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Ex14.3-(1)The following frequency distribution gives the monthly consumption of
electricity of 68 consumers of a locality. Find the median, mean and mode
of the data and compare them.
Monthly consumption (in units) | No. of consumers of electricity |
65 - 85 | 4 |
85 - 105 | 5 |
105 - 125 | 13 |
125 - 145 | 20 |
145 - 165 | 14 |
165 - 185 | 8 |
185 - 205 | 4 |
Soln.
class
65 - 85
85 - 105
105 - 125
145 - 165
165 - 185
185 - 205
4
5
13
14
8
4
4
9
56
64
68
c.f
Here
n
2
=
68
2
= 34
which lies in the class 125 - 145
∴ Median class is 125 -145
l = 125
Median =
l +
n
2
- c.f.
f
x h
= 125 +
34 - 22
20
x 20
= 125 + 12
∴ Median = 137 units.
Find the median, mean and mode
of the data and compare them
4 + 5 = 9
9 + 13 = 22
Monthly consumption (in units)
No. of consumers of electricity
frequency
42
, h = 20,
f = 20 ,
c.f. =22,
125 - 145
20
f1
22
Exercise 14.3 – Q.1
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Determination of mean and mode
h =
class
Class Mark
xi
Frequency
fi
ui =
xi -
20
fiui
65 - 85
85 - 105
105 - 125
125 - 145
145 - 165
165 - 185
185 - 205
75
95
115
135
155
175
195
4
5
13
20
14
8
4
-3
-2
-1
0
1
2
3
-12
-10
-13
0
14
∑fi =
∑fiui =
Let assumed mean, a = 135.
20
a
a
135
16
12
68
7
By step deviation method,
x
= a +
∑fiui
∑fi
X h
= 135+
7
68
X 20
= 135 +
35
17
= 135 + 2.05
∴ Mean = 137.05 units
5
17
2.05
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class
Class Mark
xi
Frequency
fi
ui =
xi -
20
fiui
65 - 85
85 - 105
105 - 125
145 - 165
165 - 185
185 - 205
75
95
115
135
155
175
195
4
5
13
20
14
8
4
-3
-2
-1
0
1
2
3
-12
-10
-13
0
14
∑fi =
∑fiui =
f0
135
16
12
f1
f2
68
7
Maximum frequency is 20
∴ Modal class is 125 - 145
which lies in the class 125 - 145
What is the Maximum frequency?
Frequency of the class
Preceeding the Modal class
Frequency of the class
succeeding the Modal class
125 - 145
l =125, h = 20, f1 = 20, f0 = 13,f2 = 14
∴ Mode =
l +
f1 – f0
2f1 – f0 – f2
x h
= 125 +
20 – 13
2(20) – 13 - 14
x 20
= 125 +
7
40 - 27
x 20
= 125 +
140
13
= 125 + 10.76
∴ Mode = 135.76 units
10.76