Page 1
1. In a hospital, the ages of diabetic patients were recorded as follows. Find the median age.
Age
(in years)
0 15 15 30 30 45 45 60 60 - 75
No. of patients 5 20 40 50 25
Sol:
We prepare the cumulative frequency table, as shown below:
Age (in years) Number of patients (fi) Cumulative Frequency (cf)
0 15 5 5
15 30 20 25
30 45 40 65
45 60 50 115
60 75 25 140
Total N = = 140
Now, N = 140 = 70.
The cumulative frequency just greater than 70 is 115 and the corresponding class is 45
60.
Thus, the median class is 45 60.
l = 45, h = 15, f = 50, N = 140 and cf = 65.
Now,
Median = l + × h
= 45 + × 15
= 45 + × 15
= 45 + 1.5
= 46.5
Page 2
1. In a hospital, the ages of diabetic patients were recorded as follows. Find the median age.
Age
(in years)
0 15 15 30 30 45 45 60 60 - 75
No. of patients 5 20 40 50 25
Sol:
We prepare the cumulative frequency table, as shown below:
Age (in years) Number of patients (fi) Cumulative Frequency (cf)
0 15 5 5
15 30 20 25
30 45 40 65
45 60 50 115
60 75 25 140
Total N = = 140
Now, N = 140 = 70.
The cumulative frequency just greater than 70 is 115 and the corresponding class is 45
60.
Thus, the median class is 45 60.
l = 45, h = 15, f = 50, N = 140 and cf = 65.
Now,
Median = l + × h
= 45 + × 15
= 45 + × 15
= 45 + 1.5
= 46.5
Hence, the median age is 46.5 years.
2. Compute mean from the following data:
Marks 0 7 7 14 14 21 21 28 28 35 35 42 42 49
Number of
Students
3 4 7 11 0 16 9
Sol:
Class Frequency (f) Cumulative Frequency (cf)
0 7 3 3
7 14 4 7
14 21 7 14
21 - 28 11 25
28 35 0 25
35 42 16 41
42 49 9 50
N = = 50
Now, N = 50 = 25.
The cumulative frequency just greater than 25 is 41 and the corresponding class is 35 42.
Thus, the median class is 35 42.
l = 35, h = 7, f = 16, cf = c.f. of preceding class = 25 and = 25.
Now,
Median = l + × h
= 35 + 7 ×
= 35 + 0
= 35
Hence, the median age is 46.5 years.
3. The following table shows the daily wages of workers in a factory:
0 100 100 200 200 300 300 400 400 500
Number of
workers
40 32 48 22 8
Find the median daily wage income of the workers.
Sol:
Class Frequency (f) Cumulative Frequency (cf)
0 100 40 40
100 200 32 72
200 300 48 120
300 400 22 142
400 500 8 150
N = = 150
Page 3
1. In a hospital, the ages of diabetic patients were recorded as follows. Find the median age.
Age
(in years)
0 15 15 30 30 45 45 60 60 - 75
No. of patients 5 20 40 50 25
Sol:
We prepare the cumulative frequency table, as shown below:
Age (in years) Number of patients (fi) Cumulative Frequency (cf)
0 15 5 5
15 30 20 25
30 45 40 65
45 60 50 115
60 75 25 140
Total N = = 140
Now, N = 140 = 70.
The cumulative frequency just greater than 70 is 115 and the corresponding class is 45
60.
Thus, the median class is 45 60.
l = 45, h = 15, f = 50, N = 140 and cf = 65.
Now,
Median = l + × h
= 45 + × 15
= 45 + × 15
= 45 + 1.5
= 46.5
Hence, the median age is 46.5 years.
2. Compute mean from the following data:
Marks 0 7 7 14 14 21 21 28 28 35 35 42 42 49
Number of
Students
3 4 7 11 0 16 9
Sol:
Class Frequency (f) Cumulative Frequency (cf)
0 7 3 3
7 14 4 7
14 21 7 14
21 - 28 11 25
28 35 0 25
35 42 16 41
42 49 9 50
N = = 50
Now, N = 50 = 25.
