Averages Relationships
In any distribution when the original items differ in size, then the value of the A.M, G.M and H.M would also differ and will be in the following order:
A.M ≥ G.M ≥ H.M
The Arithmetic mean is greater than geometric mean and the geometric mean is greater than harmonic mean. The equality signs hold only if all the number X1 X2 .............Xn are identical
Proof: prove that it if a and b are positive number then their AM ≥ GM ≥ HM.
Solution: let a and b are two positive quantities such that a # b.
Then A.M. and H.M of these two quantities are
X = (a +b)/2: G.M. = √a x√ b; H.M. = {1 / 1 / (a + 1) / b} = {2ab / (a + b)}
We have to prove that A.M > G.M. > H.M. let us first prove that A.M. > G.M. or (a + b)/2 > √a x b
(a + b)/ 2(√a x b) or (a + b) >2 √ab
(a + b) - 2√AB > 0 [since (a +b) - 2√ab = (√a - √b) 2]
(√a - √b) 2 > 0,
As the square of any real quantity is positive hence (√a - √b)2 will always be positive. And hence (a + b)/2 > √ab
Let us now prove that GM ≥ HM
Or √ab > 2ab / a + b 1 > √ab/ (a + b) or a + b > 2√ab.
This has already been proved above and hence G.M > H.M
Since we show that G.M> H.M , Therefore it is automatically proved that A.M > G.M > H.M
If a and b are equal in that case AM = G.M = H.M thus
A.M . ≥ G.M ≥ N.M
Illustration:
Calculate the arithmetic mean, median and mode from the following frequency distribution
| Variable |
frequency |
variable |
frequency |
| 10-13 |
8 |
25-28 |
54 |
| 13-16 |
15 |
28-31 |
36 |
| 16-19 |
27 |
31-34 |
18 |
| 19-22 |
51 |
34-37 |
9 |
| 22-25 |
75 |
37-40 |
7 |
Solution: Calculation of mean median & mode
| Variable |
m. p m |
(m-23.5)/3 |
F |
Fd |
c.f |
| 10-13 |
11.5 |
-4 |
8 |
-32 |
8 |
| 13-16 |
14.5 |
-3 |
15 |
-45 |
23 |
| 16-19 |
17.5 |
-2 |
24 |
-54 |
50 |
| 19-22 |
20.5 |
-1 |
51 |
-51 |
101 |
| 22-25 |
23.5 |
0 |
75 |
0 |
176 |
| 25-28 |
26.5 |
+1 |
54 |
+54 |
230 |
| 28-31 |
29.5 |
+2 |
36 |
+72 |
266 |
| 31-34 |
32.5 |
+3 |
18 |
+54 |
284 |
| 34-37 |
35.5 |
+4 |
9 |
+36 |
293 |
| 37-40 |
38.5 |
+5 |
7 |
+35 |
300 |
| |
|
|
N = 300 |
Σƒd = 69 |
|
X = A + Σƒd / N x I = 23.5 + 69 / 300 x 3 = 24. 19
Med = size of N2th item = 300/2 = 150th item
Median lies in the class 22-25
Med = L + N/2 - c.f/f x I = 22 + 150 - 101 / 75 x 3 = 22 + 1.96 = 23.96
The Mode by inspection mode lies in the class 22-25.
Mo = L+ Δ1 / Δ1 + Δ2 x i
L = 22, Δ1 = 75 - 51 = 24 Δ2 = 75 - 54 = 21, I = 3
Mo = 22 + 24 / 24 + 21 x 3 = 22 + 1.6 + 23.6
X = A + Σƒd / N x I = 23.5 + 69 / 300 x 3 = 24. 19
Med = size of N2th item = 300/2 = 150th item
Median lies in the class 22-25
Med = L + N/2 - c.f/f x I = 22 + 150 - 101 / 75 x 3 = 22 + 1.96 = 23.96
The Mode by inspection mode lies in the class 22-25.
Mo = L+ Δ1 / Δ1 + Δ2 x i
L = 22, Δ1 = 75 - 51 = 24 Δ2 = 75 - 54 = 21, I = 3
Mo = 22 + 24 / 24 + 21 x 3 = 22 + 1.6 + 23.6
X = A + Σƒd / N x I = 23.5 + 69 / 300 x 3 = 24. 19
Med = size of N2th item = 300/2 = 150th item
The Median lies in the class 22-25
Med = L + N/2 - c.f/f x I = 22 + 150 - 101 / 75 x 3 = 22 + 1.96 = 23.96
The Mode by inspection mode lies in the class 22-25.
Mo = L+ Δ1 / Δ1 + Δ2 x i
L = 22, Δ1 = 75 - 51 = 24 Δ2 = 75 - 54 = 21, I = 3
Mo = 22 + 24 / 24 + 21 x 3 = 22 + 1.6 + 23.6