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RD Sharma solutions for Class 10 Mathematics chapter 15 - Statistics

10 Mathematics

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Chapters

RD Sharma 10 Mathematics

10 Mathematics

Chapter 15: Statistics

Ex. 15.1Ex. 15.2Ex. 15.3Ex. 15.4Ex. 15.5Ex. 115.5Ex. 15.6Others

Chapter 15: Statistics Exercise 15.1 solutions [Pages 5 - 6]

Ex. 15.1 | Q 1 | Page 5

Calculate the mean for the following distribution:-

x 5 6 7 8 9
f 4 8 14 11 3
Ex. 15.1 | Q 1 | Page 5

Calculate the mean for the following distribution:-

x 5 6 7 8 9
f 4 8 14 11 3
Ex. 15.1 | Q 2 | Page 5

Find the mean of the following data:-

x 19 21 23 25 27 29 31
f 13 15 16 18 16 15 13
Ex. 15.1 | Q 2 | Page 5

Find the mean of the following data:-

x 19 21 23 25 27 29 31
f 13 15 16 18 16 15 13
Ex. 15.1 | Q 3 | Page 5

If the mean of the following data is 20.6. Find the value of p.

x 10 15 P 25 35
f 3 10 25 7 5
Ex. 15.1 | Q 3 | Page 5

If the mean of the following data is 20.6. Find the value of p.

x 10 15 P 25 35
f 3 10 25 7 5
Ex. 15.1 | Q 4 | Page 5

If the mean of the following data is 15, find p.

x 5 10 15 20 25
f 6 P 6 10 5
Ex. 15.1 | Q 4 | Page 5

If the mean of the following data is 15, find p.

x 5 10 15 20 25
f 6 P 6 10 5
Ex. 15.1 | Q 5 | Page 5

Find the value of p for the following distribution whose mean is 16.6

x 8 12 15 P 20 25 30
f 12 16 20 24 16 8 4
Ex. 15.1 | Q 5 | Page 5

Find the value of p for the following distribution whose mean is 16.6

x 8 12 15 P 20 25 30
f 12 16 20 24 16 8 4
Ex. 15.1 | Q 6 | Page 5

Find the missing value of p for the following distribution whose mean is 12.58

x 5 8 10 12 P 20 25
f 2 5 8 22 7 4 2
Ex. 15.1 | Q 6 | Page 5

Find the missing value of p for the following distribution whose mean is 12.58

x 5 8 10 12 P 20 25
f 2 5 8 22 7 4 2
Ex. 15.1 | Q 7 | Page 5

Find the missing frequency (p) for the following distribution whose mean is 7.68.

x 3 5 7 9 11 13
f 6 8 15 P 8 4
Ex. 15.1 | Q 7 | Page 5

Find the missing frequency (p) for the following distribution whose mean is 7.68.

x 3 5 7 9 11 13
f 6 8 15 P 8 4
Ex. 15.1 | Q 8 | Page 5

The following table gives the number of boys of a particular age in a class of 40 students. Calculate the mean age of the students

Age (in years) 15 16 17 18 19 20
No. of students 3 8 10 10 5 4
Ex. 15.1 | Q 8 | Page 5

The following table gives the number of boys of a particular age in a class of 40 students. Calculate the mean age of the students

Age (in years) 15 16 17 18 19 20
No. of students 3 8 10 10 5 4
Ex. 15.1 | Q 9 | Page 6

Candidates of four schools appear in a mathematics test. The data were as follows:-

Schools No. of Candidates Average Score
I 60 75
II 48 80
III NA 55
IV 40 50

If the average score of the candidates of all the four schools is 66, find the number of candidates that appeared from school III.

Ex. 15.1 | Q 9 | Page 6

Candidates of four schools appear in a mathematics test. The data were as follows:-

Schools No. of Candidates Average Score
I 60 75
II 48 80
III NA 55
IV 40 50

If the average score of the candidates of all the four schools is 66, find the number of candidates that appeared from school III.

Ex. 15.1 | Q 10 | Page 6

Five coins were simultaneously tossed 1000 times and at each toss the number of heads were observed. The number of tosses during which 0, 1, 2, 3, 4 and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss.

No. of heads per toss No. of tosses
0 38
1 144
2 342
3 287
4 164
5 25
Total 1000
Ex. 15.1 | Q 10 | Page 6

Five coins were simultaneously tossed 1000 times and at each toss the number of heads were observed. The number of tosses during which 0, 1, 2, 3, 4 and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss.

No. of heads per toss No. of tosses
0 38
1 144
2 342
3 287
4 164
5 25
Total 1000
Ex. 15.1 | Q 11 | Page 6

The arithmetic mean of the following data is 14. Find the value of k

x1 5 10 15 20 25
f1 7 k 8 4 5
Ex. 15.1 | Q 11 | Page 6

The arithmetic mean of the following data is 14. Find the value of k

x1 5 10 15 20 25
f1 7 k 8 4 5
Ex. 15.1 | Q 12 | Page 6

The arithmetic mean of the following data is 25, find the value of k.

x1 5 15 25 35 45
f1 3 k 3 6 2
Ex. 15.1 | Q 12 | Page 6

The arithmetic mean of the following data is 25, find the value of k.

x1 5 15 25 35 45
f1 3 k 3 6 2
Ex. 15.1 | Q 13 | Page 6

If the mean of the following data is 18.75. Find the value of p.

x 10 15 P 25 30
f 5 10 7 8 2
Ex. 15.1 | Q 13 | Page 6

If the mean of the following data is 18.75. Find the value of p.

x 10 15 P 25 30
f 5 10 7 8 2
Ex. 15.1 | Q 14 | Page 6

Find the value of p, if the mean of the following distribution is 20.

x 15 17 19 20+P 23
f 2 3 4 5P 6
Ex. 15.1 | Q 14 | Page 6

Find the value of p, if the mean of the following distribution is 20.

x 15 17 19 20+P 23
f 2 3 4 5P 6
Ex. 15.1 | Q 15 | Page 6

Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is 50.

x 10 30 50 70 90  
f 17 f1 32 f2 19 Total 120
Ex. 15.1 | Q 15 | Page 6

Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is 50.

x 10 30 50 70 90  
f 17 f1 32 f2 19 Total 120

Chapter 15: Statistics Exercise 15.2 solutions [Pages 13 - 14]

Ex. 15.2 | Q 1 | Page 13

The number of telephone calls received at an exchange per interval for 250 successive one minute
intervals are given in the following frequency table:

No. of calls(x) 0 1 2 3 4 5 6
No. of intervals (f) 15 24 29 46 54 43 39

Compute the mean number of calls per interval.

Ex. 15.2 | Q 1 | Page 13

The number of telephone calls received at an exchange per interval for 250 successive one minute
intervals are given in the following frequency table:

No. of calls(x) 0 1 2 3 4 5 6
No. of intervals (f) 15 24 29 46 54 43 39

Compute the mean number of calls per interval.

Ex. 15.2 | Q 2 | Page 13

Five coins were simultaneously tossed 1000 times and at each toss the number of heads were observed. The number of tosses during which 0, 1, 2, 3, 4 and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss.

No. of heads per toss No. of tosses
0 38
1 144
2 342
3 287
4 164
5 25
Total 1000
Ex. 15.2 | Q 2 | Page 13

Five coins were simultaneously tossed 1000 times and at each toss the number of heads were observed. The number of tosses during which 0, 1, 2, 3, 4 and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss.

