Here you will learn how to solve mean by using step deviation method and by short method and properties of mean.

Let’s begin –

An average value or central value of a distribution is the value of variable which is representative of the entire distribution, this representative value are called the measures of central tendency.

Generally the following the measures of central tendency.

(a) Mathematical average

(i) Arithmetic mean (ii) Geometric mean (iii) Harmonic mean

(b) Positional average

(i) Median (ii) Mode

## Mean By Step Deviation Method

Sometime during the application of short method (given below) of finding the A.M. If each deviation \(d_i\) are divisible by a common number h(let)

Let \(u_i\) = \(d_i\over h\) = \(x_i – a\over h\)

\(\therefore\) \(\bar{x}\) = a + (\(\sum f_iu_i\over N\))h

Example : Find the A.M. of the following freq. dist.

\(x_i\) | 5 | 15 | 25 | 35 | 45 | 55 |

\(f_i\) | 12 | 18 | 27 | 20 | 17 | 6 |

Solution : Let assumed mean a = 35, h = 10

here N = \(\sum f_i\) = 100, \(u_i\) = \(x_i – 35\over 10\)

\(\sum f_iu_i\) = (12\(\times\)-3) + (18\(\times\)-2) + (27\(\times\)-1) + (20\(\times\)0) + (17\(\times\)1) + (6\(\times\)2)
= -70

\(\therefore\) \(\bar{x}\) = a + (\(\sum f_ix_i\over N\))h = 35 + \((-70)\over 100\)\(\times\)10 = 28

## Mean By Short Method

If the value of \(x_i\) are large, then calculation of A.M. by using mean formula is quite tedious and time consuming. In such case we take deviation of variate from an arbitrary point a.

Let \(d_i\) = \(x_i\) – a

\(\therefore\) \(\bar{x}\) = a + \(\sum f_id_i\over N\), where a is assumed mean

**Weighted Mean**

If \(w_1\), \(w_2\), ……\(w_n\) are the weights assigned to the values \(x_1\), \(x_2\), …..\(x_n\) respectively then their weighted mean is defined as

Weighted mean = \(w_1x_1 + w_2x_2 +……+ w_nx_n\over {w_1 +…….+ w_n}\) = \({\sum_{i=1}^{n}w_ix_i}\over {\sum_{i=1}^{n}w_i}\)

**Combined Mean**

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 group is given by

Combined mean = \(n_1\bar{x_1} + n_2\bar{x_2}\over {n_1 + n_2}\)

If there are more than two groups then, combined mean = \(n_1\bar{x_1} + n_2\bar{x_2} + n_3\bar{x_3} + …..\over {n_1 + n_2 + n_3 + ….}\)

**Properties of Arithmetic Mean**

(i) Sum of deviations of variate from their A.M. is always zero i.e. \(\sum\)(\(x_1 – \bar{x}\)) = 0, \(\sum\)\(f_i\)(\(x_1 – \bar{x}\)) = 0

(ii) Sum of square of deviations of variate from their A.M. is minimum i.e. \(\sum\)(\(x_1 – \bar{x})^2\) is minimum.

(iii) A.M. is independent of change of assumed mean i.e. it is not effected by any change in assumed mean.