Exercise 7.5
1. Find the mode of the following data:
(i) 3, 5, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4
(ii) 3, 3, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4
(iii) 15, 8, 26, 25, 24, 15, 18, 20, 24, 15, 19, 15
Soln:
Mode is the value that occurs the maximum number of times in a given data set.
(i)
Value (x) | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
Frequency (f) | 4 | 2 | 5 | 2 | 2 | 1 | 2 |
Mode = 5
(ii)
Value (x) | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
Frequency (f) | 5 | 2 | 4 | 2 | 2 | 1 | 2 |
Mode = 3
(iii)
Value (x) | 8 | 15 | 18 | 19 | 20 | 24 | 25 | 26 |
Frequency (f) | 1 | 4 | 1 | 1 | 1 | 2 | 1 | 1 |
Mode = 15
2. The shirt size worn by a group of 200 persons, who bought the shirt from a store, are as follows:
Shirt size: | 37 38 39 40 41 42 43 44 |
Number of persons: | 15 25 39 41 36 17 15 12 |
Find the model shirt size worn by the group.
Soln:
Shirt size | 37 | 38 | 39 | 40 | 41 | 42 | 43 | 44 |
Number of persons | 15 | 25 | 39 | 41 | 36 | 17 | 15 | 12 |
In this data 40 occurs the maximum number of times.
So, Model shirt size = 40
3. Find the mode of the following distribution.
(i)
Class interval: | 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 |
Frequency: | 5 8 7 12 28 20 10 10 |
(ii)
Class interval: | 10-15 15-20 20-25 25-30 30-35 35-40 |
Frequency: | 30 45 75 35 25 15 |
(iii)
Class interval: | 25-30 30-35 35-40 40-45 45-50 50-60 |
Frequency: | 25 34 50 42 29 15 |
Soln:
(i)
Class interval | 0 – 10 | 10–20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
Frequency | 5 | 8 | 7 | 12 | 28 | 20 | 10 | 10 |
Here, as the maximum frequency is 28, the corresponding class i.e. 40 – 52 will be the modal class
l = 40, h = 50 – 40 = 10, f = 28, f_{1} = 12, f_{2 }= 20
Mode = \(l + \frac{f – f_{1}}{2f – f_{1}-f_{2}}\times h\)
= \(40 + \frac{28 -12}{2\times 28 – 12 -20}\times 10\)
= 40 + 160/ 24
= 40 + 6.67
= 46.67
(ii)
Class interval | 10-15 | 15–20 | 20-25 | 25-30 | 30-35 | 35-40 |
Frequency | 30 | 45 | 75 | 35 | 25 | 15 |
As the maximum frequency is 75 here, the corresponding class i.e. 20 – 25 will be the modal class
So, l = 20, h = 25 – 20 = 5, f = 75, f_{1} = 45, f_{2 }= 35
Mode = \(l + \frac{f – f_{1}}{2f – f_{1}-f_{2}}\times h\)
= \(20 + \frac{75 – 45}{2\times 75 – 45 -35}\times 5\)
= 20 + 150/ 70
= 20 + 2.14
= 22.14
(iii)
Class interval | 25-30 | 30-35 | 35-40 | 40-45 | 45-50 | 50-60 |
Frequency | 25 | 34 | 50 | 42 | 38 | 14 |
As the maximum frequency is 50, its corresponding class 35 – 40 will be the modal class
l = 35, h = 40 – 35 = 5, f = 50, f_{1} = 34, f_{2 }= 42
Mode = \(l + \frac{f – f_{1}}{2f – f_{1}-f_{2}}\times h\)
= \(35 + \frac{50 – 34}{2\times 50 – 34 -42}\times 5\)
= 35 + 80/24
= 35 + 3.33
= 38.33
4. Compare the modal ages of two groups of students appearing for an entrance test:
Age (in years): | 16–18 18–20 20–22 22–24 24–26 |
Group A: | 50 78 46 28 23 |
Group B: | 54 89 40 25 17 |
Soln:
Age in years | 16-18 | 18-20 | 20-22 | 22-24 | 24-26 |
Group A | 50 | 78 | 46 | 28 | 23 |
Group B | 54 | 89 | 40 | 25 | 17 |
For Group A
Here class 18 – 20 is model class as the maximum frequency is 78. So,
l = 18, h = 20 – 18 = 2, f = 78, f_{1} = 50, f_{2 }= 46
Mode = \(l + \frac{f – f_{1}}{2f – f_{1}-f_{2}}\times h\)
= \(18 + \frac{78 – 50}{2\times 78 – 50 – 46}\times 2\)
= 18 + 56/ 60
= 18 + 0.93
= 18.93 years
For group B
Here class 18 – 20 is the modal class as the maximum frequency in the data is 89.