No. of heads per toss No. of tosses
0 38
1 144
2 342
3 287
4 164
5 25
Total 1000
Ex. 15.2 | Q 3 | Page 13

The following table gives the number of branches and number of plants in the garden of a school.

No. of branches (x) 2 3 4 5 6
No. of plants (f) 49 43 57 38 13

Calculate the average number of branches per plant.

Ex. 15.2 | Q 3 | Page 13

The following table gives the number of branches and number of plants in the garden of a school.

No. of branches (x) 2 3 4 5 6
No. of plants (f) 49 43 57 38 13

Calculate the average number of branches per plant.

Ex. 15.2 | Q 4 | Page 13

The following table gives the number of children of 150 families in a village. Find the average number of children per family.

No. of children (x) 0 1 2 3 4 5
No. of families (f) 10 21 55 42 15 7
Ex. 15.2 | Q 4 | Page 13

The following table gives the number of children of 150 families in a village. Find the average number of children per family.

No. of children (x) 0 1 2 3 4 5
No. of families (f) 10 21 55 42 15 7
Ex. 15.2 | Q 5 | Page 13

The marks obtained out of 50, by 102 students in a Physics test are given in the frequency table below:

Marks(x) 15 20 22 24 25 30 33 38 45
Frequency (f) 5 8 11 20 23 18 13 3 1

Find the average number of marks.

Ex. 15.2 | Q 5 | Page 13

The marks obtained out of 50, by 102 students in a Physics test are given in the frequency table below:

Marks(x) 15 20 22 24 25 30 33 38 45
Frequency (f) 5 8 11 20 23 18 13 3 1

Find the average number of marks.

Ex. 15.2 | Q 6 | Page 13

The number of students absent in a class were recorded every day for 120 days and the information is given in the following frequency table:

No. of students absent (x) 0 1 2 3 4 5 6 7
No. of days (f) 1 4 10 50 34 15 4 2

Find the mean number of students absent per day.

Ex. 15.2 | Q 6 | Page 13

The number of students absent in a class were recorded every day for 120 days and the information is given in the following frequency table:

No. of students absent (x) 0 1 2 3 4 5 6 7
No. of days (f) 1 4 10 50 34 15 4 2

Find the mean number of students absent per day.

Ex. 15.2 | Q 7 | Page 13

In the first proof reading of a book containing 300 pages the following distribution of misprints was obtained:

No. of misprints per page (x) 0 1 2 3 4 5
No. of pages (f) 154 95 36 9 5 1

Find the average number of misprints per page.

Ex. 15.2 | Q 7 | Page 13

In the first proof reading of a book containing 300 pages the following distribution of misprints was obtained:

No. of misprints per page (x) 0 1 2 3 4 5
No. of pages (f) 154 95 36 9 5 1

Find the average number of misprints per page.

Ex. 15.2 | Q 8 | Page 13

The following distribution gives the number of accidents met by 160 workers in a factory during a month.

No. of accidents(x) 0 1 2 3 4
No. of workers (f) 70 52 34 3 1

Find the average number of accidents per worker.

Ex. 15.2 | Q 8 | Page 13

The following distribution gives the number of accidents met by 160 workers in a factory during a month.

No. of accidents(x) 0 1 2 3 4
No. of workers (f) 70 52 34 3 1

Find the average number of accidents per worker.

Ex. 15.2 | Q 9 | Page 14

Find the mean from the following frequency distribution of marks at a test in statistics:

Marks(x) 5 10 15 20 25 30 35 40 45 50
No. of students (f) 15 50 80 76 72 45 39 9 8 6
Ex. 15.2 | Q 9 | Page 14

Find the mean from the following frequency distribution of marks at a test in statistics:

Marks(x) 5 10 15 20 25 30 35 40 45 50
No. of students (f) 15 50 80 76 72 45 39 9 8 6

Chapter 15: Statistics Exercise 15.3 solutions [Pages 22 - 25]

Ex. 15.3 | Q 1 | Page 22

The following table gives the distribution of total household expenditure (in rupees) of manual workers in a city. Find the average expenditure (in rupees) per household.

Expenditure
(in rupees) (x1)
Frequency(f1)
100 - 150 24
150 - 200 40
200 - 250 33
250 - 300 28
300 - 350 30
350 - 400 22
400 - 450 16
450 - 500 7
Ex. 15.3 | Q 1 | Page 22

The following table gives the distribution of total household expenditure (in rupees) of manual workers in a city. Find the average expenditure (in rupees) per household.

Expenditure
(in rupees) (x1)
Frequency(f1)
100 - 150 24
150 - 200 40
200 - 250 33
250 - 300 28
300 - 350 30
350 - 400 22
400 - 450 16
450 - 500 7
Ex. 15.3 | Q 2 | Page 22

A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Number of plants 0 - 2 2 - 4 4 - 6 6 - 8 8 - 10 10 - 12 12 - 14
Number of houses 1 2 1 5 6 2 3

Which method did you use for finding the mean, and why?

Ex. 15.3 | Q 2 | Page 22

A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Number of plants 0 - 2 2 - 4 4 - 6 6 - 8 8 - 10 10 - 12 12 - 14
Number of houses 1 2 1 5 6 2 3

Which method did you use for finding the mean, and why?

Ex. 15.3 | Q 3 | Page 22

Consider the following distribution of daily wages of 50 worker of a factory.

Daily wages (in Rs)

100­ − 120

120­ − 140

140 −1 60

160 − 180

180 − 200

Number of workers

12

14

8

6

10

Find the mean daily wages of the workers of the factory by using an appropriate method.

Ex. 15.3 | Q 3 | Page 22

Consider the following distribution of daily wages of 50 worker of a factory.

Daily wages (in Rs)

100­ − 120

120­ − 140

140 −1 60

160 − 180

180 − 200

Number of workers

12

14

8

6

10

Find the mean daily wages of the workers of the factory by using an appropriate method.

Ex. 15.3 | Q 4 | Page 22

Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarized as follows. Fine the mean heart beats per minute for these women, choosing a suitable method.

 

Number of heart beats per minute 65 − 68 68­ − 71 71 −74 74 − 77 77 − 80 80 − 83 83 − 86
Number of women 2 4 3 8 7 4 2
Ex. 15.3 | Q 4 | Page 22

Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarized as follows. Fine the mean heart beats per minute for these women, choosing a suitable method.

