Step-I : Locate the modal class, the class with highest frequency say, $f_m$. [If more than one class has that same frequency, any one can be chosen. In such cases, there will be more than one mode.]

Step-II : Call $f_{m-1}$ and $f_{m+1}$ to be the frequencies of the classes immediately preceding and succeeding the modal class.

Step-III : Calculate mode $\left(M_0\right)$ by using the following formula:

$$ \mathrm{M}_{\mathrm{o}}=x_l+\frac{f_m-f_{m-1}}{2 f_m-f_{m-1}-f_{m+1}} \times w $$

Where,

$x_l$ - lower limit of model class

$f_m$ - frequency of the model class

$f_{m-1}$ - frequency of the class preceding model class

$f_{m+1}$ - frequency of the class succeeding model class

$w$ - width of the modal class

For comparing two or more series for variability, we calculate the coefficient of standard deviation and the coefficient of variation.

The coefficient of standard deviation is defined as :

coefficient of standard deviation $=\frac{\sigma}{\bar{x}}$.

The coefficient of variation is defined as :

coefficient of variation $=\left(\frac{\sigma}{\bar{x}}\right) \times 100$.

where $\bar{x}$ and $$\sigma $$ as usual represent respectively the mean and the standard deviation of the data.

Let $\sigma_1$ and $\sigma_2$ be the standard deviations of the two groups containing $n_1$ and $n_2$ items respectively. Let $\bar{x}_1$ and $\bar{x}_2$ be their respective A.M. Let $\overline{\mathrm{x}}$ and $\sigma$ be the A.M. and S.D. of the combined group respectively. Then

$$ \begin{aligned} & \bar{x}=\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}, \\\\ & S.D=\sigma=\sqrt{\frac{n_1\left[\sigma_1^2+\left(\bar{x}_1-\bar{x}\right)^2\right]+n_2\left[\sigma_2^2+\left(\bar{x}_2-\bar{x}\right)^2\right]}{n_1+n_2}} \end{aligned} $$

Variance , $\sigma^2=\frac{n_1 \sigma_1^2+n_2 \sigma_2^2}{n_1+n_2}+\frac{n_1 n_2}{\left(n_1+n_2\right)^2}\left(\bar{x}_1-\bar{x}_2\right)^2$

1. $\operatorname{Var} .\left(x_i+\mathrm{b}\right)=\operatorname{Var} .\left(x_i\right)$

2. Var. $\left(a x_i\right)=\mathrm{a}^2 \cdot \operatorname{Var}\left(x_i\right)$

3. $\operatorname{Var}\left(a x_i+b\right)=a^2 \cdot \operatorname{Var}\left(x_i\right)$

where $\lambda, a, b$, are constant.

If a variable $x$ takes values $x_1,x_2,\dots ,x_n$ with respective frequencies $f_i,f_2,\dots ,f_n$ then standard deviation is given by

$\sigma =\sqrt{\frac{\sum _{i=1}^n{f}_i{\left(x_i-\bar{x}\right)}^2}{\sum _{i=1}^n{f}_i}}\text{ where }\bar{x}=\frac{\sum _{i=1}^n{f}_i{x}_i}{\sum _{i=1}^n{f}_i}.$

$\Rightarrow$ $\sigma =\sqrt{\frac{\sum _{i=1}^n{f}_i{x}_i^2}{\sum _{i=1}^n{f}_i}-{\left(\frac{\sum _{i=1}^n{f}_i{x}_i{}^2}{\sum _{i=1}^n{f}_i}\right)}^2}$

$\therefore$ Variance, ${\sigma }^2=\frac{\sum _{i=1}^n{f}_i{\left(x_i-\overline{x}\right)}^2}{\sum _{i=1}^n{f}_i}$

$=\frac{\sum _{i=1}^n{f}_i{\left(x_i\right)}^2}{\sum _{i=1}^n{f}_i}-{\left(\frac{\sum _{i=1}^n{f}_i{x}_i}{\sum _{i=1}^n{f}_i}\right)}^2$

Let us consider $n$ observations $x_1, x_2, \ldots, x_n$. Let the arithmetic mean of these observations be $\bar{x}$. Then standard deviation is given by

$$ \sigma=\sqrt{\frac{1}{n}\left[\left(x_1-\bar{x}\right)^2+\left(x_2-\bar{x}\right)^2+\ldots+\left(x_n-\bar{x}\right)^2\right]}$$

$$=\sqrt{\frac{1}{n} \sum\limits_{i=1}^n\left(x_i-\bar{x}\right)^2} \text {. } $$

And Variance, $${\sigma ^2} = {{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \overline x } \right)}^2}} } \over n}$$

To reduce number of arithmetic operations to be carried out we simplify the above equation.

