M
l
+
× h
–
c.f
n
2
E
M
E
=
Classes must be continuous.
D
A
I
N
I
A
N
D
f
l = Lower limit of median class
n = Total frequency
c.f. = Cumulative frequency of class preceding the median class
f = Frequency of median class
h = Width of the class
Steps :
Median =
l
+
× h
–
c.f
n
2
f
Total frequency (N)
2
Find the value of
(i)
(ii) Check the ‘Cumulative frequency less than type’
coloumn for the value x
(iii) The cumulative frequency less than type which is
just greater than or equal to x. The corresponding
class is the median class.
= x
14.3(5)The following table gives the distribution of the life time of 400 neon lamps :
Find the median life time of a lamp.
Life time (hours) | 1500 - 2000 | 2000 - 2500 | 2500 – 3000 | 3000 - 3500 | 3500 - 4000 | 4000 - 4500 | 4500 - 5000 |
No. of lamps | 14 | 56 | 60 | 86 | 74 | 62 | 48 |
Sol:
| | |
| | |
| | |
| | |
| | |
| | |
| | |
| | |
| | |
Class
Frequency
c. f.
2000 - 2500
2500 – 3000
3500 - 4000
1500 - 2000
4000 - 4500
4500 - 5000
14
56
60
74
62
48
14
70
130
216
290
352
400
Changing the data in continuous form
14
+
56
=
70
70
+
60
=
130
=
400
n
Total
Here
n
2
=
400
2
=
200
86
f
←
3000 – 3500
c. f.
←
∴
Median class is 3000-3500.
l
=
3000 ,
h
=
500 ,
f
=
86 ,
c.f.
=
130
Median
=
n
2
-
c. f.
f
×
h
l
+
=
3000
+
200
-
130
86
=
3000
+
17500
43
=
3000
+
406.98
∴
Median
=
3406.98
hrs.
Which c.f is greater than
or equal to 200
Which lies in the class 3000 - 3500
Exercise 14.3 – Q.5