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Exercise 14.1

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
6
2
3
Which method did you use for finding the mean, and why?



To find the class mark (xi) for each interval, the following relation is used.
Class mark (xi)








                                              
xand fixi can be calculated as follows.
Number of plants
Number of houses
(fi)
xi
fixi
0­ − 2
1
1
1 × 1 = 1
2­ − 4
2
3
2 × 3 = 6
4 − 6
1
5
1 × 5 = 5
6 − 8
5
7
5 × 7 = 35
8 − 10
6
9
6 × 9 = 54
10 − 12
2
11
2 ×11 = 22
12 − 14
3
13
3 × 13 = 39
Total
20
162
From the table, it can be observed that






Mean,


Therefore, mean number of plants per house is 8.1.
Here, direct method has been used as the values of class marks (xi) and fi are small.




Question 2:
Consider the following distribution of daily wages of 50 worker of a factory.
Daily wages (in Rs)
100­ − 120
120­ − 140
140 −1 60
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.
To find the class mark for each interval, the following relation is used.




Class size (h) of this data = 20
Taking 150 as assured mean (a), diui, and fiui can be calculated as follows.
Daily wages
(in Rs)
Number of workers (fi)
xi
di = x− 150
fiui
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, it can be observed that









Therefore, the mean daily wage of the workers of the factory is Rs 145.20.




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 (in Rs)
11­ − 13
13­ − 15
15 −17
17 − 19
19 − 21
21 − 23
23 − 25
Number of workers
7
6
9
13
f
5
4




To find the class mark (xi) for each interval, the following relation is used.






Given that, mean pocket allowance,
Taking 18 as assured mean (a), di and fidi are calculated as follows.
Daily pocket allowance
(in Rs)
Number of children
fi
Class markxi
di = x− 18
fidi
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
f
21 − 23
5
22
4
20
23 − 25
4
24
6
24
Total
2f − 40
From the table, we obtain












Hence, the missing frequency, f, is 20.




Question 4:
Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarized as follows. Fine 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



To find the class mark of each interval (xi), the following relation is used.





Class size, h, of this data = 3
Taking 75.5 as assumed mean (a), diuifiui are calculated as follows.
Number of heart beats per minute
Number of women
fi
xi
di = x− 75.5
fiui
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, we obtain








Therefore, mean hear 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?



Number of mangoes
Number of boxes fi
50 − 52
15
53 − 55
110
56 − 58
135
59 − 61
115
62 − 64
25
It can be observed that class intervals are not continuous. There is a gap of 1 between two class intervals. Therefore 1/2  has to be subtracted from the lower class limit
of each interval.
, has to be added to the upper class limit and
Class mark (xi) can be obtained by using the following relation.





Class size (h) of this data = 3
Taking 57 as assumed mean (a), diuifiui are calculated as follows.
Class interval
fi
xi
di xi − 57
fiui
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
It can be observed that






Mean number of mangoes kept in a packing box is 57.19.
Step deviation method is used here as the values of fi, di are big and also, there is a common multiple between all di.
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.
To find the class mark (xi) for each interval, the following relation is used.




Class size = 50
Taking 225 as assumed mean (a), diuifiui are calculated as follows.
Daily expenditure (in Rs)
fi
xi
di xi − 225
fiui
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, we obtain







Therefore, mean daily expenditure on food is Rs 211.




Question 7:
To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:
concentration of SO2 (in ppm)
Frequency
0.00 − 0.04
4
0.04 − 0.08
9
0.08 − 0.12
9
0.12 − 0.16
2
0.16 − 0.20
4
0.20 − 0.24
2
Find the mean concentration of SO2 in the air.
To find the class marks for each interval, the following relation is used.





Class size of this data = 0.04
Taking 0.14 as assumed mean (a), diui, fiui are calculated as follows.
Concentration of SO2(in ppm)
Frequency
fi
Class mark
xi
di x− 0.14
fiui
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, we obtain







Therefore, mean concentration of SO2 in the air is 0.099 ppm.

Exercise 14.2

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. 


To find the class marks (xi), the following relation is used.


