In this chapter, we provide RD Sharma Class 10 Ex 7.5 Solutions Chapter 7 Statistics for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 10 Ex 7.5 Solutions Chapter 7 Statistics pdf, Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 7 |

Chapter Name | Statistics |

Exercise | 7.5 |

Category | RD Sharma Solutions |

**RD Sharma Solutions for Class 10 Chapter** **7**** Statistics** Ex 7.5 Download PDF

**Statistics**Ex 7.5 Download PDF

**Chapter 7: Statistics Exercise – 7.5**

**Question: 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 |

Mode = 5 because 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 |

Mode = 3 because it occurs maximum number of times.

(iii)

Value (x) | 8 | 15 | 18 | 19 | 20 | 24 | 25 | 26 |

Frequency (f) | 1 | 4 | 1 | 1 | 1 | 2 | 1 | 1 |

Mode = 15 because it occurs maximum number of times.

**Question: 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 |

Model shirt size = 40 because it occurs maximum number of times.

**Question: 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 – 55 |

Frequency: | 25 | 34 | 50 | 42 | 29 | 15 |

**Solution:**

(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 the maximum frequency is 28 then the corresponding class 40 – 50 is the modal class

l = 40, h = 50 40 = 10, f = 28, f_{1} = 12, f_{2 }= 20

= 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 |

Here the maximum frequency is 75, then the corresponding class 20 – 25 is the modal class

l = 20, h = 25 – 20 = 5, f = 75, f_{1} = 45, f_{2 }= 35

= 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 |

Here the maximum frequency is 50 then the corresponding class 35 – 40 is the modal class

l = 35, h = 40 – 35 = 5, f = 50, f_{1} = 34, f_{2 }= 42

= 35 + 80/24

= 35 + 3.33

= 38.33

**Question: 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 Here the maximum frequency is 78, then the corresponding class 18 – 20 is model class l = 18, h = 20 – 18 = 2, f = 78, f_{1} = 50, f_{2 }= 46

= 18 + 56/60

= 18 + 0.93

= 18.93 years

For group B Here the maximum frequency is 89, then the corresponding class 18 – 20 is the modal class l = 18, h = 20 – 18 = 2, f = 89, f_{1} = 54, f_{2 }= 40 Mode

= 18 + 70/84

= 18 + 0.83

= 18.83 years

Hence the modal age for the Group A is higher than that for Group B

**Question: 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 |

Here the maximum frequency is 20, then the corresponding class 50 – 60 is the modal class l = 50, h = 60 – 50 = 10, f = 20, f_{1} = 13, f_{2 }= 5

= 50 + 70/22

= 50 + 3.18

= 53.18

**Question: 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 |

Here the maximum frequency is 142, then the corresponding class 165.5 – 168.5 is the modal class l = 165.5, h = 168.5 – 165.5 = 3, f = 142, f_{1} = 118, f_{2} = 127

= 165.5 + 72/39

= 165.5 + 1.85

= 167.35 cm

**Question: 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:**

We may compute class marks (x_{i}) as per the relation

Now taking 30 as assumed mean (a) we may calculate d_{i }and f_{i}d_{i }as follows:

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 we may observe that Σf_{i} = 80, Σf_{i }d_{i} = 430,

= 30 + 430/80

= 30 + 5.375

= 35.375

= 35.38

Clearly, mean of this data is 35.38. It represents that on an average the age of a patients admitted to hospital was 35.38 years.

As we may observe that maximum class frequency is 23 belonging to class interval 35 – 45

So, modal class = 35 – 45 Lower limit (l) of modal class = 35 Frequency (f) of modal class = 23

Class size (h) = 10 Frequency (f_{1}) of class preceding the modal class = 21 Frequency (f_{2}) of class succeeding the modal class = 14 Mode

Clearly mode is 36.8. It represents that maximum number of patients admitted in hospital were of 36.8 years.

