 Lesson: Statistics
Exercise 14.1 (9)
Question: 1
A survey was conducted by a group of students as a
part of their environment awareness programme, in
which they collected the following data regarding
the number of plants in 20 houses in a locality. Find
the mean number of plants per house.
Number
of plants
0 2
2 4
4 6
6 8
8 10
10 12
12 14
Number
of houses
1
2
1
5
2
3
Which method did you use for finding the mean,
and why?
Solution:
First, find the class mark for each interval. For
i
x
this, the following relation should be used.
i
Upper limit Lower limit
x
2
Now, calculate and .
i
x
i i
x f Number of
plants
Number of
houses ( )
i
f
i
x
i i
x f
0 2
1
1
1
2 4
2
3
6
4 6
1
5
5
6 8
5
7
35
8 10
6
9
54
10 12
2
11
22
12 14
3
13
39
Total
20
162
From the table,
and
i
f 20
i i
f x 162
Now, find the value of mean.
Mean
i i
i
f x
x
f
162
20
8.1
Therefore, mean number of plants per house is 8.1. As the values of class marks and are small,
i
x
i
f
so direct method has been used for finding the
mean.
Question: 2
Consider the following distribution of daily wages
of 50 workers of a factory.
Daily wages
(in Rs)
100 120
120 140
140 160
160 180
180 200
Number of
workers
12
14
8
6
10
Find the mean daily wages of the workers of the
factory by using an appropriate method.
Solution:
First, find the class mark for each interval. For
i
x
this, the following relation should be used.
i
Upper limit Lower limit
x
2
Class size
h 20
Take assumed mean and calculate and
a 150
i
d ,
i
u
as follows.
i i
fu Daily wages
(in Rs)
Number of
workers
i
f
i
x
i i
d x 150
i
i
d
u ,
h
h 20
i i
fu
100 120
12
110
40
2
24
120 140
14
130
20
1
14
140 160
8
150
0
0
0
160 180
6
170
20
1
6
180 200
10
190
40
2
20
Total
50
12
From the table,
and
i
f 50
i i
fu 12
So, mean is given by,
x
i i
i
fu
x a h
f
12
150 20
50
12
150 2
5
24
150
5
150 4.8
145.20 Therefore, Rs 145.20 is the mean daily wage of the
workers of the factory.
Question: 3
The following distribution shows the daily pocket
allowance of children of a locality. The mean pocket
allowance is Rs 18. Find the missing frequency .
f
Daily
pocket
allowance
11 13
13 15
15 17
17 19
19 21
21 23
23 25
Number
of
children
7
6
9
13
f
5
4
Solution:
Given: Mean pocket allowance
Rs18
Find the class mark for each interval. For this,
i
x
the following relation should be used.
i
Upper limit Lower limit
x
2
Take assumed mean and calculate and
a 18
i
d
i i
fd
as follows.
Daily pocket
allowance (in Rs)
Number of
children
i
f
Class
mark
xi
i i
d x 18
i i
fd 11 13
7
12
6
42
13 15
6
14
4
24
15 17
9
16
2
18
17 19
13
18
0
0
19 21
f
20
2
2f
21 23
5
22
4
20
23 25
4
24
6
24
Total
i
f 44 f
2f 40
From the table,
and
i
f 44 f
i i
fd 2f 40
So, mean is given by,
x
i i
i
fd
x a
f
2f 40
18 18
44 f
2f 40
18 18
44 f
2f 40
0
44 f
2f 40 0
2f 40
f 20 Hence, the missing frequency is .
f
20
Question: 4
Thirty women were examined in a hospital by a
doctor and the number of heart beats per minute
were recorded and summarised as follows. Find the
mean heart beats per minute for these women,
choosing a suitable method.
Number of
heart beats
per minute
65 68
68 71
71 74
74 77
77 80
80 83
83 86
Number of
women
2
4
3
8
7
4
2
Solution:
Find the class mark for each interval. For this,
i
x
the following relation should be used.
i
Upper limit Lower limit
x
2
Class size
h 3
Take assumed mean and calculate
a 75.5
i
d ,
i
u
and as follows.
i i
fu Number of
heart beats
per minute
Number of
women
i
f
i
x
i i
d x 75.5
i
i
d
u ,h 3
h
i i
fu
65 68
2
66.5
9
3
6
68 71
4
69.5
6
2
8
71 74
3
72.5
3
1
3
74 77
8
75.5
0
0
0
77 80
7
78.5
3
1
7
80 83
4
81.5
6
2
8
83 86
2
84.5
9
3
6
Total
30
4
From the table,
and
i
f 30
i i
fu 4
So, mean is given by,
x
i i
i
fu
x a h
f
4
75.5 3
30
4
75.5
10
75.5 0.4
75.9 Hence, mean heart beats per minute for these
women are 75.9 beats per minute.
