Calculate the mean deviation about the mean

Question:

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

Solution:

Given set of first $n$ natural numbers when $n$ is an even number.

Now we have to find the mean deviation about the mean

We know first $\mathrm{n}$ natural numbers are $1,2,3 \ldots ., \mathrm{n}$. And given $\mathrm{n}$ is even number.

So mean is,

$\overline{\mathrm{x}}=\frac{1+2+3+\cdots+\mathrm{n}}{\mathrm{n}}=\frac{\frac{\mathrm{n}(\mathrm{n}+1)}{2}}{\mathrm{n}}=\frac{(\mathrm{n}+1)}{2}$

The deviations of numbers from the mean are as shown below,

$1-\frac{(n+1)}{2}, 2-\frac{(n+1)}{2}, 3-\frac{(n+1)}{2}, \ldots, \frac{(n-2)}{2}-\frac{(n+1)}{2}, \frac{(n)}{2}$

$-\frac{(n+1)}{2}, \frac{(n+2)}{2}-\frac{(n+1)}{2}, \ldots ., n-\frac{(n+1)}{2}$

Or,

$\frac{2-(n+1)}{2}, \frac{4-(n+1)}{2}, \frac{6-(n+1)}{2}, \ldots, \frac{n-2-(n+1)}{2}, \frac{n-(n+1)}{2}, \frac{(n+2)-(n+1)}{2}, \ldots, \frac{2 n-(n+1)}{2}$

The above equation can be written as

$\frac{2-(n+1)}{2}, \frac{4-(n+1)}{2}, \frac{6-(n+1)}{2}, \ldots \ldots, \frac{-3}{2}, \frac{-1}{2}, \frac{1}{2} \ldots, \frac{2 n-(n+1)}{2}$

Or,

$\frac{1-\mathrm{n}}{2}, \frac{3-\mathrm{n}}{2}, \frac{5-\mathrm{n}}{2}, \ldots \ldots, \frac{-3}{2}, \frac{-1}{2}, \frac{1}{2}, \ldots \ldots, \frac{\mathrm{n}-1}{2}$

So the absolute values of deviation from the mean is

$\left|x_{i}-\bar{x}\right|=\frac{(n-1)}{2}, \frac{(n-3)}{2}, \frac{(n-5)}{2}, \ldots \ldots, \frac{3}{2}, \frac{1}{2}, \frac{1}{2}, \ldots \ldots, \frac{n-1}{2}$

The sum of absolute values of deviations from the mean, is

$\sum\left|x_{i}-\bar{x}\right|=\frac{(n-1)}{2}+\frac{(n-3)}{2}+\frac{(n-5)}{2}+\cdots+\frac{3}{2}+\frac{1}{2}+\frac{1}{2}+\cdots+\frac{n-1}{2}$

We can write as

$\sum\left|x_{1}-\bar{x}\right|=\left(\frac{1}{2}+\frac{3}{2}+\cdots+\frac{(n-1)}{2}\right)\left(\frac{n}{2}\right)$

Now we know sum of first $n$ natural numbers $=n^{2}$

Therefore, mean deviation about the mean is

$\mathrm{M} . \mathrm{D}=\frac{\sum\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|}{\mathrm{n}}=\frac{\left(\frac{1}{2}+\frac{3}{2}+\cdots+\frac{(\mathrm{n}-1)}{2}\right)\left(\frac{\mathrm{n}}{2}\right)}{\mathrm{n}}$

$M . D=\frac{\sum\left|x_{i}-\bar{x}\right|}{n}=\frac{\left(\frac{n}{2}\right)^{2}}{n}$

$M . D=\frac{\sum\left|x_{i}-\bar{x}\right|}{n}=\frac{n^{2}}{4 n}=\frac{n}{4}$

Hence the mean deviation about the mean of the set of first $n$ natural numbers when $n$ is an even number is $n / 4$

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