l
+
× h
f1 – f0
2f1 – f0 – f2
Classes must be continuous
l = Lower limit of modal class
l
+
× h
f1 – f0
2f1 – f0 – f2
Mode =
f1 = Maximum frequency
f0 = Frequency of class preceding modal class
f2 = Frequency of class succeeding the modal class
h = Width of the class
The class having a maximum frequency is termed as Modal class.
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Determine the modal lifetimes of the components
Ex.14.2) 2) The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:
Lifetime ( in hours) | 0 – 20 | 20 - 40 | 40 - 60 | 60 - 80 | 80 - 100 | 100 -120 |
frequency | 10 | 35 | 52 | 61 | 38 | 29 |
Lifetime ( in hours)
frequency
Sol:
Class width(h) =
20
frequency
( fi )
20 - 40
40 - 60
60 - 80
80 - 100
class interval
10
35
52
61
38
For finding mode classes must be continuous
0 - 20
Class width (h) is found by subtracting two consecutive lower limits or two consecutive upper limits
Maximum frequency is 61
100 - 120
29
Determine the modal lifetimes of the components
The corresponding class 60 – 80 is modal class.
What is the Maximum frequency?
f1
f0
f2
Frequency of the class
Preceeding the Modal class
Frequency of the class
succeeding the Modal class
∴
l
=
60,
h
=
20,
f1
=
61,
f0
=
52,
f2
38,
=
Mode
=
l
+
2f1
×
h
- f0
- f2
f1
- f0
∴
Mode
=
60
+
2(61)
×
20
- 52
- 38
61
- 52
∴
Mode
=
60
+
122
×
20
- 90
9
Mode
=
60
+
32
180
=
65.625
∴
Mode
=
∴
65.625 hours
Modal
Class
5.625
Frequency of the modal class
Exercise 14.2 – Q.2