Question: 5
In a retail market, fruit vendors were selling
mangoes kept in packing boxes. These boxes
contained varying number of mangoes. The
following was the distribution of mangoes
according to the number of boxes.
Number of
mangoes
50 52
53 55
56 58
59 61
62 64
Number of
boxes
15
110
135
115
25
Find the mean number of mangoes kept in a
packing box. Which method of finding the mean
did you choose?
Solution:
Given table:
Number of mangoes
Number of boxes
50 52
15
53 55
110
56 58
135 59 61
115
62 64
25
The class intervals are not continuous, as there is a
gap of between two class intervals.
1
So, must be added to the upper-class limit and
0.5
must be subtracted from the lower-class limit of
0.5
each interval.
Then, find the class mark for each interval. For
i
x
this, the following relation should be used.
i
Upper limit Lower limit
x
2
Class size
h 3
Take assumed mean and calculate and
a 57
i
d ,
i
u
as follows.
i i
fu
Number of
mangoes
Number
of boxes
i
f
i
x
i i
d x 57
i
i
d
u ,h 3
h
i i
fu
49.5 – 52.5
15
51
6
2
30
52.5 – 55.5
110
54
3
1
110
55.5 – 58.5
135
57
0
0
0
58.5 – 61.5
115
60
3
1
115 61.5 – 64.5
25
63
6
2
50
Total
400
25
From the table,
and
i
f 400
i i
fu 25
Now, find the mean .
x
i i
i
fu
x a h
f
25
57 3
400
1
57 3
16
3
57
16
57 0.1875
57.1875
57.19
Hence, the mean number of mangoes kept in a
packing box is 57.19.
The method used here is the step deviation method,
as the values of and are large and, also there is a
i
f
i
d
common multiple between all i.e. .
i
d Question: 6
The table below shows the daily expenditure on
food of 25 households in a locality.
Daily expenditure
(in Rs)
100 – 150
150 – 200
200 – 250
250 – 300
300 – 350
Number of
households
4
5
12
2
2
Find the mean daily expenditure on food by a
suitable method.
Solution:
First, find the class mark for each interval. For
i
x
this, the following relation should be used.
i
Upper limit Lower limit
x
2
Class size
h 50
Take assumed mean and calculate and
a 225
i
d ,
i
u
as follows.
i i
fu
Daily
expenditure
(in Rs)
Number of
household
s
i
f
i
x
i i
d x 225
i
i
d
u ,
h
h 50
i i
fu
100 – 150
4
125
100
2
8 150 – 200
5
175
50
1
5
200 – 250
12
225
0
0
0
250 – 300
2
275
50
1
2
300 – 350
2
325
100
2
4
Total
25
7
From the table,
and
i
f 25
i i
fu 7
Now, find the mean
x
i i
i
fu
x a h
f
7
225 50
25
225 7 2
225 14
211
Hence, mean daily expenditure on food is Rs 211.
Question: 7
To find out the concentration of in the air (in
2
SO
parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented
below:
Concentration of (in ppm)
2
SO
Frequency
0.00 – 0.04
0.04 – 0.08
0.08 – 0.12
0.12 – 0.16
0.16 – 0.20
0.20 – 0.24
4
9
9
2
4
2
Find the mean concentration of in the air.
2
SO
Solution:
First, find the class mark for each interval. For
i
x
this, the following relation should be used.
i
Upper limit Lower limit
x
2
Class size
h 0.04
Take assumed mean and calculate
a 0.14
i
d ,
i
u
and as follows.
i i
fu
Concentra
tion of
(in
2
SO
ppm)
Frequency
i
f
i
x
i i
d x 0.14
i
i
d
u ,
h
h 0.04
i i
fu 0.00 – 0.04
4
0.02
0.12
3
12
0.04 – 0.08
9
0.06
0.08
2
18
0.08 – 0.12
9
0.10
0.04
1
9
0.12 – 0.16
2
0.14
0
0
0
0.16 – 0.20
4
0.18
0.04
1
4
0.20 – 0.24
2
0.22
0.08
2
4
Total
30
31
From the table,
and
i
f 30
i i
fu 31
Now, find the mean .
x
i i
i
fu
x a h
f
31
0.14 0.04
30
0.14 0.04133
0.09867
0.099ppm
Hence, mean concentration of in the air is 0.099
2
SO
ppm.