 

Number of heart beats per minute 65 − 68 68­ − 71 71 −74 74 − 77 77 − 80 80 − 83 83 − 86
Number of women 2 4 3 8 7 4 2
Ex. 15.3 | Q 5 | Page 22

Find the mean of each of the following frequency distributions: (5 - 14)

Class interval 0 - 6 6 - 12 12 - 18 18 - 24 24 - 30
Frequency 6 8 10 9 7
Ex. 15.3 | Q 5 | Page 22

Find the mean of each of the following frequency distributions: (5 - 14)

Class interval 0 - 6 6 - 12 12 - 18 18 - 24 24 - 30
Frequency 6 8 10 9 7
Ex. 15.3 | Q 6 | Page 22

Find the mean of each of the following frequency distributions

Class interval 50 - 70 70 - 90 90 - 110 110 - 130 130 - 150 150 - 170
Frequency 18 12 13 27 8 22
Ex. 15.3 | Q 6 | Page 22

Find the mean of each of the following frequency distributions

Class interval 50 - 70 70 - 90 90 - 110 110 - 130 130 - 150 150 - 170
Frequency 18 12 13 27 8 22
Ex. 15.3 | Q 7 | Page 23

Find the mean of each of the following frequency distributions

Class interval 0 - 8 8 - 16 16 - 24 24 - 32 32 - 40
Frequency 6 7 10 8 9
Ex. 15.3 | Q 7 | Page 23

Find the mean of each of the following frequency distributions

Class interval 0 - 8 8 - 16 16 - 24 24 - 32 32 - 40
Frequency 6 7 10 8 9
Ex. 15.3 | Q 8 | Page 23

Find the mean of each of the following frequency distributions

Class interval 0 - 6 6 - 12 12 - 18 18 - 24 24 - 30
Frequency 7 5 10 12 6
Ex. 15.3 | Q 8 | Page 23

Find the mean of each of the following frequency distributions

Class interval 0 - 6 6 - 12 12 - 18 18 - 24 24 - 30
Frequency 7 5 10 12 6
Ex. 15.3 | Q 9 | Page 23

Find the mean of each of the following frequency distributions

Class interval 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50
Frequency 9 12 15 10 14
Ex. 15.3 | Q 9 | Page 23

Find the mean of each of the following frequency distributions

Class interval 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50
Frequency 9 12 15 10 14
Ex. 15.3 | Q 10 | Page 23

Find the mean of each of the following frequency distributions

Class interval 0 - 8 8 - 16 16 - 24 24 - 32 32 - 40
Frequency 5 9 10 8 8
Ex. 15.3 | Q 10 | Page 23

Find the mean of each of the following frequency distributions

Class interval 0 - 8 8 - 16 16 - 24 24 - 32 32 - 40
Frequency 5 9 10 8 8
Ex. 15.3 | Q 11 | Page 23

Find the mean of each of the following frequency distributions

Class interval 0 - 8 8 - 16 16 - 24 24 - 32 32 - 40
Frequency 5 6 4 3 2
Ex. 15.3 | Q 11 | Page 23

Find the mean of each of the following frequency distributions

Class interval 0 - 8 8 - 16 16 - 24 24 - 32 32 - 40
Frequency 5 6 4 3 2
Ex. 15.3 | Q 12 | Page 23

Find the mean of each of the following frequency distributions

Class interval 10 - 30 30 - 50 50 - 70 70 - 90 90 - 110 110 - 130
Frequency 5 8 12 20 3 2
Ex. 15.3 | Q 12 | Page 23

Find the mean of each of the following frequency distributions

Class interval 10 - 30 30 - 50 50 - 70 70 - 90 90 - 110 110 - 130
Frequency 5 8 12 20 3 2
Ex. 15.3 | Q 13 | Page 23

Find the mean of each of the following frequency distributions

Class interval 25 - 35 35 - 45 45 - 55 55 - 65 65 - 75
Frequency 6 10 8 12 4
Ex. 15.3 | Q 13 | Page 23

Find the mean of each of the following frequency distributions

Class interval 25 - 35 35 - 45 45 - 55 55 - 65 65 - 75
Frequency 6 10 8 12 4
Ex. 15.3 | Q 14 | Page 23

Find the mean of each of the following frequency distributions

Classes 25 - 29 30 - 34 35 - 39 40 - 44 45 - 49 50 - 54 55 - 59
Frequency 14 22 16 6 5 3 4
Ex. 15.3 | Q 14 | Page 23

Find the mean of each of the following frequency distributions

Classes 25 - 29 30 - 34 35 - 39 40 - 44 45 - 49 50 - 54 55 - 59
Frequency 14 22 16 6 5 3 4
Ex. 15.3 | Q 15 | Page 23

For the following distribution, calculate mean using all suitable methods:

Size of item 1 - 4 4 - 9 9 - 16 16 - 27
Frequency 6 12 26 20
Ex. 15.3 | Q 15 | Page 23

For the following distribution, calculate mean using all suitable methods:

Size of item 1 - 4 4 - 9 9 - 16 16 - 27
Frequency 6 12 26 20
Ex. 15.3 | Q 16 | Page 23

The weekly observations on cost of living index in a certain city for the year 2004 - 2005 are given below. Compute the weekly cost of living index.

Cost of living Index Number of Students
1400 - 1500 5
1500 - 1600 10
1600 - 1700 20
1700 - 1800 9
1800 - 1900 6
1900 - 2000 2
Ex. 15.3 | Q 16 | Page 23

The weekly observations on cost of living index in a certain city for the year 2004 - 2005 are given below. Compute the weekly cost of living index.

Cost of living Index Number of Students
1400 - 1500 5
1500 - 1600 10
1600 - 1700 20
1700 - 1800 9
1800 - 1900 6
1900 - 2000 2
Ex. 15.3 | Q 17 | Page 23

The following table shows the marks scored by 140 students in an examination of a certain paper:

Marks: 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50
Number of students: 20 24 40 36 20

Calculate the average marks by using all the three methods: direct method, assumed mean deviation and shortcut method.

Ex. 15.3 | Q 17 | Page 23

The following table shows the marks scored by 140 students in an examination of a certain paper:

Marks: 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50
Number of students: 20 24 40 36 20

Calculate the average marks by using all the three methods: direct method, assumed mean deviation and shortcut method.

Ex. 15.3 | Q 18 | Page 23

The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Compute the missing frequency f1 and f2.

Class 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 100 - 120
Frequency 5 f1 10 f2 7 8
Ex. 15.3 | Q 18 | Page 23

The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Compute the missing frequency f1 and f2.

Class 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 100 - 120
Frequency 5 f1 10 f2 7 8
Ex. 15.3 | Q 19 | Page 24

The following distribution shows the daily pocket allowance given to the children of a multistorey building. The average pocket allowance is Rs 18.00. Find out the missing frequency.

Class interval 11 - 13 13 - 15 15 - 17 17 - 19 19 - 21 21 - 23 23 - 25
Frequency 7 6 9 13 - 5 4
Ex. 15.3 | Q 19 | Page 24

The following distribution shows the daily pocket allowance given to the children of a multistorey building. The average pocket allowance is Rs 18.00. Find out the missing frequency.

Class interval 11 - 13 13 - 15 15 - 17 17 - 19 19 - 21 21 - 23 23 - 25
Frequency 7 6 9 13 - 5 4
Ex. 15.3 | Q 20 | Page 24

If the mean of the following distribution is 27, find the value of p.

Class 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50
Frequency 8 p 12 13 10
Ex. 15.3 | Q 20 | Page 24

If the mean of the following distribution is 27, find the value of p.

Class 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50
Frequency 8 p 12 13 10
Ex. 15.3 | Q 21 | Page 24

In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Number of mangoe 50 − 52 53 − 55 56 − 58 59 − 61 62 − 64
Number of boxes 15 110 135 115 25

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Ex. 15.3 | Q 21 | Page 24

In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Number of mangoe 50 − 52 53 − 55 56 − 58 59 − 61 62 − 64
Number of boxes 15 110 135 115 25

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Ex. 15.3 | Q 22 | Page 24

The table below shows the daily expenditure on food of 25 households in a locality.