Taking 30 as assumed mean (a), di and fidiare calculated as follows.
Age (in years)
Number of patients
fi
Class mark
xi
di xi − 30
fidi
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, we obtain


Mean of this data is 35.38. It represents that on an average, the age of a patient admitted to hospital was 35.38 years.
It can be observed that the maximum class frequency is 23 belonging to class interval 35 − 45.
Modal class = 35 − 45
Lower limit (l) of modal class = 35
Frequency (f1) of modal class = 23
Class size (h) = 10
Frequency (f0) of class preceding the modal class = 21
Frequency (f2) of class succeeding the modal class = 14
Mode =



Mode is 36.8. It represents that the age of maximum number of patients admitted in hospital was 36.8 years.



Question 2:
The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:
Lifetimes (in 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. 

From the data given above, it can be observed that the maximum class frequency is 61, belonging to class interval 60 − 80.
Therefore, modal class = 60 − 80
Lower class limit (l) of modal class = 60
Frequency (f1) of modal class = 61
Frequency (f0) of class preceding the modal class = 52
Frequency (f2) of class succeeding the modal class = 38
Class size (h) = 20






Therefore, 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
24
1500 − 2000
40
2000 − 2500
33
2500 − 3000
28
3000 − 3500
30
3500 − 4000
22
4000 − 4500
16
4500 − 5000
7



It can be observed from the given data that the maximum class frequency is 40, belonging to 1500 − 2000 intervals.
Therefore, modal class = 1500 − 2000
Lower limit (l) of modal class = 1500
Frequency (f1) of modal class = 40
Frequency (f0) of class preceding modal class = 24
Frequency (f2) of class succeeding modal class = 33
Class size (h) = 500






Therefore, modal monthly expenditure was Rs 1847.83.
To find the class mark, the following relation is used.


Class size (h) of the given data = 500
Taking 2750 as assumed mean (a), diui, and fiuiare calculated as follows.
Expenditure (in Rs)
Number of families
fi
xi
di = xi − 2750
fiui
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, we obtain



Therefore, mean monthly expenditure was 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
3
20 − 25
8
25 − 30
9
30 − 35
10
35 − 40
3
40 − 45
0
45 − 50
0
50 − 55
2
It can be observed from the given data that the maximum class frequency is 10 belonging to class interval 30 − 35.
Therefore, modal class = 30 − 35
Class size (h) = 5
Lower limit (l) of modal class = 30
Frequency (f1) of modal class = 10
Frequency (f0) of class preceding modal class = 9
Frequency (f2) of class succeeding modal class = 3



It represents that most of the states/U.T have a teacher-student ratio as 30.6.
To find the class marks, the following relation is used.



Taking 32.5 as assumed mean (a), diui, and fiui are calculated as follows.
Number of students per teacher
Number of states/U.T
(fi)
xi
di = xi − 32.5
fiui
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





Therefore, mean of the data is 29.2.
It represents that on an average, teacher−student ratio was 29.2.




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
4
4000 − 5000
18
5000 − 6000
9
6000 − 7000
7
7000 − 8000
6
8000 − 9000
3
9000 − 10000
1
10000 − 11000
1
Find the mode of the data.


From the given data, it can be observed that the maximum class frequency is 18, belonging to class interval 4000 − 5000.
Therefore, modal class = 4000 − 5000
Lower limit (l) of modal class = 4000
Frequency (f1) of modal class = 18
Frequency (f0) of class preceding modal class = 4
Frequency (f2) of class succeeding modal class = 9
Class size (h) = 1000


Therefore, mode of the given data is 4608.7 runs.




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


From the given data, it can be observed that the maximum class frequency is 20, belonging to 40 − 50 class intervals.
Therefore, modal class = 40 − 50
Lower limit (l) of modal class = 40
Frequency (f1) of modal class = 20
Frequency (f0) of class preceding modal class = 12
Frequency (f2) of class succeeding modal class = 11
Class size = 10







Therefore, mode of this data is 44.7 cars.


Exercise 14.3

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
4
85 − 105
5
105 − 125
13
125 − 145
20
145 − 165
14
165 − 185
8
185 − 205
4



To find the class marks, the following relation is used.