Question: 8 A class teacher has the following absentee record of
40 students of a class for the whole term. Find the
mean number of days a student was absent.
Number of
days
0 – 6
6 – 10
10 – 14
14 – 20
20 – 28
28 – 38
38 – 40
Number of
students
11
10
7
4
4
3
1
Solution:
First, find the class mark for each interval. For
i
x
this, the following relation should be used.
i
Upper limit Lower limit
x
2
Take assumed mean and calculate and
a 17
i
d
i i
fd
as follows.
Number of
days
Number of
students
i
f
i
x
i i
d x 17
i i
fd
0 – 6
11
3
14
154
6 – 10
10
8
9
90
10 – 14
7
12
5
35
14 – 20
4
17
0
0
20 – 28
4
24
7
28
28 – 38
3
33
16
48 38 – 40
1
39
22
22
Total
40
181
From the table,
and b
i
f 40
i i
fu 181
Now, find the mean .
x
i i
i
fd
x a
f
181
17
40
17 4.525
12.475
12.48
Hence, the mean number of days for which a
student was absent are 12.48.
Question: 9
The following table gives the literacy rate (in
percentage) of 35 cities. Find the mean literacy rate.
Literacy
rate (in %)
45 – 55
55 – 65
65 – 75
75 – 85
85 – 95
Number of
3
10
11
8
3 cities
Solution:
First, find the class mark for each interval. For
i
x
this, the following relation should be used.
i
Upper limit Lower limit
x
2
Class size
h 10
Take assumed mean and calculate and
a 70
i
d ,
i
u
as follows.
i i
fu
Literacy
rate (in %)
Number
of cities
i
f
i
x
i i
d x 70
i
i
d
u ,
h
h 10
i i
fu
45 – 55
3
50
20
2
6
55 – 65
10
60
10
1
10
65 – 75
11
70
0
0
0
75 – 85
8
80
10
1
8
85 – 95
3
90
20
2
6
Total
35
2
From the table,
and
i
f 35
i i
fu 2 Now, find the mean .
x
i i
i
fu
x a h
f
2
70 10
35
20
70
35
70 0.57
69.43
Hence, mean literacy rate is 69.43%.
Exercise 14.2 (6)
Question: 1
The following table shows the ages of the patients
admitted in a hospital during a year:
Age (in
years)
5 – 15
15 – 25
25 – 35
35 – 45
45 – 55
55 – 65
Number
of patients
6
11
21
23
14
5 Find the mode and the mean of the data given
above. Compare and interpret the two measures of
central tendency.
Solution:
First, find the class mark for each interval. For
i
x
this, the following relation should be used.
i
Upper limit Lower limit
x
2
Take assumed mean and calculate and
a 30
i
d
i i
fd
as follows.
Age (in
years)
Number of
patients
i
f
i
x
i i
d x 30
i i
fd
5 – 15
6
10
20
120
15 – 25
11
20
10
110
25 – 35
21
30
0
0
35 – 45
23
40
10
230
45 – 55
14
50
20
280
55 – 65
5
60
30
150
Total
80
430
From the table,
and
i
f 80
i i
fu 430 Now, find the mean .
x
i i
i
fd
x a
f
430
30
80
30 5.375
35.375
35.38
So, mean of the given data is 35.38.
This represents that, on an average, the age of a
patient admitted to hospital was years.
35.38
Now, the maximum class frequency is 23 belonging
to class interval .
35 40
Modal class
35 40
Lower limit of modal class
l
35
Frequency of modal class
1
f
23
Class size
h 10
Frequency of class preceding the modal class
0
f
21
Frequency of class succeeding the modal class
2
f
14
Therefore, mode is given by, 1 0
1 0 2
f f
Mode l h
2f f f
23 21
35 10
2 23 21 14
2
35 10
46 35
20
35
11
35 1.81
36.8
Hence, mode is 36.8 and it represents the age of
maximum number of patients admitted in hospital
i.e. 36.8 years.
