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Calculate the mean deviation about the mean of the set of first $n$ natural numbers when $n$ is an odd number.
Solution:
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Verified Answer
Let $\mathrm{n}=2 \mathrm{k}+1($ odd $) \Rightarrow \mathrm{k}=\frac{(\mathrm{n}-1)}{2}$
$$
\bar{x}=\frac{1+2+3+\ldots+n}{n}=\frac{n(n+1)}{2 n}
$$
$$
=\frac{\mathrm{n}+1}{2}=\frac{2 \mathrm{k}+1+1}{2}=(\mathrm{k}+1)
$$
$$
\begin{aligned}
&\therefore \quad \Sigma|\mathrm{x}-\overline{\mathrm{x}}|=2(1+2+3+\ldots .+\mathrm{k}) \\
&=2\left\{\frac{\mathrm{k}(\mathrm{k}+1)}{2}\right\}=2\left\{\frac{\left(\frac{\mathrm{n}-1}{2}\right)\left(\frac{\mathrm{n}-1}{2}+1\right)}{2}\right\}
\end{aligned}
$$
$$
=\left(\frac{\mathrm{n}-1}{2}\right)\left(\frac{\mathrm{n}+1}{2}\right)=\frac{\mathrm{n}^2-1}{4}
$$
$\therefore \quad$ Required mean deviation
$$
=\frac{1}{n} \Sigma|x-\bar{x}|=\frac{n^2-1}{4 n}
$$
$$
\bar{x}=\frac{1+2+3+\ldots+n}{n}=\frac{n(n+1)}{2 n}
$$
$$
=\frac{\mathrm{n}+1}{2}=\frac{2 \mathrm{k}+1+1}{2}=(\mathrm{k}+1)
$$
$$
\begin{aligned}
&\therefore \quad \Sigma|\mathrm{x}-\overline{\mathrm{x}}|=2(1+2+3+\ldots .+\mathrm{k}) \\
&=2\left\{\frac{\mathrm{k}(\mathrm{k}+1)}{2}\right\}=2\left\{\frac{\left(\frac{\mathrm{n}-1}{2}\right)\left(\frac{\mathrm{n}-1}{2}+1\right)}{2}\right\}
\end{aligned}
$$
$$
=\left(\frac{\mathrm{n}-1}{2}\right)\left(\frac{\mathrm{n}+1}{2}\right)=\frac{\mathrm{n}^2-1}{4}
$$
$\therefore \quad$ Required mean deviation
$$
=\frac{1}{n} \Sigma|x-\bar{x}|=\frac{n^2-1}{4 n}
$$
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