**Question: 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 as above we may observe that maximum class frequency is 61 belonging to class interval 60 – 80

So, modal class limit (l) of modal class = 60 Frequency (f) of modal class = 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 Mode

So, modal lifetime of electrical components is 65.625 hours

**Question: 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 have, N = 50 Then, N/2 = 50/2 = 25

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

Here the maximum frequency is 14, then the corresponding class 120 – 140 is the modal class l = 120, h = 140 – 120 = 20, f = 14, f_{1 }= 12, f_{2 }= 8

= 120 + 5

= 125

**Question: 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 |

**Solution:**

We may observe from the given data that maximum class frequency is 10 belonging to class interval 30 – 35.

So, modal class = 30 – 35 Class size (h) = 5 Lower limit (l) of modal class = 30 Frequency (f) of modal class = 10 Frequency (f_{1}) of class preceding modal class = 9

Frequency (f_{2}) of class succeeding modal class = 3

Mode = 30.6 It represents that most of states/ U.T have a teacher- student ratio as 30.6 Now we may find class marks by using the relation Class mark = (upper class limit + lower class limit) /2.

Now taking 32.5 as assumed mean (a) we may calculate d_{i}, u_{i}, and f_{i}u_{i }as following

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,

So mean of data is 29.2 It represents that on an average teacher-student ratio was 29.2

**Question: 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 |

**Solution:**

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 | Σfx = 4225 |

We have, N = 25

Then, N/2

= 25/2

= 12.5

The cumulative frequency just greater than N/2 is 16, then the median class is 150 – 200 such that l = 150, h = 200 – 150 = 50, f = 6, F = 10

= 150 + 125/6

= 150 + 20.83

= 170.83

Here the maximum frequency is 6, then the corresponding class 150 – 200 is the modal class l = 150, h = 200 – 150 = 50, f = 6, f_{1 }= 5, f_{2 }= 5

= 150 + 50/2

= 150 + 25

= 175

**Question: 12**

A students noted the number of cars pass through a spot on a road for 100 periods each of 3 minute and summarized it in the table given below. Find the mode of the data.

**Solution:**

From the given data we may observe that maximum class frequency is 20 belonging to 40 – 50 class intervals.

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

= 40 + 80/17

= 40 + 4.7

= 44.7

So mode of this data is 44.7 cars

**Question: 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 |

**Solution:**

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 | Σfx = 9320 |

We have, N = 68 N/2 = 68/2 = 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

= 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

= 125 + 140/13

= 135.77

**Question: 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.

**Solution:**

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 | Σfx = 832 |

We have, N = 100 N/2 = 100/2 = 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

= 7 + 1.05

= 8.05

Here the maximum frequency is 40, then the corresponding class 7 – 10 is the modal class l = 7, h = 10 – 7 = 3, f = 40, f_{1 }= 30, f_{2 }= 16

= 7 + 30/34

= 7 + 0.88

= 7.88

**Question: 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 |

**Solution:**

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 | Σfx = 3120 |

We have, N = 50 Then, N/2 = 50/2 = 25

The cumulative frequency just greater than N/2 is 36, then the median class is 60 – 80 such that l = 60, h = 80 – 60 = 20, f = 12, F = 24

= 60 + 20/12

= 60 + 1.67

= 61.67

Here the maximum frequency is 12, then the corresponding class 60 – 80 is the modal class l = 60, h = 80 – 60 = 20, f = 12, f_{1 }= 10, f_{2 }= 6

**Question: 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 |

**Solution:**

We may observe from the given data that maximum class frequency is 40 belonging to 1500 -200 intervals. So, modal class = 1500 -2000

Lower limit (l) of modal class = 1500 Frequency (f) of modal class = 40 Frequency (f_{1}) of class preceding modal class = 24

Frequency (f_{2}) of class succeeding modal class = 33 Class size (h) = 500

= 1500 + 347.826

= 1847.826

1847.83

So modal monthly expenditure was Rs. 1847.83

Now we may find class mark as Class mark

Class size (h) of given data = 500 Now taking 2750 as assumed mean (a) we may calculate d_{i }u_{i }as follows:

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 |

Now from table may observe that Σf_{i} = 200, Σf_{i}d_{i} = -35

So mean monthly expenditure was Rs. 2662.5

**Question: 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.

**Solution:**

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. Now

= 4000 + 608.695

= 4608.695

So mode of given data is 4608.7 runs

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