Question: 2
The following data gives the information on the
observed lifetimes (in hours) of 225 electrical
components:
hours)
0 – 20
20 – 40
40 – 60
60 – 80
80 – 100
100 – 120
Frequency
10
35
52
61
38
29
Determine the modal lifetimes of the components.
Solution: The maximum class frequency is 61 belonging to
class interval .
60 80
Lower limit of modal class
l
60
Frequency of modal class
1
f
61
Class size
h 20
Frequency of class preceding the modal class
0
f
52
Frequency of class succeeding the modal class
2
f
38
Therefore, mode is given by,
1 0
1 0 2
f f
Mode l h
2f f f
61 52
60 20
2 61 52 38
9
60 20
122 90
9 20
60
32
180
60
32
60 5.625
65.625 Hence, modal lifetime of electrical components is
65.625 hours.
Question: 3
The following data gives the distribution of total
monthly household expenditure of 200 families of a
village. Find the modal monthly expenditure of the
families. Also, find the mean monthly expenditure:
Expenditure (in Rs)
Number of families
1000 – 1500
1500 – 2000
2000 – 2500
2500 – 3000
3000 – 3500
3500 – 4000
4000 – 4500
4500 – 5000
24
40
33
28
30
22
16
7
Solution:
The maximum class frequency is 40 belonging to
class interval .
1500 2000
Lower limit of modal class
l
1500
Frequency of modal class
1
f
40 Class size
h 500
Frequency of class preceding the modal class
0
f
24
Frequency of class succeeding the modal class
2
f
33
Now, find the mode.
1 0
1 0 2
f f
Mode l h
2f f f
40 24
1500 500
2 40 24 33
16
1500 500
80 57
16 500
1500
23
8000
1500
23
1500 347.826
1847.826
1847.83
So, modal monthly expenditure was Rs 1847.83.
Now, find the class mark for each interval. For
i
x
this, the following relation should be used.
i
Upper limit Lower limit
x
2 Take assumed mean and calculate
a 2750
i
d ,
i
u
and as follows.
i i
fu
Expenditur
e (in Rs)
Number
of
families
i
f
i
x
i i
d x 2750
i
i
d
u ,
h
h 500
i i
fu
1000 – 1500
24
1250
1500
3
72
1500 – 2000
40
1750
1000
2
80
2000 – 2500
33
2250
500
1
33
2500 – 3000
28
2750
0
0
0
3000 – 3500
30
3250
500
1
30
3500 – 4000
22
3750
1000
2
44
4000 – 4500
16
4250
1500
3
48
4500 – 5000
7
4750
2000
4
28
Total
200
35
From the table,
and
i
f 200
i i
fu 35
Now, find the mean .
x
i i
i
fu
x a h
f 35
2750 500
200
35 5
2750
2
175
2750
2
2750 87.5
2662.5
Hence, mean monthly expenditure is Rs 2662.50.
Question: 4
The following distribution gives the state-wise
teacher-student ratio in higher secondary schools of
India. Find the mode and mean of this data.
Interpret the two measures.
Number of students
per teacher
Number of
states/U.T.
15 – 20
20 – 25
25 – 30
30 – 35
35 – 40
40 – 45
3
8
9
10
3
0 45 – 50
50 – 55
0
2
Solution:
The maximum class frequency is 10 belonging to
class interval .
30 35
Lower limit of modal class
l
30
Frequency of modal class
1
f
10
Class size
h 5
Frequency of class preceding the modal class
0
f
9
Frequency of class succeeding the modal class
2
f
3
Now, find the mode.
1 0
1 0 2
f f
Mode l h
2f f f
10 9
30 5
2 10 9 3 1
30 5
20 12
5
30
8
30 0.625
30.6
So, mode represents that, most of the states/U.T.
have a teacher-student ratio as 30.6.
Now, find the class mark for each interval. For
i
x
this, the following relation should be used.
i
Upper limit Lower limit
x
2
Take assumed mean and calculate
a 32.5
i
d ,
i
u
and as follows.
i i
fu
Number of
students per
teacher
Number of
states/U.T.
i
f
i
x
i i
d x 32.5
i
i
d
u ,
h
h 5
i i
fu
15 – 20
3
17.5
15
3
9
20 – 25
8
22.5
10
2
16
25 – 30
9
27.5
5
1
9
30 – 35
10
32.5
0
0
0
35 – 40
3
37.5
5
1
3
40 – 45
0
42.5
10
2
0 45 – 50
0
47.5
15
3
0
50 – 55
2
52.5
20
4
8
Total
35
23
From the table,
and
i
f 35
i i
fu 23
Now, find the mean .
x
i i
i
fu
x a h
f
23
32.5 5
35
23
32.5
7
32.5 3.28
29.2
Hence, mean of data is Rs 29.2 which represents the
average teacher-student ratio.