Daily expenditure (in Rs) 100 − 150 150 − 200 200 − 250 250 − 300 300 − 350
Number of households 4 5 12 2 2

Find the mean daily expenditure on food by a suitable method.

Ex. 15.3 | Q 22 | Page 24

The table below shows the daily expenditure on food of 25 households in a locality.

Daily expenditure (in Rs) 100 − 150 150 − 200 200 − 250 250 − 300 300 − 350
Number of households 4 5 12 2 2

Find the mean daily expenditure on food by a suitable method.

Ex. 15.3 | Q 23 | Page 24

To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

concentration of SO2 (in ppm) Frequency
0.00 − 0.04 4
0.04 − 0.08 9
0.08 − 0.12 9
0.12 − 0.16 2
0.16 − 0.20 4
0.20 − 0.24 2

Find the mean concentration of SO2 in the air.

Ex. 15.3 | Q 23 | Page 24

To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

concentration of SO2 (in ppm) Frequency
0.00 − 0.04 4
0.04 − 0.08 9
0.08 − 0.12 9
0.12 − 0.16 2
0.16 − 0.20 4
0.20 − 0.24 2

Find the mean concentration of SO2 in the air.

Ex. 15.3 | Q 24 | Page 24

A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Number of days 0 − 6 6 − 10 10 − 14 14 − 20 20 − 28 28 − 38 38 − 40
Number of students 11 10 7 4 4 3 1

 

Ex. 15.3 | Q 24 | Page 24

A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Number of days 0 − 6 6 − 10 10 − 14 14 − 20 20 − 28 28 − 38 38 − 40
Number of students 11 10 7 4 4 3 1

 

Ex. 15.3 | Q 25 | Page 24

The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate

Literacy rate (in %) 45 − 55 55 − 65 65 − 75 75 − 85 85 − 95
Number of cities 3 10 11 8 3

 

Ex. 15.3 | Q 25 | Page 24

The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate

Literacy rate (in %) 45 − 55 55 − 65 65 − 75 75 − 85 85 − 95
Number of cities 3 10 11 8 3

 

Ex. 15.3 | Q 26 | Page 24

 The following is the cumulative frequency distribution ( of less than type ) of 1000 persons each of age 20 years and above . Determine the mean age .

Age below (in years): 30 40 50 60 70 80
Number of persons : 100 220 350 750 950 1000

 

Ex. 15.3 | Q 27 | Page 25

If the mean of the following frequency distribution is 18 , find the missing frequency .

Class interval : 11-13 13-15 15-17 17-19 19-21 21-23 23-25
Frequency : 3 6 9 13 f 5 4

 

Ex. 15.3 | Q 28 | Page 25

Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is 50.

x 10 30 50 70 90  
f 17 f1 32 f2 19 Total 120
Ex. 15.3 | Q 28 | Page 25

Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is 50.

x 10 30 50 70 90  
f 17 f1 32 f2 19 Total 120
Ex. 15.3 | Q 29 | Page 25

The daily income of a sample of 50 employees are tabulated as follows:

Income (in Rs.): 1-1200 201 -400 401-600 601 - 800
No.of employees : 14 15 14 7

Find the mean daily income of employees.

Chapter 15: Statistics Exercise 15.4 solutions [Pages 34 - 36]

Ex. 15.4 | Q 1 | Page 34

Following are the lives in hours of 15 pieces of the components of aircraft engine. Find the median:

715, 724, 725, 710, 729, 745, 694, 699, 696, 712, 734, 728, 716, 705, 719.

Ex. 15.4 | Q 1 | Page 34

Following are the lives in hours of 15 pieces of the components of aircraft engine. Find the median:

715, 724, 725, 710, 729, 745, 694, 699, 696, 712, 734, 728, 716, 705, 719.

Ex. 15.4 | Q 2 | Page 34

The following is the distribution of height of students of a certain class in a certain city:

Height (in cm): 160 - 162 163 - 165 166 - 168 169 - 171 172 - 174
No. of students: 15 118 142 127 18

Find the median height.

Ex. 15.4 | Q 2 | Page 34

The following is the distribution of height of students of a certain class in a certain city:

Height (in cm): 160 - 162 163 - 165 166 - 168 169 - 171 172 - 174
No. of students: 15 118 142 127 18

Find the median height.

Ex. 15.4 | Q 3 | Page 34

Following is the distribution of I.Q. of loo students. Find the median I.Q.

I.Q.: 55 - 64 65 - 74 75 - 84 85 - 94 95 - 104 105 - 114 115 - 124 125 - 134 135 - 144
No of Students: 1 2 9 22 33 22 8 2 1
Ex. 15.4 | Q 3 | Page 34

Following is the distribution of I.Q. of loo students. Find the median I.Q.

I.Q.: 55 - 64 65 - 74 75 - 84 85 - 94 95 - 104 105 - 114 115 - 124 125 - 134 135 - 144
No of Students: 1 2 9 22 33 22 8 2 1
Ex. 15.4 | Q 4 | Page 34

Calculate the median from the following data:

Rent (in Rs.): 15 - 25 25 - 35 35 - 45 45 - 55 55 - 65 65 - 75 75 - 85 85 - 95
No. of Houses: 8 10 15 25 40 20 15 7
Ex. 15.4 | Q 4 | Page 34

Calculate the median from the following data:

Rent (in Rs.): 15 - 25 25 - 35 35 - 45 45 - 55 55 - 65 65 - 75 75 - 85 85 - 95
No. of Houses: 8 10 15 25 40 20 15 7
Ex. 15.4 | Q 5 | Page 34

Calculate the median from the following data:

Marks below: 10 20 30 40 50 60 70 80
No. of students: 15 35 60 84 96 127 198 250
Ex. 15.4 | Q 5 | Page 34

Calculate the median from the following data:

Marks below: 10 20 30 40 50 60 70 80
No. of students: 15 35 60 84 96 127 198 250
Ex. 15.4 | Q 6 | Page 34

Calculate the missing frequency from the following distribution, it being given that the median of the distribution is 24.

Age in years 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50
No. of persons 5 25 ? 18 7
Ex. 15.4 | Q 6 | Page 34

Calculate the missing frequency from the following distribution, it being given that the median of the distribution is 24.

Age in years 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50
No. of persons 5 25 ? 18 7
Ex. 15.4 | Q 7 | Page 34

The following table gives the frequency distribution of married women by age at marriage:

Age (in years) Frequency
15-19 53
20-24 140
25-29 98
30-34 32
35-39 12
40-44 9
45-49 5
50-54 3
55-59 3
60 and above 2

Calculate the median and interpret the results.

Ex. 15.4 | Q 7 | Page 34

The following table gives the frequency distribution of married women by age at marriage:

Age (in years) Frequency
15-19 53
20-24 140
25-29 98
30-34 32
35-39 12
40-44 9
45-49 5
50-54 3
55-59 3
60 and above 2

Calculate the median and interpret the results.

Ex. 15.4 | Q 8 | Page 34

Find the following table gives the distribution of the life time of 400 neon lamps

Life time (in hours) Number of lamps
1500 − 2000 14
2000 − 2500 56
2500 − 3000 60
3000 − 3500 86
3500 − 4000 74
4000 − 4500 62
4500 − 5000 48

Find the median life time of a lamp.

