In statistics, the mode is one important parameter. This exercise completely deals with finding the mode/modal value of the given continuous frequency distribution. Suitable methods are introduced for computing the modal value, and this can be clearly understood in the RD Sharma Solutions Class 10. If you want to get a grip on finding mode, the RD Sharma Solutions for Class 10 Maths Chapter 7 Statistics Exercise 7.5 PDF is available for free to download.
RD Sharma Solutions for Class 10 Maths Chapter 7 Statistics Exercise 7.5
Access RD Sharma Solutions for Class 10 Maths Chapter 7 Statistics 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
Solution:
(i)
| Value (x) | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| Frequency (f) | 4 | 2 | 5 | 2 | 2 | 1 | 2 |
Thus, the mode = 5 since it occurs the maximum number of times.
(ii)
| Value (x) | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| Frequency (f) | 5 | 2 | 4 | 2 | 2 | 1 | 2 |
Thus, the mode = 3 since it occurs the maximum number of times.
(iii)
| Value (x) | 8 | 15 | 18 | 19 | 20 | 24 | 25 |
| Frequency (f) | 1 | 4 | 1 | 1 | 1 | 2 | 1 |
Thus, the mode = 15 since it occurs the maximum number of times.
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.
Solution:
| Shirt size: | 37 | 38 | 39 | 40 | 41 | 42 | 43 | 44 |
| Number of persons: | 15 | 25 | 39 | 41 | 36 | 17 | 15 | 12 |
From the data, it is observed that,
Model shirt size = 40 since it was the size which occurred for the maximum number of times.
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 |
Solution:
| 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 |
It’s seen that the maximum frequency is 28.
So, the corresponding class, i.e. 40 – 50, is the modal class.
And,
l = 40, h = 50 40 = 10, f = 28, f1 = 12, f2 = 20
Using the formula for finding mode, we get
= 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 |
Solution:
| Class interval | 10 – 15 | 15 – 20 | 20 – 25 | 25 – 30 | 30 – 35 | 35 – 40 |
| Frequency | 30 | 45 | 75 | 35 | 25 | 15 |
It’s seen that the maximum frequency is 75.
So, the corresponding class, i.e. 20 – 25, is the modal class.
And,
l = 20, h = 25 – 20 = 5, f = 75, f1 = 45, f2 = 35
Using the formula for finding mode, we get
= 20 + 150/70
= 20 + 2.14
= 22.14
(iii)
| Class interval | 25 – 30 | 30 – 35 | 35 – 40 | 40 – 45 | 45 – 50 | 50 – 55 |
| Frequency | 25 | 34 | 50 | 42 | 38 | 14 |
Solution:
| Class interval | 25 – 30 | 30 – 35 | 35 – 40 | 40 – 45 | 45 – 50 | 50 – 55 |
| Frequency | 25 | 34 | 50 | 42 | 38 | 14 |
It’s seen that the maximum frequency is 50.
So, the corresponding class, i.e., 35 – 40, is the modal class.
And,
l = 35, h = 40 – 35 = 5, f = 50, f1 = 34, f2 = 42
Using the formula for finding mode, we get
= 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 |
Solution:
| 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:
It’s seen that the maximum frequency is 78.
So, the corresponding class 18 – 20 is the model class.
And,
l = 18, h = 20 – 18 = 2, f = 78, f1 = 50, f2 = 46
Using the formula for finding mode, we get
= 18 + 56/60
= 18 + 0.93
= 18.93 years
For group B:
It’s seen that the maximum frequency is 89
So, the corresponding class 18 – 20 is the modal class.
And,
l = 18, h = 20 – 18 = 2, f = 89, f1 = 54, f2 = 40
Using the formula for finding mode, we get
Mode
= 18 + 70/84
= 18 + 0.83
= 18.83 years
Therefore, the modal age of Group A is higher than that of Group B.
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 |
Solution:
| 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 |
It’s seen that the maximum frequency is 20.
So, the corresponding class 50 – 60 is the modal class.
And,
l = 50, h = 60 – 50 = 10, f = 20, f1 = 13, f2 = 5
Using the formula for finding mode, we get
= 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 maximum number of students.
Solution:
| 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 |
It’s seen that the maximum frequency is 142.
So, the corresponding class 165.5 – 168.5 is the modal class.
And,
l = 165.5, h = 168.5 – 165.5 = 3, f = 142, f1 = 118, f2 = 127
Using the formula for finding mode, we get
= 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.
Solution:
To find the mean:
For the given data, let the assumed mean (A) = 30
| Age (in years) | Number of patients fi | Class marks xi | di = xi – 275 | 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 |
| N = 80 | Σfi di = 430 |
It’s observed from the table that Σfi = N = 80 and Σfi di = 430.
Using the formula for mean,
= 30 + 430/80
= 30 + 5.375
= 35.375
= 35.38
Thus, the mean of this data is 35.38. It can also be interpreted that, on average, the age of patients admitted to the hospital was 35.38 years.
It is also observed that the maximum class frequency is 23, and it belongs to the class interval 35 – 45
So, the modal class is 35 – 45 with the Lower limit (l) of modal class = 35
And, Frequency (f) of modal class = 23
Class size (h) = 10
Frequency (f1) of class preceding the modal class = 21
Frequency (f2) of class succeeding the modal class = 14
Mode
Therefore, the mode is 36.8. This represents that the maximum number of patients admitted to the hospital was of 36.8 years.
Hence, it’s seen that mode is greater than the mean.
8. 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 |
| No. of components: | 10 | 35 | 52 | 61 | 38 | 29 |
Determine the modal lifetimes of the components.
Solution:
From the data given above it is observed that the maximum class frequency is 61, which belongs to the class interval 60 – 80.
So, modal class limit (l) of modal class = 60
Frequency (f) of modal class = 61
Frequency (f1) of class preceding the modal class = 52
Frequency (f2) of class succeeding the modal class = 38
Class size (h) = 20
Using the formula for find mode, we have
Mode
Thus, the modal lifetime of electrical components is 65.625 hours
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.
Solution:
| 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 | Σfx = 7260 |
We know that,
Mean = Σfx / N
= 7260/ 50
= 145.2
Then,
We have, N = 50
⇒ N/2 = 50/2 = 25
So, the cumulative frequency just greater than N/2 is 26, then the median class is 120 – 140
Such that l = 120, h = 140 – 120 = 20, f = 14, F = 12
= 120 + 260/14
= 120 + 18.57
= 138.57
From the data, its observed that maximum frequency is 14, so the corresponding class 120 – 140 is the modal class
And,
l = 120, h = 140 – 120 = 20, f = 14, f1 = 12, f2 = 8
= 120 + 5
= 125
Therefore, mean = 145.2, median = 138.57 and mode = 125
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