Question: 5
The given distribution shows the number of runs
scored by some top batsmen of the world in one-
day international cricket matches. Runs scored
Number of batsmen
3000 – 4000
4000 – 5000
5000 – 6000
6000 – 7000
7000 – 8000
8000 – 9000
9000 – 10000
10000 – 11000
4
18
9
7
6
3
1
1
Find the mode of the data.
Solution:
The maximum class frequency is 18 belonging to
class interval .
4000 5000
Lower limit of modal class
l
4000
Frequency of modal class
1
f
18
Class size
h 1000
Frequency of class preceding the modal class
0
f
4
Frequency of class succeeding the modal class
2
f
9
Now, find the mode. 1 0
1 0 2
f f
Mode l h
2f f f
 
18 4
4000 1000
2 18 4 9
14
4000 1000
23
4000 608.695
4608.695
Hence, mode of the given data is 4608.695.
Question: 6
A student noted the number of cars passing
through a spot on a road for 100 periods each of 3
minutes and summarised it in the table given
below. Find the mode of the data:
Number
of cars
0 –
10
10 –
20
20 –
30
30 –
40
40 –
50
50 –
60
60 –
70
70 –
80
Frequency
7
14
13
12
20
11
15
8
Solution:
The maximum class frequency is 20 belonging to
class interval .
40 50 Lower limit of modal class
l
40
Frequency of modal class
1
f
20
Class size
h 10
Frequency of class preceding the modal class
0
f
12
Frequency of class succeeding the modal class
2
f
11
Now, find the mode.
1 0
1 0 2
f f
Mode l h
2f f f
20 12
40 10
2 20 12 11
8
40 10
17
80
40
17
40 4.7
44.7
Hence, mode of the given data is 44.7.
Exercise 14.3 (7) Question: 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)
Number of consumers
65 – 85
85 – 105
105 – 125
125 – 145
145 – 165
165 – 185
185 – 205
4
5
13
20
14
8
4
Solution:
First, find the class mark for each interval. For
i
x
this, the following relation should be used.
i
Upper limit Lower limit
x
2
Take assumed mean and calculate and
a 135
i
d ,
i
u
as follows.
i i
fu Monthly
consumptio
n (in units)
Number of
consumers
i
f
i
x
i i
d x 135
i
i
d
u ,
h
h 20
i i
fu
65 – 85
4
75
60
3
12
85 – 105
5
95
40
2
10
105 – 125
13
115
20
1
13
125 – 145
20
135
0
0
0
145 – 165
14
155
20
1
14
165 – 185
8
175
40
2
16
185 – 205
4
195
60
3
12
Total
68
7
From the table,
and
i
f 68
i i
fu 7
Now, find the mean .
x
i i
i
fu
x a h
f
7
135 20
68
140
135
68
135 2.058 137.058
So, mean of the given data is 137.058.
Now, the maximum class frequency is 20 belonging
to class interval .
125 145
Modal class
125 145
Lower limit of modal class
l
125
Frequency of modal class
1
f
20
Class size
h 20
Frequency of class preceding the modal class
0
f
13
Frequency of class succeeding the modal class
2
f
14
Therefore, mode is given by,
1 0
1 0 2
f f
Mode l h
2f f f
20 13
125 20
2 20 13 14
7
125 20
13
140
125
13
125 10.76
135.76 Hence, the mode is 135.76.
Now, find the median of the given data, and for
this, cumulative frequency is calculated as follows.
Monthly
consumption (in
units)
Number of
consumers
Cumulative
frequency
65 – 85
85 – 105
105 – 125
125 – 145
145 – 165
165 – 185
185 – 205
4
5
13
20
14
8
4
4
4 5 9
9 13 22
22 20 42
42 14 56
56 8 64
64 4 68
From the table,
n 68
Cumulative frequency just greater than
cf
is 42, belonging to interval .
n 68
34
2 2
125 145
Median class
125 145
Lower limit of median class
l
125
Class size
h 20
Frequency of median class
f
20 Cumulative frequency of class preceding
cf
median class
22
Now, find the median.
n
cf
2
Median l h
f
34 22
125 20
20
12
125 20
20
125 12
137
Therefore, median, mode and mean of the given
data are 137, 135.76 and 137.05 respectively.
