. No calculation is needed — it's a matter of recognizing the terminology.

Step-by-Step Reasoning

Identify what is being described The phrase "mean of the absolute deviations of these observations from $\overline{x}$ " translates directly to:

$\frac{1}{n} \sum_{i=1}^{n} |x_i - \overline{x}|$
This is a standard statistical formula.

Match the formula to the options

(A) Variance: $\frac{1}{n}\sum (x_i - \overline{x})^2$ — uses squares, not absolute values.

(B) Mean proportion: Not a standard measure of dispersion; irrelevant here.

(C) Standard deviation: $\sqrt{\frac{1}{n}\sum (x_i - \overline{x})^2}$ — again, squares and a square root. …
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TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQ
$M_1$
›Reveal solutionSolution
We first identify the dataset as the first 15 even integers. We then calculate the mean and median, finding both to be 16. Subsequently, we compute the mean deviation from the mean ( $M_1$ ) and the mean deviation from the median ( $M_2$ ), which are both $\frac{112}{15}$ . Their sum is $\boxed{\frac{224}{15}}$ .

The problem asks us to calculate the sum of the mean deviation from the mean ( $M_1$ ) and the mean deviation from the median ( $M_2$ ) for the first 15 even integers.

Concept and Intuition

Mean deviation is a measure of dispersion, indicating the average absolute difference between each data point and a central value (either the mean or the median). It helps us understand how spread out the data points are around that central tendency.

Mean ( $\bar{x}$ ): The arithmetic average of all observations. It is calculated as the sum of all observations divided by the number of observations.

Median ( $M$ ): The middle value of a dataset when arranged in ascending or descending order. For an odd number of observations, it is the single middle value.

Mean Deviation from Mean ( $M_1$ ): This is the average of the absolute differences between each observation and the mean of the dataset.

Mean Deviation from Median ( $M_2$ ): This is the average of the absolute differences between each observation and the median of the dataset.

For a symmetric distribution, such as an arithmetic progression, the mean and median often coincide. If they do, the mean deviation from the mean and the mean deviation from the median will be identical.

Step-by-Step Calculation

Identify the Data Set

The first 15 even integers are $2, 4, 6, \dots$ .

This is an arithmetic progression with the first term $a = 2$ , common difference $d = 2$ , and number of terms $n = 15$ .

The last term is $l = a + (n-1)d = 2 + (15-1)2 = 2 + 14 \times 2 = 2 + 28 = 30$ .

So, the data set is $\{2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30\}$ .

Calculate the Mean ( $\bar{x}$ )

The sum of an arithmetic progression is $S_n = \frac{n}{2}(a + l)$ .

$S_{15} = \frac{15}{2}(2 + 30) = \frac{15}{2}(32) = 15 \times 16 = 240$ .

The mean is $\bar{x} = \frac{S_n}{n} = \frac{240}{15} = 16$ .

Calculate $M_1$ (Mean Deviation from Mean)

The formula for mean deviation from the mean is:
$M_1 = \frac{\sum_{i=1}^{n} |x_i - \bar{x}|}{n}$
Sum of absolute deviations $\sum |x_i - \bar{x}| = (14+12+10+8+6+4+2) + 0 + (2+4+6+8+10+12+14)$ . …
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$x_i$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P$	$0.1$	$0.1$	$0.2$	$0.2$	$0.3$	$0.1$

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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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