l = 18, h = 20 – 18 = 2, f = 89, f_{1} = 54, f_{2 }= 40
Mode = \(l + \frac{f – f_{1}}{2f – f_{1}-f_{2}}\times h\)
= \(18 + \frac{89 – 54}{2\times 89 – 54 – 40}\times 2\)
= 18 + 70/ 84
= 18+ 0.83
= 18.83 years
So, the modal age for the Group B is lower than that of Group A.
5. The marks in science of 80 students of class X are given below. Find the mode of the marks obtained by the students in science.
Marks: | 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 |
Frequency: | 3 5 16 12 13 20 5 4 1 1 |
Soln:
Marks | 0 – 10 | 10–20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 | 90-100 |
Frequency | 3 | 5 | 16 | 12 | 13 | 20 | 5 | 4 | 1 | 1 |
From the data, it can be seen that the maximum frequency is 20. So, the corresponding class 50 – 60 will be the modal class.
l = 50, h = 60 – 50 = 10, f = 20, f_{1} = 13, f_{2 }= 5
Mode = \(l + \frac{f – f_{1}}{2f – f_{1}-f_{2}}\times h\)
= \(50 + \frac{20 – 13}{2\times 20 – 13 – 5}\times 10\)
= 50 + 70/ 22
= 50 + 3.18
= 53.18
6. The following is the distribution of height of students of a certain class in a city:
Height (in cm): | 160-162 163-165 166-168 169-171 172-174 |
No of students: | 15 118 142 127 18 |
Find the average height of the maximum number of students.
Soln:
Heights(exclusive) | 160-162 | 163-165 | 166-168 | 169-171 | 172-174 |
Heights (inclusive) | 159.5-162.5 | 162.5-165.5 | 165.5-168.5 | 168.5-171.5 | 171.5-174.5 |
No of students | 15 | 118 | 142 | 127 | 18 |
From the data, it can be seen that the maximum frequency is 142 and so, 165.5 – 168.5 will be the modal class
l = 165.5, h = 168.5 – 165.5 = 3, f = 142, f_{1} = 118, f_{2 }= 127
Mode = \(l + \frac{f – f_{1}}{2f – f_{1}-f_{2}}\times h\)
= \(165.5 + \frac{142 – 118}{2\times 142 – 118 – 127}\times 3\)
= 165.5 + 72/ 39
= 165.5 + 1.85
= 167.35 cm
7.The following table shows the ages of the patients admitted in a hospital during a year:
Ages (in years): | 5-15 15-25 25-35 35-45 45-55 55-65 |
No of students: | 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.
Soln:
The class marks (x_{i}) can be computed using the following formula:
x_{i} = \( \frac {upper\, class\, limit\; +\; lower\, class\, limit}{2} \)
Now, consider 30 as assumed mean (a) and calculate d_{i }and f_{i}d_{i }as given in the table below:
Age (in years) | Number of patients f_{i} | Class marks x_{i} | d_{i }= x_{i} – 275 | f_{i}d_{i} |
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 it can be observed that:
Σf_{i} = 80
Σf_{i}d_{i} = 430
Mean \(\overline{x}= a + \frac {\Sigma f_{i} d_{i}}{\Sigma f_{i}}\)
30 + 430/ 80
= 35.375
≈ 35.38
Now, as the mean of this data is 35.38, it can be concluded that 35.38 is the average age of the patients admitted to the hospital.