Ex. 15.4 | Q 8 | Page 34

Find the following table gives the distribution of the life time of 400 neon lamps

Life time (in hours) Number of lamps
1500 − 2000 14
2000 − 2500 56
2500 − 3000 60
3000 − 3500 86
3500 − 4000 74
4000 − 4500 62
4500 − 5000 48

Find the median life time of a lamp.

Ex. 15.4 | Q 9 | Page 35

The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Weight (in kg) 40 − 45 45 − 50 50 − 55 55 − 60 60 − 65 65 − 70 70 − 75
Number of students 2 3 8 6 6 3 2
Ex. 15.4 | Q 9 | Page 35

The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Weight (in kg) 40 − 45 45 − 50 50 − 55 55 − 60 60 − 65 65 − 70 70 − 75
Number of students 2 3 8 6 6 3 2
Ex. 15.4 | Q 10 | Page 35

Find the missing frequencies and the median for the following distribution if the mean is 1.46.

No. of accidents: 0 1 2 3 4 5 Total
Frequency (No. of days): 46 ? ? 25 10 5 200
Ex. 15.4 | Q 10 | Page 35

Find the missing frequencies and the median for the following distribution if the mean is 1.46.

No. of accidents: 0 1 2 3 4 5 Total
Frequency (No. of days): 46 ? ? 25 10 5 200
Ex. 15.4 | Q 11 | Page 35

An incomplete distribution is given below:

Variable: 10-20 20-30 30-40 40-50 50-60 60-70 70-80
Frequency: 12 30 - 65 - 25 18

You are given that the median value is 46 and the total number of items is 230.

(i) Using the median formula fill up missing frequencies.

(ii) Calculate the AM of the completed distribution.

Ex. 15.4 | Q 11 | Page 35

An incomplete distribution is given below:

Variable: 10-20 20-30 30-40 40-50 50-60 60-70 70-80
Frequency: 12 30 - 65 - 25 18

You are given that the median value is 46 and the total number of items is 230.

(i) Using the median formula fill up missing frequencies.

(ii) Calculate the AM of the completed distribution.

Ex. 15.4 | Q 12 | Page 35

If the median of the distribution is given below is 28.5, find the values of x and y

Class interval Frequency
0 - 10 5
10 - 20 x
20 - 30 20
30 - 40 15
40 - 50 y
50 - 60 5
Total 60
Ex. 15.4 | Q 12 | Page 35

If the median of the distribution is given below is 28.5, find the values of x and y

Class interval Frequency
0 - 10 5
10 - 20 x
20 - 30 20
30 - 40 15
40 - 50 y
50 - 60 5
Total 60
Ex. 15.4 | Q 13 | Page 35

The median of the following data is 525. Find the missing frequency, if it is given that there are 100 observations in the data:

Class interval Frequency
0 - 100 2
100 - 200 5
200 - 300 f1
300 - 400 12
400 - 500 17
500 - 600 20
600 - 700 f2
700 - 800 9
800 - 900 7
900 - 1000 4
Ex. 15.4 | Q 13 | Page 35

The median of the following data is 525. Find the missing frequency, if it is given that there are 100 observations in the data:

Class interval Frequency
0 - 100 2
100 - 200 5
200 - 300 f1
300 - 400 12
400 - 500 17
500 - 600 20
600 - 700 f2
700 - 800 9
800 - 900 7
900 - 1000 4
Ex. 15.4 | Q 14 | Page 35

If the median of the following data is 32.5, find the missing frequencies.

Class interval: 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 Total
Frequency: f1 5 9 12 f2 3 2 40
Ex. 15.4 | Q 14 | Page 35

If the median of the following data is 32.5, find the missing frequencies.

Class interval: 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 Total
Frequency: f1 5 9 12 f2 3 2 40
Ex. 15.4 | Q 15.1 | Page 35

Compute the median for the following data:

Marks No. of students
Less than 10 0
Less than 30 10
Less than 50 25
Less than 70 43
Less than 90 65
Less than 110 87
Less than 130 96
Less than 150 100
Ex. 15.4 | Q 15.1 | Page 35

Compute the median for the following data:

Marks No. of students
Less than 10 0
Less than 30 10
Less than 50 25
Less than 70 43
Less than 90 65
Less than 110 87
Less than 130 96
Less than 150 100
Ex. 15.4 | Q 15.2 | Page 35

Compute the median for the following data:

Marks No. of students
More than 150 0
More than 140 12
More than 130 27
More than 120 60
More than 110 105
More than 100 124
More than 90 141
More than 80 150
Ex. 15.4 | Q 15.2 | Page 35

Compute the median for the following data:

Marks No. of students
More than 150 0
More than 140 12
More than 130 27
More than 120 60
More than 110 105
More than 100 124
More than 90 141
More than 80 150
Ex. 15.4 | Q 16 | Page 36

A survey regarding the height (in cm) of 51 girls of class X of a school was conducted and the following data was obtained:

Height in cm Number of Girls
Less than 140 4
Less than 145 11
Less than 150 29
Less than 155 40
Less than 160 46
Less than 165 51

Find the median height.

Ex. 15.4 | Q 16 | Page 36

A survey regarding the height (in cm) of 51 girls of class X of a school was conducted and the following data was obtained:

Height in cm Number of Girls
Less than 140 4
Less than 145 11
Less than 150 29
Less than 155 40
Less than 160 46
Less than 165 51

Find the median height.

Ex. 15.4 | Q 17 | Page 36

A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.

Age (in years) Number of policy holders
Below 20 2
Below 25 6
Below 30 24
Below 35 45
Below 40 78
Below 45 89
Below 50 92
Below 55 98
Below 60 100
Ex. 15.4 | Q 18 | Page 36

The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:

Length (in mm)

Number or leaves fi

118 − 126

3

127 − 135

5

136 − 144

9

145 − 153

12

154 − 162

5

163 − 171

4

172 − 180

2

Find the median length of the leaves.

(Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 − 126.5, 126.5 − 135.5… 171.5 − 180.5)

Ex. 15.4 | Q 18 | Page 36

The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:

Length (in mm)

Number or leaves fi

118 − 126

3

127 − 135

5

136 − 144

9

145 − 153

12

154 − 162

5

163 − 171

4

172 − 180

2

Find the median length of the leaves.

(Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 − 126.5, 126.5 − 135.5… 171.5 − 180.5)

Ex. 15.4 | Q 19 | Page 36

An incomplete distribution is given as follows:

Variable: 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70
Frequency: 10 20 ? 40 ? 25 15

You are given that the median value is 35 and the sum of all the frequencies is 170. Using the median formula, fill up the missing frequencies.

Ex. 15.4 | Q 19 | Page 36

An incomplete distribution is given as follows:

Variable: 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70
Frequency: 10 20 ? 40 ? 25 15

You are given that the median value is 35 and the sum of all the frequencies is 170. Using the median formula, fill up the missing frequencies.

Ex. 15.4 | Q 20 | Page 36

The median of the distribution given below is 14.4 . Find the values of x and y , if the total frequency is 20.

Class interval : 0-6 6-12 12-18 18-24  24-30
Frequency : 4 5 y 1
Ex. 15.4 | Q 21 | Page 36

The median of the following data is 50. Find the values of p and q  , if the sum of all  the frequencies is 90 .

