Agriculture Reference
In-Depth Information
To meet the other charges and to have some
profit by selling the same product, he/she should
sell the tea mixture at more than Rs 228.57 per
kilogram of tea.
4. Use the above values in the formula for
median.
5. Median will be having the same unit as that of
the variable.
Example 8.11.
Thenumberofinsectsperplant
for the 10 plants of a particular variety of rose
is given as follows: 17, 22, 21, 13, 29, 23, 15,
16, 25, 27. The objective is to find out the
median value of the variable number of insect
per plant.
8.2.1.2 Median
The median divides the whole set of data into two
equal halves; below and above the median, there
are equal numbers of observations. The median of a
set of observations is defined as the value of the
middle-most observation when the observations are
arranged either in ascending or descending order.
Solution. Now if we arrange the data in ascend-
ing order, then it will become 13, 15, 16, 17, 21,
22, 23, 25, 27, and 29. The middle-most
observations are the 5th and 6th observations;
hence, the median value for the above data set
is 21.5, that is, (21 + 22)/2
Median
<
:
n þ
1
2
th observation when
n
is odd
h
th observation
i
¼
21.5.
1
2
2
th observation
þ
2
þ
1
¼
Example 8.12.
Find the median value of yield
(q/ha) from the following frequency table.
when
n
is even
:
Of course, at first, the observations should be
arranged in increasing or decreasing order. Then,
the median value is to be worked out.
For grouped data, the median is calculated as
Yield
classes
Frequency
(
Mid
value (
Cumulative
frequency (CF
f
i
)
x
i
)
<
)
10-20
16
15
16
20-30
54
25
70
30-40
14
35
84
¼ x
l
þ
N=
2
F
me
1
Me
:
CI
;
40-50
11
45
95
f
me
50-60
10
55
105
60-70
10
65
115
where
x
l
is the lower class boundary of the
median class,
N
70-80
8
75
123
80-90
7
85
130
F
me
1
is the cumula-
tive frequency (less than type) of the class pre-
ceding the median class,
is the total frequency,
1. The total number of observation is 130;
N
/
f
me
is the frequency of
the median class, and CI is the width of the
median class.
65.
2. The median class is 20-30.
3. The lower boundary (
2
¼
x
l
) is 20, the class width
(CI) is 10, and the frequency (
Steps in Calculating the Median:
1. Identify the median class, that is, the class
having the
f
me
) is 54 of the
median class.
4. The cumulative frequency (less than the type)
of
N
/2th observation from the cumu-
lative frequency (less than the type) column.
2. Identify the lower class boundary (
x
l
), the
class width (CI), and the frequency (
the class preceding the median class
F
me
1
) is 16.
So the median of the above frequency distri-
bution is
(
f
me
)
values of the median class.
3. Identify the cumulative frequency (less than
the type ) of the class preceding the median
class (
65
16
Me
¼
20
þ
10
¼
29
:
074 q
=
ha
:
54
F
me
1
).