The three quantities are approximately the same in
this case.
Question: 2
If the median of the distribution given below is 28.5,
find the values of and .
x
y Class interval
Frequency
0 – 10
10 – 20
20 – 30
30 – 40
40 – 50
50 – 60
5
x
20
15
y
5
Total
60
Solution:
Cumulative frequency is given by,
Class interval
Frequency
Cumulative
frequency
0 – 10
10 – 20
20 – 30
30 – 40
40 – 50
50 – 60
5
x
20
15
y
5
5
5 x
25 x
40 x
40 x y
45 x y
Total
n
60
From the table,
n 60
45 x y 60 x y 15 ......(1)
Median of the data is given as , belonging to
28.5
interval .
20 30
Median class
20 30
Lower limit of median class
l
20
Class size
h 10
Frequency of median class
f
20
Cumulative frequency of class preceding
cf
median class
5 x
Put the values in the formula of median.
n
cf
2
Median l h
f
60
5 x
2
28.5 20 10
20
30 5 x
28.5 20 10
20
25 x
8.5
2
17 25 x
x 8
Put the value of in equation .
x
1 8 y 15
y 7
Hence, .
x 8,
y 7
Question: 3
A life insurance agent found the following data for
distribution of ages of 100 policy holders. Calculate
the median age, if policies are given only to persons
having age 18 years onwards but less than 60 years.
Age (in years)
Number of policy holders
Below 20
Below 25
Below 30
Below 35
Below 40
Below 45
Below 50
Below 55
Below 60
2
6
24
45
78
89
92
98
100
Solution: It is not essential to adjust the frequencies
according to class intervals, as class width is not the
same.
According to the question, the frequency table is of
less than type represented with upper class limits.
The policies were given only to persons with age 18
years onwards but less than 60 years.
So, class intervals with their respective cumulative
frequency is given by,
Age (in
years)
Number of policy
holders
Cumulative
frequency
cf
18 – 20
2
2
20 – 25
6 2 4
2 4 6
25 – 30
24 6 18
6 18 24
30 – 35
45 24 21
24 21 45
35 – 40
78 45 33
45 33 78
40 – 45
89 78 11
78 11 89
45 – 50
92 89 3
89 3 92
50 – 55
98 92 6
92 6 98
55 – 60
100 98 2
98 2 100
From the table,
n 100 Cumulative frequency just greater than
cf
is 78, belonging to interval .
n 100
50
2 2
35 40
Median class
35 40
Lower limit of median class
l
35
Class size
h 5
Frequency of median class
f
33
Cumulative frequency of class preceding
cf
median class
45
Now, find the median.
n
cf
2
Median l h
f
50 45
35 5
33
5
35 5
33
25
35
33
35 0.76
35.76
Hence, the median age is 35.76 years. Question: 4
The lengths of 40 leaves of a plant are measured
correct to the nearest millimetre, and the data
obtained is represented in the following table:
Length (in mm)
Number of leaves
118 – 126
127 – 135
136 – 144
145 – 153
154 – 162
163 – 171
172 – 180
3
5
9
12
5
4
2
Find the median length of the leaves.
Solution:
The class intervals are not continuous, as there is a
gap of between two class intervals.
1
So, must be added to the upper-class limit and
0.5
must be subtracted from the lower-class limit of
0.5
each interval.
So, class intervals with their respective cumulative
frequency is given by,
Length (in mm)
Number of
Cumulative leaves
frequency
cf
117.5 – 126.5
3
3
126.5 – 135.5
5
3 5 8
135.5 – 144.5
9
8 9 17
144.5 – 153.5
12
17 12 29
153.5 – 162.5
5
29 5 34
162.5 – 171.5
4
34 4 38
171.5 – 180.5
2
38 2 40
From the table,
n 40
Cumulative frequency just greater than
cf
is 29, belonging to interval .
n 40
20
2 2
144.5 153.5
Median class
144.5 153.5
Lower limit of median class
l
144.5
Class size
h 9
Frequency of median class
f
12
Cumulative frequency of class preceding
cf
median class
17
Now, find the median. n
cf
2
Median l h
f
20 17
144.5 9
12
3
144.5 9
12
9
144.5
4
144.5 2.25
146.75
Hence, the median length of leaves is 146.75 mm.