Marks : 20 -30 30-40 40-50 50-60 60-70 70-80 80-90
Frequency : P 15 25 20 q 8 10
Ex. 15.4 | Q 21 | Page 36

The median of the following data is 50. Find the values of p and q  , if the sum of all  the frequencies is 90 .

Marks : 20 -30 30-40 40-50 50-60 60-70 70-80 80-90
Frequency : P 15 25 20 q 8 10

Chapter 15: Statistics Exercise 15.5, 115.5 solutions [Pages 45 - 48]

Ex. 15.5 | Q 1.1 | Page 45

Find the mode of the following data:

3, 5, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4

Ex. 15.5 | Q 1.2 | Page 45

Find the mode of the following data:

3, 3, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4

Ex. 15.5 | Q 1.3 | Page 45

Find the mode of the following data:

15, 8, 26, 25, 24, 15, 18, 20, 24, 15, 19, 15

Ex. 15.5 | Q 2 | Page 45

The shirt sizes 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.

Ex. 15.5 | Q 3.1 | Page 45

Find the mode of the following distribution.

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

 

Ex. 15.5 | Q 3.2 | Page 45

Find the mode of the following distribution.

Class-interval: 10 - 15 15 - 20 20 - 25 25 - 30 30 - 35 35 - 40
Frequency: 30 45 75 35 25 15

 

Ex. 15.5 | Q 3.3 | Page 45

Find the mode of the following distribution.

Class-interval: 25 - 30 30 - 35 35 - 40 40 - 45 45 - 50 50 - 55
Frequency: 25 34 50 42 38 14

 

Ex. 15.5 | Q 4 | Page 45

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
Ex. 15.5 | Q 5 | Page 46

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

 

Ex. 15.5 | Q 6 | Page 46

The following is the distribution of height of students of a certain class in a certain 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.

Ex. 15.5 | Q 7 | Page 46

The following table shows the ages of the patients admitted in a hospital during a year:

age (in years)

5 − 15

15 − 25

25 − 35

35 − 45

45 − 55

55 − 65

Number of patients

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.

Ex. 15.5 | Q 8 | Page 46

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
Frequency 10 35 52 61 38 29

Determine the modal lifetimes of the components.

Ex. 115.5 | Q 9 | Page 46

The following table gives the daily income of 50 workers of a factory:

Daily income (in Rs) 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.

Ex. 15.5 | Q 10 | Page 46

The following distribution gives the state-wise teacher-student 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
15 − 20 3
20 − 25 8
25 − 30 9
30 − 35 10
35 − 40 3
40 − 45 0
45 − 50 0
50 − 55 2
Ex. 15.5 | Q 11 | Page 46

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

 

Ex. 15.5 | Q 12 | Page 46

A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data:

Number of cars 0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80
Frequency 7 14 13 12 20 11 15 8
Ex. 15.5 | Q 13 | Page 46

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 (in units) Number of consumers
65 - 85 4
85 - 105 5
105 - 125 13
125 - 145 20
145 - 165 14
165 - 185 8
185 - 205 4
Ex. 15.5 | Q 14 | Page 47

100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

 

Number of letters Number of surnames
1 - 4 6
4 − 7 30
7 - 10 40
10 - 13 6
13 - 16 4
16 − 19 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.

Ex. 15.5 | Q 14 | Page 47

100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

 

Number of letters Number of surnames
1 - 4 6
4 − 7 30
7 - 10 40
10 - 13 6
13 - 16 4
16 − 19 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.

Ex. 15.5 | Q 15 | Page 47

Find the mean, median and mode of the following data:

Classes: 0-20 20-40 40-60 40-60 80-100 100-120 120-140
Frequency: 6 8 10 12 6 5 3

 

Ex. 15.5 | Q 16 | Page 47

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 (in Rs) Number of families
1000 − 1500 24
1500 − 2000 40
2000 − 2500 33
2500 − 3000 28
3000 − 3500 30
3500 − 4000 22
4000 − 4500 16
4500 − 5000 7
Ex. 15.5 | Q 17 | Page 47

The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches

Runs scored Number of batsmen
3000 − 4000 4
4000 − 5000 18
5000 − 6000 9
6000 − 7000 7
7000 − 8000 6
8000 − 9000 3
9000 − 10000 1
10000 − 11000 1

Find the mode of the data.

Ex. 15.5 | Q 18 | Page 47

The frequency distribution table of agriculture holdings in a village is given below:

Area of land (in hectares) :  1-3  3-5   5-7 7-9 9-11 11-13
Number of families : 20 45 80 55 40 12

Find the modal agricultural holdings of the village .

Ex. 15.5 | Q 19 | Page 48

 The monthly income of 100 families are given as below :

   Income in ( in  ₹)   Number of families
  0-5000   8
5000-10000  26
10000-15000 41
15000-20000 16
20000-25000  3
25000-30000 3
30000-35000   2
35000-40000      1

Calculate the modal income.

Chapter 15: Statistics Exercise 15.6 solutions [Pages 62 - 64]

Ex. 15.6 | Q 1 | Page 62

Draw an ogive by less than method for the following data:

No. of rooms: 1 2 3 4 5 6 7 8 9 10
No. of houses: 4 9 22 28 24 12 8 6 5 2

 

Ex. 15.6 | Q 1 | Page 62

Draw an ogive by less than method for the following data:

No. of rooms: 1 2 3 4 5 6 7 8 9 10
No. of houses: 4 9 22 28 24 12 8 6 5 2

 

Ex. 15.6 | Q 2 | Page 62

The marks scored by 750 students in an examination are given in the form of a frequency distribution table:

Marks No. of students
600 - 640 16
640 - 680 45
680 - 720 156
720 - 760 284
760 - 800 172
800 - 840 59
840 - 880 18
Ex. 15.6 | Q 2 | Page 62

The marks scored by 750 students in an examination are given in the form of a frequency distribution table:

Marks No. of students
600 - 640 16
640 - 680 45
680 - 720 156
720 - 760 284
760 - 800 172
800 - 840 59
840 - 880 18
Ex. 15.6 | Q 3 | Page 63

Draw an ogive to represent the following frequency distribution:

Class-interval: 0 - 4 5 - 9 10 - 14 15 - 19 20 - 24
Frequency: 2 6 10 5 3
Ex. 15.6 | Q 3 | Page 63

Draw an ogive to represent the following frequency distribution:

Class-interval: 0 - 4 5 - 9 10 - 14 15 - 19 20 - 24
Frequency: 2 6 10 5 3
Ex. 15.6 | Q 4 | Page 63

The monthly profits (in Rs.) of 100 shops are distributed as follows:

Profits per shop: 0 - 50 50 - 100 100 - 150 150 - 200 200 - 250 250 - 300
No. of shops: 12 18 27 20 17 6

Draw the frequency polygon for it.

Ex. 15.6 | Q 5 | Page 63

The following distribution gives the daily income of 50 workers of a factory.

Daily income (in Rs 100 − 120 120 − 140 140 − 160 160 − 180 180 − 200
Number of workers 12 14 8 6 10

Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.

Ex. 15.6 | Q 6 | Page 63

The following table gives production yield per hectare of wheat of 100 farms of a village:

Production yield in kg per hectare: 50 - 55 55 - 60 60 - 65 65 - 70 70 - 75 75 - 80
Number of farms: 2 8 12 24 38 16

Draw ‘less than’ ogive and ‘more than’ ogive.

