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 xi 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 di = xi – a
∴ ˉx = a + ∑fidiN, 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 |
fi | 17 | 35 | 43 | 40 | 21 | 24 |
Solution : Let assumed mean a = 175
Class Interval | mid value (xi) | di = xi–175 | frequency fi | fidi |
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 |
∑fi = 180 | ∑fidi = -4750 |
Now, a = 175 and N = ∑fi = 180
∴ ˉx = a + (∑fidiN) = 175 + (−4750)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 di are divisible by a common number h(let)
Let ui = dih = xi–ah, where a is assumed mean.
∴ ˉx = a + (∑fiuiN)h
Example : Find the A.M. of the following freq. dist.
xi | 5 | 15 | 25 | 35 | 45 | 55 |
fi | 12 | 18 | 27 | 20 | 17 | 6 |
Solution : Let assumed mean a = 35, h = 10
here N = ∑fi = 100, ui = xi–3510
∑fiui = (12×-3) + (18×-2) + (27×-1) + (20×0) + (17×1) + (6×2)
= -70
∴ ˉx = a + (∑fiuiN)h = 35 + (−70)100×10 = 28
Properties of Arithmetic Mean
(i) Sum of deviations of variate from their A.M. is always zero i.e. ∑(x1–ˉx) = 0, ∑fi(x1–ˉx) = 0
(ii) Sum of square of deviations of variate from their A.M. is minimum i.e. ∑(x1–ˉx)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.