$ \text { We have } \sigma=\sqrt{\frac{1}{n} \sum\limits_{i=1}^n\left(x_i-\bar{x}\right)^2} $

$=\sqrt{\frac{1}{n} \sum\limits_{i=1}^n\left(x_i^2-2 x_i \bar{x}+\bar{x}^2\right)}$

$ = \sqrt{\frac{1}{n} \sum x_i^2-(2 \bar{x}) \frac{1}{n} \sum x_i+(\bar{x})^2 \frac{1}{n} \sum 1}$

$=\sqrt{\frac{1}{n} \sum x_i^2-2(\bar{x})^2+(\bar{x})^2}$

$ = \sqrt{\frac{1}{n} \sum x_i^2-(\bar{x})^2}$

$ =\sqrt{\frac{1}{n}\left[\sum x_i^2-\frac{\left(\sum x_i\right)^2}{n}\right]}$

And Variance, $${\sigma ^2} = {{\sum\limits_{i = 1}^n {{{\left( {{x_i}} \right)}^2}} } \over n} - {\left( {\overline x } \right)^2}$$

Standard deviation of a given set of observations is defined as the positive square root of the average of squared deviations of all observations taken from their arithmetic mean. It is generally denoted by Greek alphabet $\sigma$ or s.

The square of the standard deviation is called variance and is denoted by $\sigma^2$.

Let $x_1, x_2, x_3, \ldots, x_n$ occur with frequencies $f_1, f_2, f_3, \ldots, f_n$ respectively and let $\sum\limits_{i=1}^n f_i=N$.

Then Mean deviation from mean $=\frac{\sum\limits_{i=1}^n f_i\left|x_i-\bar{x}\right|}{\sum\limits_{i=1}^n f_i}$

where $\bar{x}=$ mean.

Mean deviation from median $=\frac{\sum\limits_{i=1}^n f_i\left|x_i-M\right|}{\sum\limits_{i=1}^n f_i}$

where $M=$ median.

Let $\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \ldots, \mathrm{x}_{\mathrm{n}}$ be $\mathrm{n}$ observations. Then we have the following formulas depending upon our choice of ' $a$ '.

M.D. $(a)=\frac{\sum\limits_{i=1}^n\left|x_i-a\right|}{n}$;

M.D. $(\bar{x})=\frac{\sum\limits_{i=1}^n\left|x_i-\bar{x}\right|}{n}$, where $\bar{x}$ is the mean of given observations;

M.D. $(M)=\frac{\sum\limits_{i=1}^n\left|x_i-M\right|}{n}$, where $M$ is the median of the given observations.

The difference between the greatest and least values taken by a discrete variable of a distribution, are called the range of that distribution.

If the distribution is grouped distribution, then its range is the difference between upper limit, of the maximum class and lower limit of the minimum class.

Note :

Coefficient of Range $=\frac{\text { difference of extreme values }}{\text { sum of extreme values }}=\frac{L-S}{L+S}$

where $L=$ largest value and $S=$ smallest value

• In symmetric distribution,

mean = mode = median

• In skew (moderately asymmetrical) distribution,

median divides mean and mode internally in 1 : 2 ratio.

$\Rightarrow \quad$ median $=\frac{2(\text { Mean })+(\text { Mode })}{3}$

i. For ungrouped distribution : If $x_1, x_2, \ldots x_n$ are $n$ values of variable $x_i$ then their A.M. $\bar{x}$ is defined as

$$ \begin{aligned} & \bar{x}=\frac{x_1+x_2+\ldots+x_n}{n}=\frac{\sum\limits_{i=1}^n x_i}{n} \\\\ & \Rightarrow \sum x_i=n \bar{x} \end{aligned} $$

ii. For ungrouped and grouped frequency distribution : If $x_1, x_2, \ldots . x_n$ are values of variable with corresponding frequencies $f_1, f_2, \ldots f_n$ then their A.M. is given by

$$ \bar{x}=\frac{f_1 x_1+f_2 x_2+\ldots+f_n x_n}{f_1+f_2+\ldots .+f_n}=\frac{\sum\limits_{i=1}^n f_i x_i}{N}, \text { where } N=\sum\limits _{i=1}^n f_i $$

iii. Assumed Mean Method : If the value of $x_i$ are large, then the calculation of A.M. by using previous formula is quite time consuming. In such case we take deviation of variable from an arbitrary point $a$.

Let $d_i=x_i-a$

$\therefore \bar{x}=a+\frac{\sum f_i d_i}{N}$, where $a$ is assumed mean.