Ex. 15.6 | Q 6 | Page 63

The following table gives production yield per hectare of wheat of 100 farms of a village:

Production yield in kg per hectare: 50 - 55 55 - 60 60 - 65 65 - 70 70 - 75 75 - 80
Number of farms: 2 8 12 24 38 16

Draw ‘less than’ ogive and ‘more than’ ogive.

Ex. 15.6 | Q 7 | Page 63

During the medical check-up of 35 students of a class, their weights were recorded as follows:

Weight (in kg Number of students
Less than 38 0
Less than 40 3
Less than 42 5
Less than 44 9
Less than 46 14
Less than 48 28
Less than 50 32
Less than 52 35

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph verify the result by using the formula.

Ex. 15.6 | Q 8 | Page 63

The annual rainfall record of a city for 66 days is given in the following table :

Rainfall (in cm ):   0-10 10-20   20-30   30-40   40-50 50-60
Number of days : 22 10 8 15 5 6

Calculate the median rainfall using ogives of more than type and less than type.

Ex. 15.6 | Q 8 | Page 63

The annual rainfall record of a city for 66 days is given in the following table :

Rainfall (in cm ):   0-10 10-20   20-30   30-40   40-50 50-60
Number of days : 22 10 8 15 5 6

Calculate the median rainfall using ogives of more than type and less than type.

Ex. 15.6 | Q 9 | Page 64

The following table gives the height of trees:
 

Height No. of trees
Less than 7
Less than 14
Less than 21
Less than 28
Less than 35
Less than 42
Less than 49
Less than 56
26
57
92
134
216
287
341
360


Draw 'less than' ogive and 'more than' ogive.

 

Ex. 15.6 | Q 9 | Page 64

The following table gives the height of trees:
 

Height No. of trees
Less than 7
Less than 14
Less than 21
Less than 28
Less than 35
Less than 42
Less than 49
Less than 56
26
57
92
134
216
287
341
360


Draw 'less than' ogive and 'more than' ogive.

 

Ex. 15.6 | Q 10 | Page 64

The annual profits earned by 30 shops of a shopping complex in a locality give rise to the following distribution:
 

Profit (in lakhs in Rs) Number of shops (frequency)
More than or equal to 5
More than or equal to 10
More than or equal to 15
More than or equal to 20
More than or equal to 25
More than or equal to 30
More than or equal to 35
30
28
16
14
10
7
3


Draw both ogives for the above data and hence obtain the median.

Ex. 15.6 | Q 10 | Page 64

The annual profits earned by 30 shops of a shopping complex in a locality give rise to the following distribution:
 

Profit (in lakhs in Rs) Number of shops (frequency)
More than or equal to 5
More than or equal to 10
More than or equal to 15
More than or equal to 20
More than or equal to 25
More than or equal to 30
More than or equal to 35
30
28
16
14
10
7
3


Draw both ogives for the above data and hence obtain the median.

Chapter 15: Statistics solutions [Pages 65 - 66]

Q 1 | Page 65

Define mean.

Q 1 | Page 65

Define mean.

Q 2 | Page 65

What is the algebraic sum of deviation of a frequency distribution about its mean?

Q 2 | Page 65

What is the algebraic sum of deviation of a frequency distribution about its mean?

Q 3 | Page 65

Which measure of central tendency is given by the x-coordinate of the point of intersection of the 'more than' ogive and 'less than' ogive?

Q 3 | Page 65

Which measure of central tendency is given by the x-coordinate of the point of intersection of the 'more than' ogive and 'less than' ogive?

Q 4 | Page 65

What is the value of the median of the data using the graph in the following figure of less than ogive and more than ogive?

Q 4 | Page 65

What is the value of the median of the data using the graph in the following figure of less than ogive and more than ogive?

Q 5 | Page 65

Write the empirical relation between mean, mode and median.

Q 5 | Page 65

Write the empirical relation between mean, mode and median.

Q 6 | Page 65

Which measure of central tendency can be determine graphically?

Q 6 | Page 65

Which measure of central tendency can be determine graphically?

Q 7 | Page 65

Write the modal class for the following frequency distribution:

Class-interval: 10−15 15−20 20−25 25−30 30−35 35−40
Frequency: 30 35 75 40 30 15

 

Q 8 | Page 65

A student draws a cumulative frequency curve for the marks obtained by 40 students of a class as shown below. Find the median marks obtained by the students of the class.

Q 8 | Page 65

A student draws a cumulative frequency curve for the marks obtained by 40 students of a class as shown below. Find the median marks obtained by the students of the class.

Q 9 | Page 65

Write the median class for the following frequency distribution:

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
Q 10 | Page 66

In the graphical representation of a frequency distribution, if the distance between mode and mean is ktimes the distance between median and mean, then write the value of k.

Q 10 | Page 66

In the graphical representation of a frequency distribution, if the distance between mode and mean is ktimes the distance between median and mean, then write the value of k.

Q 11 | Page 66

Find the class marks of classes 10−25 and 35−55.

Q 12 | Page 66

Write the median class of the following distribution:

Class-interval: 0−10 10−20 20−30 30−40 40−50 50−60 60−70
Frequency: 4 4 8 10 12 8 4
Q 12 | Page 66

Write the median class of the following distribution:

Class-interval: 0−10 10−20 20−30 30−40 40−50 50−60 60−70
Frequency: 4 4 8 10 12 8 4

Chapter 15: Statistics solutions [Pages 66 - 69]

Q 1 | Page 66

Which of the following is not a measure of central tendency?

  • Mean

  • Median

  •  Mode

  • Standard deviation

Q 2 | Page 66

The algebraic sum of the deviations of a frequency distribution from its mean is

  •  always positive

  • always negative

  • 0

  •  a non-zero number

Q 2 | Page 66

The algebraic sum of the deviations of a frequency distribution from its mean is

  •  always positive

  • always negative

  • 0

  •  a non-zero number

Q 3 | Page 66

The arithmetic mean of 1, 2, 3, ... , n is

  • \[\frac{n + 1}{2}\]

  • \[\frac{n - 1}{2}\]

  • \[\frac{n}{2}\]

  • \[\frac{n}{2} + 1\]

Q 3 | Page 66

The arithmetic mean of 1, 2, 3, ... , n is

  • \[\frac{n + 1}{2}\]

  • \[\frac{n - 1}{2}\]

  • \[\frac{n}{2}\]

  • \[\frac{n}{2} + 1\]

Q 4 | Page 66

For a frequency distribution, mean, median and mode are connected by the relation

  •  Mode = 3 Mean − 2 Median

  • Mode = 2 Median − 3 Mean

  • Mode = 3 Median − 2 Mean

  • Mode = 3 Median + 2 Mean

Q 5 | Page 66

Which of the following cannot be determined graphically?

  • Mean

  • Median

  • Mode

  • None of these

Q 5 | Page 66

Which of the following cannot be determined graphically?