The cumulative frequency just greater than 25 is 41 and the corresponding class is 35 42.
Thus, the median class is 35 42.
l = 35, h = 7, f = 16, cf = c.f. of preceding class = 25 and = 25.
Now,
Median = l + × h
= 35 + 7 ×
= 35 + 0
= 35
Hence, the median age is 46.5 years.
3. The following table shows the daily wages of workers in a factory:
0 100 100 200 200 300 300 400 400 500
Number of
workers
40 32 48 22 8
Find the median daily wage income of the workers.
Sol:
Class Frequency (f) Cumulative Frequency (cf)
0 100 40 40
100 200 32 72
200 300 48 120
300 400 22 142
400 500 8 150
N = = 150
Now, N = 150
= 75.
The cumulative frequency just greater than 75 is 120 and the corresponding class is 200
300.
Thus, the median class is 200 300.
l = 200, h = 100, f = 48, cf = c.f. of preceding class = 72 and = 75.
Now,
Median, M = l +
= 200 +
= 200 + 6.25
= 206.25
Hence, the median daily wage income of the workers is Rs 206.25.
4. Calculate the median from the following frequency distribution table:
Class 5 10 10 15 15 20 20 25 25 30 30 35 35 40 40 45
Frequency 5 6 15 10 5 4 2 2
Sol:
Class Frequency (f) Cumulative Frequency
(cf)
5 10 5 5
10 15 6 11
15 20 15 26
20 25 10 36
25 30 5 41
30 35 4 45
35 40 2 47
40 45 2 49
N = = 49
Now, N = 49
= 24.5.
The cumulative frequency just greater than 24.5 is 26 and the corresponding class is 15 -
20.
Thus, the median class is 15 20.
l = 15, h = 5, f = 15, cf = c.f. of preceding class = 11 and = 24.5.
Now,
Median, M = l +
= 15 +
Page 4
1. In a hospital, the ages of diabetic patients were recorded as follows. Find the median age.
Age
(in years)
0 15 15 30 30 45 45 60 60 - 75
No. of patients 5 20 40 50 25
Sol:
We prepare the cumulative frequency table, as shown below:
Age (in years) Number of patients (fi) Cumulative Frequency (cf)
0 15 5 5
15 30 20 25
30 45 40 65
45 60 50 115
60 75 25 140
Total N = = 140
Now, N = 140 = 70.
The cumulative frequency just greater than 70 is 115 and the corresponding class is 45
60.
Thus, the median class is 45 60.
l = 45, h = 15, f = 50, N = 140 and cf = 65.
Now,
Median = l + × h
= 45 + × 15
= 45 + × 15
= 45 + 1.5
= 46.5
Hence, the median age is 46.5 years.
2. Compute mean from the following data:
Marks 0 7 7 14 14 21 21 28 28 35 35 42 42 49
Number of
Students
3 4 7 11 0 16 9
Sol:
Class Frequency (f) Cumulative Frequency (cf)
0 7 3 3
7 14 4 7
14 21 7 14
21 - 28 11 25
28 35 0 25
35 42 16 41
42 49 9 50
N = = 50
Now, N = 50 = 25.
The cumulative frequency just greater than 25 is 41 and the corresponding class is 35 42.
Thus, the median class is 35 42.
l = 35, h = 7, f = 16, cf = c.f. of preceding class = 25 and = 25.
Now,
Median = l + × h
= 35 + 7 ×
= 35 + 0
= 35
Hence, the median age is 46.5 years.
3. The following table shows the daily wages of workers in a factory:
0 100 100 200 200 300 300 400 400 500
Number of
workers
40 32 48 22 8
Find the median daily wage income of the workers.
