Agriculture Reference
In-Depth Information
1
=n
Y
Solution. There are eight numbers of
observations. If we denote the geometric mean
by
1
=n
X
g
¼ðx
1
:x
2
...x
n
Þ
¼
x
i
Taking logarithm of both the sides, we have
G
, then
G ¼
(20
24
28
32
36
48)
1/8
40
1/8[Log(20) +
Log(24) + Log(28) + Log(32) + Log(36) + Log
(40) + Log(44) + Log(48)]
44
¼ >
Log(
G
)
¼
log
Y
1
n
log
ðX
g
Þ¼
x
i
1
n
:
x
2
þ
...
þ
¼
½
log
x
1
þ
log
log
x
n
x
Log(
x
)
20
1.3010
X
n
1
n
24
1.3802
¼
log
x
i
¼ Að
say
Þ
28
1.4472
i¼
1
32
1.5051
so,
ðAÞ:
Thus, the logarithm of a geometric mean is the
arithmetic mean of logarithm of the observations.
For grouped/classified data on variable
X
g
¼
Antilog
ð
log
x
g
Þ¼
Antilog
36
1.5563
40
1.6021
44
1.6435
48
1.6812
X
hav-
ing
x
n
as class mid values with the
respective frequencies of
x
1
,
x
2
,
...
,
Thus, log(
G
)
¼
1/8 (12.1166)
¼
1.5146
f
1,
f
2,
f
3,
...
,
f
n
,
the
So,
the antilog of
log(
G
)
¼ G ¼
antilog
geometric mean is given by
(1.5146)
32.7039. So the geometric mean of
the given numbers is 32.7039.
¼
1
=
P
n
1
=
P
n
Y
i¼
1
f
i
i¼
1
f
i
X
g
¼ x
f
1
:x
f
2
...x
f
n
x
i
f
i
¼
:
Example 8.5.
Find the geometric mean from the
following frequency distribution.
For grouped frequency data,
x
i
is taken as the
mid value of the
i
th class.
With the help of a log conversion or scientific
calculator, one can easily find out the geometric
mean.
Variable values 15 20 20 25 30
Frequency 3 4 3 4 6
Solution. This is a simple frequency dist-
ribution and we have the geometric mean
1
=
P
n
i¼
Example 8.4.
Find the geometric mean of the
following observations: 20,24,28,32,36,40,44,
and 48.
1
f
i
. Taking the logarithm
of both sides, we have Log(
X
g
¼ x
f
1
:x
f
2
...x
f
n
x
g
)
!
log
Y
n
1
P
n
i¼
1
f
i
1
20
3 log
i¼
1
x
1
fi
¼
½
ð
15
Þþ
4 log
ð
20
Þþ
3 log
ð
20
Þþ
4 log
ð
25
Þþ
6 log
ð
30
Þ
1
20
3
¼
½
1
:
1760
þ
4
1
:
3010
þ
3
1
:
3010
þ
4
1
:
3979
þ
6
1
:
4771
Þ
1
20
27
¼
½
:
0892
¼
1
:
35446
:
X
g
¼
Alog 1
ð
:
35446
Þ ¼
22
:
618
:
Thus, the geometric mean of the above simple
frequency distribution is 22.618.
Example 8.6.
Find the geometric mean from the
following frequency distribution of the stem borer.