Question: 5
The following table gives the distribution of the life
time of 400 neon lamps.
Life time (in hours)
Number of lamps 1500 – 2000
2000 – 2500
2500 – 3000
3000 – 3500
3500 – 4000
4000 – 4500
4500 – 5000
14
56
60
86
74
62
48
Find the median life time of a lamp.
Solution:
Class intervals with their respective cumulative
frequency is given by,
Life time
Number of
lamps
i
f
Cumulative
frequency
cf
1500 – 2000
14
14
2000 – 2500
56
14 56 70
2500 – 3000
60
70 60 130
3000 – 3500
86
130 86 216
3500 – 4000
74
216 74 290
4000 – 4500
62
290 62 352
4500 – 5000
48
352 48 400
Total
n
400 From the table,
n 400
Cumulative frequency just greater than
cf
is 216, belonging to interval
n 400
200
2 2
.
3000 3500
Median class
3000 3500
Lower limit of median class
l
3000
Class size
h 500
Frequency of median class
f
86
Cumulative frequency of class preceding
cf
median class
130
Now, find the median.
n
cf
2
Median l h
f
200 130
3000 500
86
70
3000 500
86
35000
3000
86
3000 406.976 3406.976
Hence, the median lifetime of lamps is
3406.98
hours.
Question: 6
100 surnames were randomly picked up from a local
telephone directory and the frequency distribution
of the number of letters in the English alphabets in
the surnames was obtained as follows:
Number of
letters
1 – 4
4 – 7
7 – 10
10 – 13
13 – 16
16 – 19
Number of
surnames
6
30
40
16
4
4
Determine the median number of letters in the
surnames. Find the mean number of letters in the
surnames? Also, find the modal size of the
surnames
Solution:
Class intervals with their respective cumulative
frequency is given by,
Number of
letters
Number of
surnames
i
f
Cumulative
frequency
cf 1 4
6
6
4 7
30
6 30 36
7 10
40
36 40 76
10 13
16
76 16 92
13 16
4
92 4 96
16 19
4
96 4 100
Total
n
100
From the table,
n 100
Cumulative frequency just greater than
cf
is 76, belonging to interval .
n 100
50
2 2
7 10
Median class
7 10
Lower limit of median class
l
7
Class size
h 3
Frequency of median class
f
40
Cumulative frequency of class preceding
cf
median class
36
Now, find the median. n
cf
2
Median l h
f
50 36
7 3
40
14
7 3
40
42
7
40
7 1.05
8.05
So, the median number of letters is .
8.05
Now, find the class mark for each interval. For
i
x
this, the following relation should be used.
i
Upper limit Lower limit
x
2
Take assumed mean and calculate
a 11.5
i
d ,
i
u
and as follows.
i i
fu
Number of
letters
Number of
surnames
i
f
i
x
i i
d x 11.5
i
i
d
u ,
h
h 3
i i
fu
1 4
6
2.5
9
3
18
4 7
30
5.5
6
2
60 7 10
40
8.5
3
1
40
10 13
16
11.5
0
0
0
13 16
4
14.5
3
1
4
16 19
4
17.5
6
2
8
Total
100
106
From the table,
i i i
f 100and fu 106
Now, find the mean .
x
i i
i
fu
x a h
f
106
11.5 3
100
318
11.5
100
11.5 3.18
8.32
So, mean of the given data is 8.32.
Now, the maximum class frequency is 40 belonging
to class interval .
7 10
Modal class
7 10
Lower limit of modal class
l
7 Frequency of modal class
1
f
40
Class size
h 3
Frequency of class preceding the modal class
0
f
30
Frequency of class succeeding the modal class
2
f
16
Now, find the mode.
1 0
1 0 2
f f
Mode l h
2f f f
40 30
7 3
2 40 30 16
10
7 3
34
30
7
34
7 0.88
7.88
Hence, median number of letters in surnames is
8.05, mean number of letters in surnames is 8.32,
and modal size of surnames is 7.88.
Question: 7 The distribution below gives the weights of 30
students of a class. Find the median weight of the
students.