Sol:
Class Frequency (f) Cumulative Frequency (cf)
0 100 40 40
100 200 32 72
200 300 48 120
300 400 22 142
400 500 8 150
N = = 150
Now, N = 150
= 75.
The cumulative frequency just greater than 75 is 120 and the corresponding class is 200
300.
Thus, the median class is 200 300.
l = 200, h = 100, f = 48, cf = c.f. of preceding class = 72 and = 75.
Now,
Median, M = l +
= 200 +
= 200 + 6.25
= 206.25
Hence, the median daily wage income of the workers is Rs 206.25.
4. Calculate the median from the following frequency distribution table:
Class 5 10 10 15 15 20 20 25 25 30 30 35 35 40 40 45
Frequency 5 6 15 10 5 4 2 2
Sol:
Class Frequency (f) Cumulative Frequency
(cf)
5 10 5 5
10 15 6 11
15 20 15 26
20 25 10 36
25 30 5 41
30 35 4 45
35 40 2 47
40 45 2 49
N = = 49
Now, N = 49
= 24.5.
The cumulative frequency just greater than 24.5 is 26 and the corresponding class is 15 -
20.
Thus, the median class is 15 20.
l = 15, h = 5, f = 15, cf = c.f. of preceding class = 11 and = 24.5.
Now,
Median, M = l +
= 15 +
= 15 + 4.5
= 19.5
Hence, the median = 19.5.
5. Given below is the number of units of electricity consumed in a week in a certain locality:
Calculate the median.
Sol:
Class Frequency (f) Cumulative Frequency (cf)
65- 85 4 4
85 105 5 9
105 125 13 22
125 145 20 42
145 165 14 56
165 185 7 63
185 205 4 67
N = = 67
Now, N = 67
= 33.5.
The cumulative frequency just greater than 33.5 is 42 and the corresponding class is 125 -
145.
Thus, the median class is 125 145.
l = 125, h = 20, f = 20, cf = c.f. of preceding class = 22 and = 33.5.
Now,
Median, M = l +
= 125 +
= 125 + 11.5
= 136.5
Hence, the median = 136.5.
6. Calculate the median from the following data:
Class 65 85 85 105 105 125 125 145 145 165 165 185 185 200
Frequency 4 5 13 20 14 7 4
Height(in
cm)
135 -
140
140 -
145
145 -
150
150 -
155
155 -
160
160 -
165
165 -
170
170 -
175
Frequency 6 10 18 22 20 15 6 3
Page 5
1. In a hospital, the ages of diabetic patients were recorded as follows. Find the median age.
Age
(in years)
0 15 15 30 30 45 45 60 60 - 75
No. of patients 5 20 40 50 25
Sol:
We prepare the cumulative frequency table, as shown below:
Age (in years) Number of patients (fi) Cumulative Frequency (cf)
0 15 5 5
15 30 20 25
30 45 40 65
45 60 50 115
60 75 25 140
Total N = = 140
Now, N = 140 = 70.
The cumulative frequency just greater than 70 is 115 and the corresponding class is 45
60.
Thus, the median class is 45 60.
l = 45, h = 15, f = 50, N = 140 and cf = 65.
Now,
Median = l + × h
= 45 + × 15
= 45 + × 15
= 45 + 1.5
= 46.5
Hence, the median age is 46.5 years.
2. Compute mean from the following data:
Marks 0 7 7 14 14 21 21 28 28 35 35 42 42 49
Number of
Students
3 4 7 11 0 16 9
Sol:
Class Frequency (f) Cumulative Frequency (cf)
0 7 3 3
7 14 4 7
14 21 7 14
21 - 28 11 25
28 35 0 25
35 42 16 41
42 49 9 50
N = = 50
Now, N = 50 = 25.
The cumulative frequency just greater than 25 is 41 and the corresponding class is 35 42.
Thus, the median class is 35 42.
l = 35, h = 7, f = 16, cf = c.f. of preceding class = 25 and = 25.