In a frequency distribution the mode is the value of that variate which has the maximum frequency.

Method for determining mode :

(i) For ungrouped distribution : The value of that variate which is repeated maximum number of times.

(ii) For ungrouped frequency distribution : The value of that variate which have maximum frequency.

Prepare cumulative frequency (c.f.) column and find value of then find the class which contain value of cumulative frequency (c.f.) which is equal or just greater to $N / 2$, this is median class.

$$ \begin{aligned} & \therefore \text { Median(M) }=l+\frac{\left(\frac{N}{2}-F\right)}{f} h \\\\ & \text { where } \\\\ & l=\text { lower limit of median class } \\\\ & f=\text { frequency of median class } \\\\ & F=\text { cumulative frequency (c.f.) of the class preceding median class } \\\\ & h=\text { class interval of median class } \end{aligned} $$

For ungrouped frequency distribution : First we prepare the cumulative frequency (c.f.) column and find value of $N$ then

$$ \text { Median }=\left[\begin{array}{l} \left(\frac{N+1}{2}\right)^{\text {th }} \text { term, (when } N \text { is odd) } \\\\ \text { Mean of }\left(\frac{N}{2}\right)^{\text {th }} \text { and }\left(\frac{N}{2}+1\right)^{\text {th }} \text { terms, (when } N \text { is even) } \end{array}\right. $$

The median of a series is the value of middle term of the series when the values are written in ascending order.

Formulae of median

For ungrouped distribution : Let $n$ be the number of variate in a series then

$$ \text { Median }=\left[\begin{array}{l} \left(\frac{n+1}{2}\right)^{\text {th }} \text { term, (when } n \text { is odd) } \\\\ \text { Mean of }\left(\frac{n}{2}\right)^{\text {th }} \text { and }\left(\frac{n}{2}+1\right)^{\text {th }} \text { terms, (when } n \text { is even) } \end{array}\right. $$

If $\bar{x}_1$ and $\bar{x}_2$ be the means of two groups having $n_1$ and $n_2$ terms respectively then the mean (combined mean) of their composite groups is given by combined mean $=\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}$

If there are more than two groups then, combined mean $=\frac{n_1 \bar{x}_1+n_2 \bar{x}_2+n_3 \bar{x}_3+\ldots}{n_1+n_2+n_3+\ldots}$

1. Sum of deviations of variable $x_i$ from their AM is always zero

$$ \text { i.e. } \sum\left(x_i-\bar{x}\right)=0, \quad \sum f_i\left(x_i-\bar{x}\right)=0 $$

2. Sum of square of deviations of variate from their AM is minimum i.e. $\sum\left(x_i-\bar{x}\right)^2$ is minimum

3. If $\bar{x}$ is the mean of variable $x_i$ then

$\mathrm{AM}$ of $\left(x_i+\lambda\right)=(\bar{x}+\lambda)$

$\operatorname{AM}$ of $\left(\lambda x_i\right)=\lambda \bar{x}$

$\mathrm{AM}$ of $\left(a x_i+b\right)=a \bar{x}+b$

(where $\lambda, a, b$ are constants)

Let $a_1, a_2, \ldots . ., a_n$ be $n$-positive real numbers and $x_1, x_2, \ldots \ldots, x_n$ be $n$-positive rational numbers,

$$ \begin{aligned} & \text { then Weighted Arithmetic Mean }=A=\frac{x_1 a_1+x_2 a_2+\ldots . .+x_n a_n}{x_1+x_2+\ldots . .+x_n} \\\\ & \text { Weighted Geometric Mean }=G=\left(a_1^{x_1} \cdot a_2^{x_2} \ldots . . a_n^{x_n}\right) ^\frac{1}{x_1+x_2+\ldots .+x_n} \\\\ & \text { Weighted Harmonic Mean }=H=\frac{x_1+x_2+\ldots . .+x_n}{\frac{x_1}{a_1}+\frac{x_2}{a_2}+\ldots . .+\frac{x_n}{a_n}} \end{aligned} $$

(i) For ungrouped distribution : If $x_1, x_2, \ldots . x_n$ are $n$ non-zero values of variable then their harmonic mean $H$ is defined as

$$ H=\frac{n}{\frac{1}{x_1}+\frac{1}{x_2}+\ldots .+\frac{1}{x_n}}=\frac{n}{\sum\limits _{i=1}^n \frac{1}{x_i}} $$

(i) For ungrouped distribution : If $x_1, x_2, \ldots . x_n$ are $n$ positive values of variable then their geometric mean $G$ is given by

$$ G=\left(x_1 \cdot x_2 \cdot x_3 \ldots \ldots x_n\right)^{1 / n} $$