Here, the maximum class frequency is 23 which corresponds to the class interval 35 – 45. This means that the modal class will be 35 – 45
Now,
l= 35, f = 23, h = 10, f_{1} = 21, and f_{2} = 14
So, using mode formula
Mode = \(l + \frac{f – f_{1}}{2f – f_{1}-f_{2}}\times h\)
= \(35 + \frac{23 – 21}{2\times 23 – 21 – 14}\times 10\)
= \(35 + \frac{2}{46 – 35}\times 10\)
= 35 + 1.81
= 36.8
Here, it can be seen that the mode is 36.8. So, the maximum number of patients admitted in the hospital were of 36.8 years.
8. The following data gives 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 |
No of components: | 10 35 52 61 38 29 |
Determine the modal lifetimes of the components.
Soln:
It is seen that the maximum class frequency is 61 which corresponds to the class interval 60 – 80
∴ Modal class limit (l) = 60
Frequency (f) = 61
Frequency (f_{1}) of class preceding the modal class=52
Frequency (f_{2}) of class succeeding the modal class=38
Class size (h) = 20
Now, using the mode formula
Mode = \(l + \frac{f – f_{1}}{2f – f_{1}-f_{2}}\times h\)
= \(60 + \frac{61 – 52}{2\times 61 – 52 – 38}\times 20\)
= \(60 + \frac{9}{122 – 90}\times 20\)
= \(60 + \frac{9\times20}{32}\)
= 60 + 90/16
= 60 + 5.625
= 65.625
Thus, 65.625 is the modal lifetime of electrical components.
9. The following table gives the daily income of 50 workers of a factory:
Daily income | 100 – 120 | 120 – 140 | 140 – 160 | 160 – 180 | 180 – 200 |
Number of workers | 12 | 14 | 8 | 6 | 10 |
Find the mean, mode and median of the above data.
Soln:
Make the cumulative frequency table as shown below:
Class interval | Mid value (x) | Frequency (f) | fx | Cumulative frequency |
100 – 120 | 110 | 12 | 1320 | 12 |
120 – 140 | 130 | 14 | 1820 | 26 |
140 – 160 | 150 | 8 | 1200 | 34 |
160 – 180 | 170 | 6 | 1000 | 40 |
180 – 200 | 190 | 10 | 1900 | 50 |
N = 50 | \(\Sigma fx\) = 7260 |
Here, Mean = Σfx/ 50 = 145.2
As N = 50, N/2 = 25
Now, the cumulative frequency which is just greater than N/ 2 is 26. So, the median class will be 120 – 140 and
l = 120, h = 140 – 120 = 20, f = 14, F = 12
Now, using median formula
Median = \(l+\frac{\frac{N}{2}-F}{f}\times h\)
= \(120+\frac{25-12}{14}\times 20\)
= 120 + 260/ 14
= 120 + 18.57
= 138.57
In the data set, the maximum frequency = 14. So, the corresponding class 120 – 140 will be the modal class
l = 120, h = 140 – 120 = 20, f = 14, f_{1 }= 12, f_{2 }= 8
Now, using the mode formila
Mode = \(l + \frac{f – f_{1}}{2f – f_{1}-f_{2}}\times h\)
= \(120 + \frac{14 – 12}{2\times 14 – 12 – 8}\times 20\)
= 120 + 40/ 8
= 120 + 5
= 125
10. The following distribution gives the state-wise teachers-students 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 |
Number of students per teacher |
Number of states/ U.T |
15 – 20 | 3 | 35 – 40 | 3 |
20 – 25 | 8 | 40 – 45 | 0 |
25 – 30 | 9 | 45 – 50 | 0 |
30 – 35 | 10 | 50 – 55 | 2 |
Soln:
Here, the maximum class frequency is 10 which corresponds to the class interval 30 – 35.
