$$\begin{array}{ll} \text { Now, } & \bar{x}=\frac{\sum f_f x_i}{\sum f_i}=\frac{433}{20}=21.65 \\ \therefore & \mathrm{MD}=\frac{\sum f_i\left|x_i-\bar{x}\right|}{\sum f_i}=\frac{25}{20}=1.25 \end{array}$$

$$\begin{array}{ll} \text { Now, } & M_e=\left(\frac{20+1}{2}\right) \text { th item }=\left(\frac{21}{2}\right)=10.5 \text { th item } \\ \therefore & M_e=12 \\ \therefore & M D=\frac{\sum f_i d_i}{\sum f_i}=\frac{25}{20}=1.25 \end{array}$$

Consider first natural number when $n$ is an odd i.e., 1, 2, 3, 4, ...n, [odd].

$$\begin{aligned} & \text { Mean } \bar{x}= \frac{1+2+3+\ldots+n}{n}=\frac{n(n+1)}{2 n}=\frac{n+1}{2} \\ \therefore \quad & \mathrm{MD}= \left|1-\frac{n+1}{2}\right|+\left|2-\frac{n+1}{2}\right|+\left|3-\frac{n+1}{2}\right|+\ldots+\left|n-\frac{n+1}{2}\right| \end{aligned}$$

$$=\frac{\left|-\frac{n+1}{2}\right|+\left|2-\frac{n+1}{2}\right|+\ldots+\left|\frac{n-1}{2}-\frac{n+1}{2}\right|+\left|\frac{n+1}{2}-\frac{n+1}{2}\right|+\left|\frac{n+3}{2}-\frac{n+1}{2}\right|+\ldots+\left|\frac{n-2}{2}-\frac{n+1}{2}\right|+\left|n-\frac{n+1}{2}\right|}{n}$$

$$\begin{aligned} & =\frac{2}{n}\left[1+2+\ldots+\frac{n-3}{2}+\frac{n-1}{2}\right] \quad\left(\frac{n-1}{2}\right) \text { terms } \\ & =\frac{2}{n}\left[\frac{\left(\frac{n-1}{2}\right)\left(\frac{n-1}{2}+1\right)}{2}\right] \quad\left[\because \text { sum of first } n \text { natural numbers }=\frac{n(n+1)}{2}\right] \\ & =\frac{2}{n} \cdot \frac{1}{2}\left[\left(\frac{n-1}{2}\right)\left(\frac{n+1}{2}\right)\right]=\frac{1}{n}\left(\frac{n^2-1}{4}\right)=\frac{n^2-1}{4 n} \end{aligned}$$

Consider first $n$ natural number, when $n$ is even i.e., $1,2,3,4,\quad$ [even]

$$\begin{aligned} & \therefore \quad \text { Mean } \bar{x}=\frac{1+2+3+\ldots+n}{n}=\frac{n(n+1)}{2 n}=\frac{n+1}{2} \\ & \mathrm{MD}=\frac{1}{n}\left[\left|1-\frac{n+1}{2}\right|+\left|2-\frac{n+1}{2}\right|+\left|3-\frac{n+1}{2}\right|\right]+\left|\frac{n-2}{2}-\frac{n+1}{2}\right|+\left|\frac{n}{2}-\frac{n+1}{2}\right| \\ &+\left|\frac{n+2}{2}-\frac{n+1}{2}\right|+\ldots+\left|n-\frac{n+1}{2}\right| \\ &=\frac{1}{n}\left[\left|\frac{1-n}{2}\right|+\left|\frac{3-n}{2}\right|+\left|\frac{5-n}{2}\right|+\ldots .+\left|\frac{-3}{2}\right|+\left|\frac{1}{2}\right|+\ldots+\left|\frac{n-1}{2}\right|\right] \\ &=\frac{2}{n}\left[\frac{1}{2}+\frac{3}{2}+\ldots+\frac{n-1}{2}\right]\left(\frac{n}{2}\right) \text { terms } \\ &=\frac{1}{n} \cdot\left(\frac{n}{2}\right)^2 \quad\left[\because \text { sum of first } n \text { natural numbers }=n^2\right]\\ &=\frac{1}{n} \cdot \frac{n^2}{4}=\frac{n}{4} \\ \end{aligned}$$

Now, $\quad \sum x_i=1+2+3+4+\ldots+n=\frac{n(n+1)}{2}$

and $\quad \sum x_i^2=1^2+2^2+3^2+\ldots+n^2=\frac{n(n+1)(2 n+1)}{6}$

$$\begin{aligned} \therefore & =\sqrt{\frac{\sum x_i^2}{N}-\left(\frac{\sum x_i}{N}\right)^2} \\ & =\sqrt{\frac{n(n+1)(2 n+1)}{6 n}-\frac{n^2(n+1)^2}{4 n^2}} \\ & =\sqrt{\frac{(n+1)(2 n+1)}{6}-\frac{(n+1)^2}{4}} \\ & =\sqrt{\frac{2\left(2 n^2+3 n+1\right)-3\left(n^2+2 n+1\right)}{12}} \\ & =\sqrt{\frac{4 n^2+6 n+2-3 n^2-6 n-3}{12}} \\ & =\sqrt{\frac{n^2-1}{12}} \end{aligned}$$