Taking 135 as assumed mean (a), diui, fiui are calculated according to step deviation method as follows.
Monthly consumption (in units)
Number of consumers (fi)
xiclass mark
di=xi− 135
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, we obtain





From the table, it can be observed that the maximum class frequency is 20, belonging to class interval 125 − 145.
Modal class = 125 − 145
Lower limit (l) of modal class = 125
Class size (h) = 20
Frequency (f1) of modal class = 20
Frequency (f0) of class preceding modal class = 13
Frequency (f2) of class succeeding the modal class = 14


To find the median of the given data, cumulative frequency is calculated as follows.
Monthly consumption (in units)
Number of consumers
Cumulative frequency
65 − 85
4
4
85 − 105
5
4 + 5 = 9
105 − 125
13
9 + 13 = 22
125 − 145
20
22 + 20 = 42
145 − 165
14
42 + 14 = 56
165 − 185
8
56 + 8 = 64
185 − 205
4
64 + 4 = 68
From the table, we obtain
n = 68
Cumulative frequency (cf) just greater than
is 42, belonging to interval 125 − 145.
Therefore, median class = 125 − 145
Lower limit (l) of median class = 125
Class size (h) = 20
Frequency (f) of median class = 20
Cumulative frequency (cf) of class preceding median class = 22


Therefore, median, mode, mean of the given data is 137, 135.76, and 137.05 respectively.
The three measures are approximately the same in this case.




Question 2:
If the median of the distribution is given below is 28.5, find the values of x and y.
Class interval
Frequency
0 − 10
5
10 − 20
x
20 − 30
20
30 − 40
15
40 − 50
y
50 − 60
5
Total
60


The cumulative frequency for the given data is calculated as follows.
Class interval
Frequency
Cumulative frequency
0 − 10
5
5
10 − 20
x
5+ x
20 − 30
20
25 + x
30 − 40
15
40 + x
40 − 50
y
40+ x + y
50 − 60
5
45 + x + y
Total (n)
60
From the table, it can be observed that = 60
45 + x + y = 60
x + y = 15 (1)
Median of the data is given as 28.5 which lies in interval 20 − 30.
Therefore, median class = 20 − 30
Lower limit (l) of median class = 20
Cumulative frequency (cf) of class preceding the median class = 5 + x
Frequency (f) of median class = 20
Class size (h) = 10



From equation (1),
8 + y = 15
y = 7
Hence, the values of x and y are 8 and 7 respectively.




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 year.
Age (in years)
Number of policy holders
Below 20
2
Below 25
6
Below 30
24
Below 35
45
Below 40
78
Below 45
89
Below 50
92
Below 55
98
Below 60
100



Here, class width is not the same. There is no requirement of adjusting the frequencies according to class intervals. The given 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. Therefore, class intervals with their respective cumulative frequency can be defined as below.
Age (in years)
Number of policy holders (fi)
Cumulative frequency (cf)
18 − 20
2
2
20 − 25
6 − 2 = 4
6
25 − 30
24 − 6 = 18
24
30 − 35
45 − 24 = 21
45
35 − 40
78 − 45 = 33
78
40 − 45
89 − 78 = 11
89
45 − 50
92 − 89 = 3
92
50 − 55
98 − 92 = 6
98
55 − 60
100 − 98 = 2
100
Total (n)
From the table, it can be observed that n = 100.
Cumulative frequency (cf) just greater than
is 78, belonging to interval 35 − 40.
Therefore, median class = 35 − 40
Lower limit (l) of median class = 35
Class size (h) = 5
Frequency (f) of median class = 33
Cumulative frequency (cf) of class preceding median class = 45


Therefore, median age is 35.76 years.




Question 4:
The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:
Length (in mm)
Number or leaves fi
118 − 126
3
127 − 135
5
136 − 144
9
145 − 153
12
154 − 162
5
163 − 171
4
172 − 180
2
Find the median length of the leaves.
(Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 − 126.5, 126.5 − 135.5… 171.5 − 180.5)


The given data does not have continuous class intervals. It can be observed that the difference between two class intervals is 1. Therefore,1/2=0.5 has to be added and subtracted to upper class limits and lower class limits respectively.
Continuous class intervals with respective cumulative frequencies can be represented as follows.
Length (in mm)
Number or leaves fi
Cumulative frequency
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, it can be observed that the cumulative frequency just greater than
is 29, belonging to class interval 144.5 − 153.5.
Median class = 144.5 − 153.5
Lower limit (l) of median class = 144.5
Class size (h) = 9
Frequency (f) of median class = 12
Cumulative frequency (cf) of class preceding median class = 17
Median




Therefore, median length of leaves is 146.75 mm.
Question 5:
Find the following table gives the distribution of the life time of 400 neon lamps:
Life time (in hours)
Number of lamps
1500 − 2000
14
2000 − 2500
56
2500 − 3000
60
3000 − 3500
86
3500 − 4000
74
4000 − 4500
62
4500 − 5000
48
Find the median life time of a lamp.


