Here you will learn how to solve mean by using step deviation method and by short method and properties of mean.
Let’s begin –
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
Example : Find the A.M. of the following freq. dist.
| Class Interval | 0-50 | 50-100 | 100-150 | 150-200 | 200-250 | 250-300 |
| \(f_i\) | 17 | 35 | 43 | 40 | 21 | 24 |
Solution : Let assumed mean a = 175
| Class Interval | mid value \((x_i)\) | \(d_i\) = \(x_i – 175\) | frequency \(f_i\) | \(f_id_i\) |
| 0-50 | 25 | -150 | 17 | -2550 |
| 50-100 | 75 | -100 | 35 | -3500 |
| 100-150 | 125 | -50 | 43 | -2150 |
| 150-200 | 175 | 0 | 40 | 0 |
| 200-250 | 225 | 50 | 21 | 1050 |
| 250-300 | 275 | 100 | 24 | 2400 |
| \(\sum f_i\) = 180 | \(\sum f_id_i\) = -4750 |
Now, a = 175 and N = \(\sum f_i\) = 180
\(\therefore\) \(\bar{x}\) = a + (\(\sum f_id_i\over N\)) = 175 + \((-4750)\over 180\) = 175 – 26.39 = 148.61
Mean By Step Deviation Method
Sometime during the application of short method (given above) of finding the A.M. If each deviation \(d_i\) are divisible by a common number h(let)
Let \(u_i\) = \(d_i\over h\) = \(x_i – a\over h\), where a is assumed mean.
\(\therefore\) \(\bar{x}\) = 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 + (\(\sum f_iu_i\over N\))h = 35 + \((-70)\over 100\)\(\times\)10 = 28
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.
