RD Sharma Solutions Class 10 Statistics Exercise 7.5

RD Sharma Class 10 Solutions Chapter 7 Ex 7.5 PDF Free Download

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, f1 = 12, f2 = 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, f1 = 45, f2 = 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, f1 = 34, f2 = 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, f1 = 50, f2 = 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, f1 = 54, f2 = 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, f1 = 13, f2 = 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, f1 = 118, f2 = 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 (xi) can be computed using the following formula:

xi = \( \frac {upper\, class\, limit\; +\; lower\, class\, limit}{2} \)

Now, consider 30 as assumed mean (a) and calculate di and fidi as given in the table below:

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
Total 80 430

From the table it can be observed that:

Σfi = 80

Σfidi = 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, f1 = 21, and f2 = 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 (f1) of class preceding the modal class=52

Frequency (f2) 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, f1 = 12, f2 = 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, f1 = 9, f2 = 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 di, ui, and fiui as given below:

Number of students per teacher Number of states/ U.T (fi) xi di = xi – 32.5 Ui 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

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, f1 = 5, f2 = 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 (f1) of class preceding modal class=12

Frequency (f2) 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 xi Frequency fi 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, f1 = 13, f2 = 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, f1 = 30, f2 = 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, f1 = 10, f2 = 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

f1 of class preceding modal class=24

f2 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 di ui as given:

Expenditure (in Rs) Number of families fi xi di = xi – 2750 Ui 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 data table, it can be observed that

Σfi= 200

Σfidi = -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

\(\overline{x}\) = 2662.5

Hence, 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 (f1) of class preceding modal class=4

Frequency (f2) 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

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