Weight (in
kg)
40 –
45
45 –
50
50 –
55
55 –
60
60 –
65
65 –
70
70 –
75
Number of
students
2
3
8
6
6
3
2
Solution:
Class intervals with their respective cumulative
frequency is given by,
Number of
letters
Number of
surnames
i
f
Cumulative
frequency
cf
40 45
2
2
45 50
3
2 3 5
50 55
8
5 8 13
55 60
6
13 6 19
60 65
6
19 6 25
65 70
3
25 3 28
70 75
2
28 2 30
Total
n
30
From the table, n 30
Cumulative frequency just greater than
cf
is 19, belonging to interval .
n 30
15
2 2
55 60
Median class
55 60
Lower limit of median class
l
55
Class size
h 5
Frequency of median class
f
6
Cumulative frequency of class preceding
cf
median class
13
Now, find the median.
n
cf
2
Median l h
f
15 13
55 5
6
2
55 5
6
10
55
6
55 1.67
56.67
Hence, median weight is 56.7. Exercise 14.4 (3)
Question: 1
The following distribution gives the daily income of
50 workers of a factory.
Daily income
(in Rs)
100 – 120
120 – 140
140 – 160
160 – 180
180 – 200
Number of
workers
12
14
8
6
10
Convert the distribution above to a less than type
cumulative frequency distribution, and draw its
ogive.
Solution:
The frequency distribution table of less than type
along with the cumulative frequency is given by,
Daily income (in Rs)
Upper class limit
Cumulative frequency
Less than 120
12
Less than 140
12 14 26
Less than 160
26 8 34
Less than 180
34 6 40 Less than 200
40 10 50
Now, take upper class limits of class intervals on
and their respective frequencies on ,
x axis
y axis
draw its ogive as follows.
Fig. Exc_14.4_1
Question: 2
During the medical check-up of 35 students of a
class, their weights were recorded as follows:
Weight (in kg)
Number of students
Less than 38
Less than 40
0
3 Less than 42
Less than 44
Less than 46
Less than 48
Less than 50
Less than 52
5
9
14
28
32
35
Draw a less than type ogive for the given data.
Hence obtain the median weight from the graph
and verify the result by using the formula.
Solution:
The frequency distribution table of less than type
along with the cumulative frequency is given by,
Weight (in kg)
Upper class limits
Number of students
(Cumulative
frequency)
Less than 38
Less than 40
Less than 42
Less than 44
Less than 46
Less than 48
Less than 50
0
3
5
9
14
28
32 Less than 52
35
Now, take upper class limits of class intervals on
and their respective frequencies on ,
x axis
y axis
draw its ogive as follows.
Fig. Exc_14.4_2 (i)
From the table,
n 35
So,
n 35
17.5
2 2
Now, mark the point A whose coordinate is 17.5
y
and its coordinate is 46.5.
x Therefore, median of this data is 46.5.
Fig. Exc_14.4_2 (ii)
The difference between two consecutive upper class
limits is 2. The class marks with their respective
frequencies are given by,
Weight is kg
Frequency
i
f
Cumulative
frequency
cf
Less than 38
0
0
38 40
3 0 3
3
40 42
5 3 2
5
42 44
9 5 4
9
44 46
14 9 5
14 46 48
28 14 14
28
48 50
32 28 4
32
50 52
35 32 3
35
Total
n
35
Cumulative frequency just greater than
cf
is 28, belonging to interval .
n 35
17.5
2 2
46 48
Median class
46 48
Lower limit of median class
l
46
Class size
h 2
Frequency of median class
f
14
Cumulative frequency of class preceding
cf
median class
14
Now, find the median.
n
cf
2
Median l h
f
17.5 14
46 2
14
3.5
46 2
14 3.5
46
7
46 0.5
46.5
The median of the given data is 46.5.
Hence verified.
Question: 3
The following table gives production yield per
hectare of wheat of 100 farms of a village.
Production
yield (in kg/ha)
50 – 55
55 – 60
60 – 65
65 – 70
70 – 75
75 – 80
Number of
farms
2
8
12
24
38
16
Change the distribution to a more than type
distribution, and draw its ogive.
Solution:
The frequency distribution table of more than type
along with the cumulative frequency is given by,
Production yield
Lower class limits
Number of farms
(Cumulative frequency)
More than or equal to 50
100 More than or equal to 55
100 2 98
More than or equal to 60
98 8 90
More than or equal to 65
90 12 78
More than or equal to 70
78 24 54
More than or equal to 75
54 38 16
Now, take lower class limits of class intervals on
and their respective frequencies on ,
x axis
y axis
draw its ogive as follows.
Fig. Exc_14.4_3