Thecumulative frequencies with their respective class intervals are as follows.
Life time
Number of lamps (fi)
Cumulative frequency
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
It can be observed that the cumulative frequency just greater than
is 216, belonging to class interval 3000 − 3500.
Median class = 3000 − 3500
Lower limit (l) of median class = 3000
Frequency (f) of median class = 86
Cumulative frequency (cf) of class preceding median class = 130
Class size (h) = 500
Median




= 3406.976
Therefore, median life time 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
6
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.
The cumulative frequencies with their respective class intervals are as follows.
Number of letters
Frequency (fi)
Cumulative frequency
1 − 4
6
6
4 − 7
30
30 + 6 = 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
It can be observed that the cumulative frequency just greater than
is 76, belonging to class interval 7 − 10.
Median class = 7 − 10
Lower limit (l) of median class = 7
Cumulative frequency (cf) of class preceding median class = 36
Frequency (f) of median class = 40
Class size (h) = 3
Median




= 8.05
To find the class marks of the given class intervals, the following relation is used.


Taking 11.5 as assumed mean (a), diui, and fiui are calculated according to step deviation method as follows.
Number of letters
Number of surnames
fi
xi
di = xi− 11.5
fiui
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, we obtain
fiui = −106
fi = 100
Mean,





= 11.5 − 3.18 = 8.32
The data in the given table can be written as
Number of letters
Frequency (fi)
1 − 4
6
4 − 7
30
7 − 10
40
10 − 13
16
13 − 16
4
16 − 19
4
Total (n)
100
From the table, it can be observed that the maximum class frequency is 40 belonging to class interval 7 − 10.
Modal class = 7 − 10
Lower limit (l) of modal class = 7
Class size (h) = 3
Frequency (f1) of modal class = 40
Frequency (f0) of class preceding the modal class = 30
Frequency (f2) of class succeeding the modal class = 16


Therefore, median number and mean number of letters in surnames is 8.05 and 8.32 respectively while 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



The cumulative frequencies with their respective class intervals are as follows.
Weight (in kg)
Frequency (fi)
Cumulative frequency
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
Cumulative frequency just greater than
is 19, belonging to class interval 55 − 60.
Median class = 55 − 60
Lower limit (l) of median class = 55
Frequency (f) of median class = 6
Cumulative frequency (cf) of median class = 13
Class size (h) = 5
Median




= 56.67
Therefore, median weight is 56.67 kg.


Exercise 14.4

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. 


The frequency distribution table of less than type is as follows.
Daily income (in Rs)
(upper class limits)
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
Taking upper class limits of class intervals on x-axis and their respective frequencies on y-axis, its ogive can be drawn as follows.









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
0
Less than 40
3
Less than 42
5
Less than 44
9
Less than 46
14
Less than 48
28
Less than 50
32
Less than 52
35
Draw a less than type ogive for the given data. Hence obtain the median weight from the graph verify the result by using the formula. 



The given cumulative frequency distributions of less than type are
Weight (in kg)
upper class limits
Number of students
(cumulative frequency)
Less than 38
0
Less than 40
3
Less than 42
5
Less than 44
9
Less than 46
14
Less than 48
28
Less than 50
32
Less than 52
35
Taking upper class limits on x-axis and their respective cumulative frequencies on y-axis, its ogive can be drawn as follows.











Here, n = 35   So 


     
           = 17.5



,
Mark the point A whose ordinate is 17.5 and its x-coordinate is 46.5. Therefore, median of this data is 46.5.











It can be observed that the difference between two consecutive upper class limits is 2. The class marks with their respective frequencies are obtained as below.
Weight (in kg)
Frequency (f)
Cumulative frequency
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
The cumulative frequency just greater than






is 28, belonging to class interval 46 − 48.


Median class = 46 − 48
Lower class limit (l) of median class = 46
Frequency (f) of median class = 14
Cumulative frequency (cf) of class preceding median class = 14
Class size (h) = 2












Therefore, median of this data is 46.5.
Hence, the value of median is 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 ogive.

The cumulative frequency distribution of more than type can be obtained as follows.
Production yield
(lower class limits)
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
Taking the lower class limits on x-axis and their respective cumulative frequencies ony-axis, its ogive can be obtained as follows.





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