∴ Modal class=30 – 35, h = 5, l = 30, f = 10, f_{1} = 9, f_{2} = 3
Using mode formula
Mode = \(l + \frac{f – f_{1}}{2f – f_{1}-f_{2}}\times h\)
= \(30 + \frac{10 – 9}{2\times 10 – 9 – 3}\times 5\)
= \(30 + \frac{1}{20 – 12}\times 5\)
= 30 + 5/8
= 30.625
⇒ Mode = 30.6
The mode here represents that most of states/ U.T have a teacher- student ratio as 30.6
For finding the class marks, the following formula can be used
Class mark = \( \frac {upper class limit + lower class limit}{2} \)
Now, take 32.5 as assumed mean (a), calculate d_{i}, u_{i}, and f_{i}u_{i }as given below:
Number of students per teacher | Number of states/ U.T (f_{i}) | x_{i} | d_{i }= x_{i} – 32.5 | U_{i} | f_{i}u_{i} |
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 |
Now using mean formula
Mean \(\overline{x}= a + \frac {\Sigma f_{i} d_{i}}{\Sigma f_{i}}\times h\)
= \( 32.5 + \frac{-23}{35}\times 5\)
= 32.5 – 23/7
= 32.5 – 3.28
= 29.22
∴ The mean of data is 29.2
So, 29.2 is the average teacher-student ratio.
11. Find the mean, median and mode of the following data:
Classes: | 0 – 50 | 50 – 100 | 100 – 150 | 150 – 200 | 200 – 250 | 250 – 300 | 300 – 350 |
Frequency: | 2 | 3 | 5 | 6 | 5 | 3 | 1 |
Soln:
Class interval | Mid value (x) | Frequency (f) | fx | Cumulative frequency |
0 – 50 | 35 | 2 | 50 | 2 |
50 – 100 | 75 | 3 | 225 | 5 |
100 – 150 | 125 | 5 | 625 | 10 |
150 – 200 | 175 | 6 | 1050 | 16 |
200 – 250 | 225 | 5 | 1127 | 21 |
250 – 300 | 275 | 3 | 825 | 24 |
300 – 350 | 325 | 1 | 325 | 25 |
N = 25 | \(\Sigma fx\) = 4225 |
Now, using the mean formula
Mean = Σfx/ N = 4225/ 25 = 169
From the data,
N = 25 and so, N/ 2 = 12.5
Hence, the median class will be 150 – 200 as the cumulative frequency just greater than N/2 is 16
⇒ l = 150, f = 6, F = 10, and h = 200 – 150 = 50
Using the median formula-
Median = \(l+\frac{\frac{N}{2}-F}{f}\times h\)
= \(150+\frac{12.5-10}{6}\times 50\)
= 150 + 125/ 6
= 150 + 20.83
= 170.83
As the maximum frequency in the data is 6, the corresponding class 150 – 200 will be the modal class
So, f_{1 }= 5, f_{2 }= 5 (value of l, f, and h is already mentioned below)
Againg using mode formula
Mode = \(l + \frac{f – f_{1}}{2f – f_{1}-f_{2}}\times h\)
= \(150 + \frac{6 – 5}{2\times 6 – 5 – 5}\times 50\)
= 150 + 50/ 2
= 150 + 25
= 175
12. A student noted the number of cars pass through a spot on a road for 100 periods each of 3 minutes and summarized it in the table given below. Find the mode of the data.
Soln:
From the given data it can be observed that maximum class frequency is 20 which belongs to the class interval 40 – 50.