Now,
Median = l + × h
= 35 + 7 ×
= 35 + 0
= 35
Hence, the median age is 46.5 years.
3. The following table shows the daily wages of workers in a factory:
0 100 100 200 200 300 300 400 400 500
Number of
workers
40 32 48 22 8
Find the median daily wage income of the workers.
Sol:
Class Frequency (f) Cumulative Frequency (cf)
0 100 40 40
100 200 32 72
200 300 48 120
300 400 22 142
400 500 8 150
N = = 150
Now, N = 150
= 75.
The cumulative frequency just greater than 75 is 120 and the corresponding class is 200
300.
Thus, the median class is 200 300.
l = 200, h = 100, f = 48, cf = c.f. of preceding class = 72 and = 75.
Now,
Median, M = l +
= 200 +
= 200 + 6.25
= 206.25
Hence, the median daily wage income of the workers is Rs 206.25.
4. Calculate the median from the following frequency distribution table:
Class 5 10 10 15 15 20 20 25 25 30 30 35 35 40 40 45
Frequency 5 6 15 10 5 4 2 2
Sol:
Class Frequency (f) Cumulative Frequency
(cf)
5 10 5 5
10 15 6 11
15 20 15 26
20 25 10 36
25 30 5 41
30 35 4 45
35 40 2 47
40 45 2 49
N = = 49
Now, N = 49
= 24.5.
The cumulative frequency just greater than 24.5 is 26 and the corresponding class is 15 -
20.
Thus, the median class is 15 20.
l = 15, h = 5, f = 15, cf = c.f. of preceding class = 11 and = 24.5.
Now,
Median, M = l +
= 15 +
= 15 + 4.5
= 19.5
Hence, the median = 19.5.
5. Given below is the number of units of electricity consumed in a week in a certain locality:
Calculate the median.
Sol:
Class Frequency (f) Cumulative Frequency (cf)
65- 85 4 4
85 105 5 9
105 125 13 22
125 145 20 42
145 165 14 56
165 185 7 63
185 205 4 67
N = = 67
Now, N = 67
= 33.5.
The cumulative frequency just greater than 33.5 is 42 and the corresponding class is 125 -
145.
Thus, the median class is 125 145.
l = 125, h = 20, f = 20, cf = c.f. of preceding class = 22 and = 33.5.
Now,
Median, M = l +
= 125 +
= 125 + 11.5
= 136.5
Hence, the median = 136.5.
6. Calculate the median from the following data:
Class 65 85 85 105 105 125 125 145 145 165 165 185 185 200
Frequency 4 5 13 20 14 7 4
Height(in
cm)
135 -
140
140 -
145
145 -
150
150 -
155
155 -
160
160 -
165
165 -
170
170 -
175
Frequency 6 10 18 22 20 15 6 3
Sol:
Class Frequency (f) Cumulative Frequency (cf)
135 140 6 6
140 145 10 16
145 150 18 34
150 155 22 56
155 160 20 76
160 165 15 91
165 170 6 97
170 175 3 100
N = = 100
Now, N = 100
= 50.
The cumulative frequency just greater than 50 is 56 and the corresponding class is 150 -
155.
Thus, the median class is 150 155.
l = 150, h = 5, f = 22, cf = c.f. of preceding class = 34 and = 50.
Now,
Median, M = l +
= 150 +
= 150 + 3.64
= 153.64
Hence, the median = 153.64.
7. Calculate the missing frequency from the following distribution, it being given that the
median of distribution is 24.
Sol:
Class Frequency (fi) Cumulative Frequency (cf)
0 10 5 5
10 20 25 30
20 30 x x + 30
30 40 18 x + 48
40 50 7 x + 55
Median is 24 which lies in 20 30
Median class = 20 30
Let the unknown frequency be x.
Class 0 10 10 20 20 30 30 40 40 - 50
Frequency 5 25 ? 18 7
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