So, modal class=40 – 50
Lower limit (l) of modal class=40
Frequency (f) of modal class=20
Frequency (f_{1}) of class preceding modal class=12
Frequency (f_{2}) of class succeeding modal class=11
Class size = 10
Mode = \(l + \frac{f – f_{1}}{2f – f_{1}-f_{2}}\times h\)
= \(40 + \frac{20 – 12 }{2(20) – 12 – 11}\times 10\)
= \(40 + \frac {80}{40 – 23}\)
= 40 + 80/ 17
= 40 + 4.7
= 44.7
So mode of this data is 44.7 cars
13. 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: | 65-85 85-105 105-125 125-145 145-165 165-185 185-205 |
No of consumers: | 4 5 13 20 14 8 4 |
Soln:
Make the cumulative frequency table as shown below:
Class interval | Mid value x_{i} | Frequency f_{i} | Fx | Cumulative frequency |
65 – 85 | 75 | 4 | 300 | 4 |
85 – 105 | 95 | 5 | 475 | 9 |
105 – 125 | 115 | 13 | 1495 | 22 |
125 – 145 | 135 | 20 | 2700 | 42 |
145 – 165 | 155 | 14 | 2170 | 56 |
165 – 185 | 175 | 8 | 1400 | 64 |
185 – 205 | 195 | 4 | 780 | 68 |
N = 68 | \(\Sigma fx\) = 9320 |
Mean = \(\frac{\Sigma fx}{N}= \frac{9320}{68} = 137.05\)
Given, N = 68
So, N/ 2 = 34
The cumulative frequency just greater than N/ 2 is 42 then the median class is 125 – 145 such that
l = 125, h = 145 – 125 = 20, f = 20, F = 22
Median = \(l+\frac{\frac{N}{2}-F}{f}\times h\)
= \(125+\frac{34-22}{20}\times 20\)
= 125 + 12
= 137
Here the maximum frequency is 20, then the corresponding class 125 – 145 is the modal class
l = 125, h = 145 – 125 = 20, f = 20, f_{1} = 13, f_{2 }= 14
Mode = \(l + \frac{f – f_{1}}{2f – f_{1}-f_{2}}\times h\)
= \(125 + \frac{20 – 13 }{2(20) – 13 – 14}\times 20\)
= 125 + 140/ 13
= 135.77
14. 100 surnames were randomly picked up from a local telephone directly and the frequency distribution of the number of letter 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 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.
Soln:
Make the cumulative frequency table as shown below:
Class interval | Mid value (x) | Frequency (f) | fx | Cumulative frequency |
1 – 4 | 2.5 | 6 | 15 | 6 |
4 – 7 | 5.5 | 30 | 165 | 36 |
7 – 10 | 8.5 | 40 | 340 | 76 |
10 – 13 | 11.5 | 16 | 184 | 92 |
13 – 16 | 14.5 | 4 | 58 | 96 |
16 – 19 | 17.5 | 4 | 70 | 100 |
N = 100 | \(\Sigma fx\) = 832 |
Mean = \(\frac{\Sigma fx}{N}= \frac{832}{100} = 8.32\)
Given N = 100
So, N/ 2 = 50
The cumulative frequency just greater than N/ 2 is 76, then the median class is 7 – 10 such that
l = 7, h = 10 – 7 = 3, f = 40, F = 36
Now, using the median formula
Median = \(l+\frac{\frac{N}{2}-F}{f}\times h\)
= \(7+\frac{50-36}{40}\times 3\)
= 7 + 52/ 40
= 7 + 1.05
= 8.05
As the maximum frequency is 40, the corresponding class 7 – 10 will be the modal class
l = 7, h = 10 – 7 = 3, f = 40, f_{1 }= 30, f_{2 }= 16
Now, using the mode formula
Mode = \(l + \frac{f – f_{1}}{2f – f_{1}-f_{2}}\times h\)
= \(7 + \frac{40 – 30}{2\times 40 – 30 – 16}\times 3\)
= 7 + 30/ 34
= 7 + 0.88
= 7.88
15. Find the mean, median and mode of the following data:
Class | 0 – 20 | 20 – 40 | 40 – 60 | 60 – 80 | 80 – 100 | 100 – 120 | 120 – 140 |
Frequency | 6 | 8 | 10 | 12 | 6 | 5 | 3 |
Soln:
Make the table as given:
Class interval | Mid value (x) | Frequency (f) | fx | Cumulative frequency |
0 – 20 | 10 | 6 | 60 | 6 |
20 – 40 | 30 | 8 | 240 | 17 |
40 – 60 | 50 | 10 | 500 | 24 |
60 – 80 | 70 | 12 | 840 | 36 |
80 – 100 | 90 | 6 | 540 | 42 |
100 – 120 | 110 | 5 | 550 | 47 |
120 – 140 | 130 | 3 | 390 | 50 |
N = 50 | \(\Sigma fx\) = 3120 |
Here, Mean = \(\frac{\Sigma fx}{N}= \frac {3120}{50}= 62.4\)
Given N = 50
So, N/ 2 = 25
The cumulative frequency which is just greater than N/ 2 is 36. So, the median class will 60 – 80 and
l = 60, h = 80 – 60 = 20, f = 12, F = 24
Now, using the median formula
Median = \(l+\frac{\frac{N}{2}-F}{f}\times h\)
= \(60+\frac{25-24}{12}\times 20\)
= 60 + 20/ 12
= 60 + 1.67
= 61.67
Since the maximum frequency is 12, the corresponding class 60 – 80 will be the modal class
l = 60, h = 80 – 60 = 20, f = 12, f_{1 }= 10, f_{2 }= 6
Mode = \(l + \frac{f – f_{1}}{2f – f_{1}-f_{2}}\times h\)
= \(60 + \frac{12 – 10}{2\times 12 – 10 – 6}\times 20\)
= 60 + 40/ 8
= 65
16. 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 | Frequency | Expenditure | Frequency |
1000-1500 | 24 | 3000-3500 | 30 |
1500-2000 | 40 | 3500-4000 | 22 |
2000-2500 | 33 | 4000-4500 | 16 |
2500-3000 | 28 | 4500-5000 | 7 |
Soln:
It is seen that the maximum class frequency in the above data is 40 which belongs to the class interval 1500 -200
Hence, modal class=1500 -2000
l of modal class=1500
f of modal class=40
f_{1} of class preceding modal class=24
f_{2} of class succeeding modal class=33
h = 500
Now, using the mode formula
Mode = \(l + \frac{f – f_{1}}{2f – f_{1}-f_{2}}\times h\)
= \(1500 + \frac{40 – 24}{2\times 40 – 24 – 33}\times 500\)
= \(1500 + \frac{16}{80 – 57}\times 500\)
= 1500 + 347.826
= 1847.826 ≈ 1847.83
∴ The modal monthly expenditure = Rs. 1847.83
Now, to find class mark,
Class mark = \( \frac {upper class limit + lower class limit}{2} \)
Given class size (h) = 500
Now, take 2750 as assumed mean (a) calculate d_{i }u_{i }as given:
Expenditure (in Rs) | Number of families f_{i} | x_{i} | d_{i }= x_{i} – 2750 | U_{i} | f_{i}u_{i} |
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 data table, it can be observed that
Σf_{i}= 200
Σf_{i}d_{i} = -35
So,
Mean \(\overline{x}= a + \frac {\Sigma f_{i} d_{i}}{\Sigma f_{i}}\times h\) \(\overline{x}= 2750+ \frac {-35}{200}\times 500\) \(\overline{x}\) = 2750 – 87.5
\(\overline{x}\) = 2662.5Hence, the mean monthly expenditure is Rs. 2662.5
17. The given distribution shows the number of runs scored by some top batsmen of the world in one day international cricket matches.
Runs scored | No of batsmen | Runs scored | No of batsmen |
3000 – 4000 | 4 | 7000 – 8000 | 6 |
4000 – 5000 | 18 | 8000 – 9000 | 3 |
5000 – 6000 | 9 | 9000 – 10000 | 1 |
6000 – 7000 | 7 | 10000 – 11000 | 1 |
Find the mode of the data.
Soln:
From the given data we may observe that maximum class frequency is 18 belonging to class interval 4000 – 5000.
So, modal class=4000 – 5000
Lower limit (l) of modal class=4000
Frequency (f) of modal class=18
Frequency (f_{1}) of class preceding modal class=4
Frequency (f_{2}) of class succeeding modal class=9
Class size (h) = 1000
Using the mode formula
mode = \(l + \frac{f – f_{1}}{2f – f_{1}-f_{2}}\times h\)
= \(4000 + \frac{18 – 4 }{2(18) – 4 – 9}\times 1000\)
= 4000 + (14000/ 23)
= 4000 + 608.695
= 4608.695
∴ The mode of given data is 4608.7 runs