  • Mean

  • Median

  • Mode

  • None of these

Q 6 | Page 66

The median of a given frequency distribution is found graphically with the help of

  •  Histogram

  • Frequency curve

  • Frequency polygon

  • Ogive

Q 7 | Page 66

The mode of a frequency distribution can be determined graphically from

  • Histogram

  •  Frequency polygon

  • Ogive

  • Frequency curve

Q 8 | Page 66

Mode is

  •  least frequency value

  • middle most value

  • most frequent value

  • None of these

Q 9 | Page 66

The mean of n observation is `overlineX` . If the first item is increased by 1, second by 2 and so on, then the new mean is

  •  `overlineX+n`

  •  `overlineX+n/2`

  •  `overlineX+(n+1)/2`

  • None of these

Q 9 | Page 66

The mean of n observation is `overlineX` . If the first item is increased by 1, second by 2 and so on, then the new mean is

  •  `overlineX+n`

  •  `overlineX+n/2`

  •  `overlineX+(n+1)/2`

  • None of these

Q 10 | Page 66

One of the methods of determining mode is

  •  Mode = 2 Median − 3 Mean

  •  Mode = 2 Median + 3 Mean

  • Mode = 3 Median − 2 Mean

  • Mode = 3 Median + 2 Mean

Q 11 | Page 67

If the mean of the following distribution is 2.6, then the value of y is

Variable (x): 1 2 3 4 5
Frequency 4 5 y 1 2
  • 3

  • 8

  • 13

  • 24

Q 12 | Page 67

The relationship between mean, median and mode for a moderately skewed distribution is

  • Mode = 2 Median − 3 Mean

  •  Mode = Median − 2 Mean

  • Mode = 2 Median − Mean

  •  Mode = 3 Median −2 Mean

Q 13 | Page 67

The mean of a discrete frequency distribution xi / fi, i = 1, 2, ......, n is given by

  • `overlineX = (sum f_ix_i)/(sumf_i)`

  • \[\frac{1}{n} \sum^n_{i = 1} f_i x_i\]

  • \[\frac{\sum^n_{i = 1} f_i x_i}{\sum^n_{i = 1} x_i}\]

  • \[\frac{\sum^n_{i = 1} f_i x_i}{\sum^n_{1 = 1} i}\]

Q 14 | Page 67

If the arithmetic mean of x, x + 3, x + 6, x + 9, and x + 12 is 10, the x =

  • 1

  • 2

  • 6

  • 4

Q 14 | Page 67

If the arithmetic mean of x, x + 3, x + 6, x + 9, and x + 12 is 10, the x =

  • 1

  • 2

  • 6

  • 4

Q 15 | Page 67

If the median of the data: 24, 25, 26, x + 2, x + 3, 30, 31, 34 is 27.5, then x =

  • 27

  • 25

  • 28

  • 30

Q 15 | Page 67

If the median of the data: 24, 25, 26, x + 2, x + 3, 30, 31, 34 is 27.5, then x =

  • 27

  • 25

  • 28

  • 30

Q 16 | Page 67

If the median of the data: 6, 7, x − 2, x, 17, 20, written in ascending order, is 16. Then x=

  • 15

  • 16

  • 17

  • 18

Q 16 | Page 67

If the median of the data: 6, 7, x − 2, x, 17, 20, written in ascending order, is 16. Then x=

  • 15

  • 16

  • 17

  • 18

Q 17 | Page 67

The median of first 10 prime numbers is

  • 11

  • 12

  • 13

  • 14

Q 17 | Page 67

The median of first 10 prime numbers is

  • 11

  • 12

  • 13

  • 14

Q 18 | Page 67

If the mode of the data: 64, 60, 48, x, 43, 48, 43, 34 is 43, then x + 3 =

  • 44

  • 45

  • 46

  • 48

Q 19 | Page 67

If the mode of the data: 16, 15, 17, 16, 15, x, 19, 17, 14 is 15, then x =

  • 15

  • 16

  • 17

  • 19

Q 20 | Page 67

The mean of 1, 3, 4, 5, 7, 4 is m. The numbers 3, 2, 2, 4, 3, 3, p have mean m − 1 and median q. Then, p + q =

  • 4

  • 5

  • 6

  • 7

Q 20 | Page 67

The mean of 1, 3, 4, 5, 7, 4 is m. The numbers 3, 2, 2, 4, 3, 3, p have mean m − 1 and median q. Then, p + q =

  • 4

  • 5

  • 6

  • 7

Q 21 | Page 67

If the mean of frequency distribution is 8.1 and Σfixi = 132 + 5k, Σfi = 20, then k =

  • 3

  • 4

  • 5

  • 6

Q 21 | Page 67

If the mean of frequency distribution is 8.1 and Σfixi = 132 + 5k, Σfi = 20, then k =

  • 3

  • 4

  • 5

  • 6

Q 22 | Page 67

If the mean of 6, 7, x, 8, y, 14 is 9, then

  •  x + y = 21

  • x + y = 19

  • x − y = 19

  • x − y = 21

Q 22 | Page 67

If the mean of 6, 7, x, 8, y, 14 is 9, then

  •  x + y = 21

  • x + y = 19

  • x − y = 19

  • x − y = 21

Q 23 | Page 67

The mean of n observation is `overlineX`.  If the first observation is increased by 1, the second by 2, the third by 3, and so on, then the new mean is

  • `overlineX`\[ + \left( 2n + 1 \right)\]
  • `overlineX`\[+ \frac{n + 1}{2}\]
  • `overlineX`\[ + \left( n + 1 \right)\]
  • `overlineX`\[ - \frac{n + 1}{2}\]

Q 23 | Page 67

The mean of n observation is `overlineX`.  If the first observation is increased by 1, the second by 2, the third by 3, and so on, then the new mean is

  • `overlineX`\[ + \left( 2n + 1 \right)\]
  • `overlineX`\[+ \frac{n + 1}{2}\]
  • `overlineX`\[ + \left( n + 1 \right)\]
  • `overlineX`\[ - \frac{n + 1}{2}\]

Q 24 | Page 67

If the mean of first n natural numbers is \[\frac{5n}{9}\], then n =

  • 5

  • 4

  • 9

  • 10

Q 24 | Page 67

If the mean of first n natural numbers is \[\frac{5n}{9}\], then n =

  • 5

  • 4

  • 9

  • 10

Q 25 | Page 68

The arithmetic mean and mode of a data are 24 and 12 respectively, then its median is

  • 25

  • 18

  • 20

  • 22

Q 25 | Page 68

The arithmetic mean and mode of a data are 24 and 12 respectively, then its median is

  • 25

  • 18

  • 20

  • 22

Q 26 | Page 68

The mean of first n odd natural number is

  • \[\frac{n + 1}{2}\]

  • \[\frac{n}{2}\]

  • n

  • `n^2`

Q 26 | Page 68

The mean of first n odd natural number is

  • \[\frac{n + 1}{2}\]

  • \[\frac{n}{2}\]

  • n

  • `n^2`

Q 27 | Page 68

The mean of first n odd natural numbers is \[\frac{n^2}{81}\],then n =

  • 9

  • 81

  • 27

  • 18

Q 27 | Page 68

The mean of first n odd natural numbers is \[\frac{n^2}{81}\],then n =

  • 9

  • 81

  • 27

  • 18

Q 28 | Page 68

If the difference of mode and median of a data is 24, then the difference of median and mean is

  • 12

  • 24

  • 8

  • 36

Q 28 | Page 68

If the difference of mode and median of a data is 24, then the difference of median and mean is

  • 12

  • 24

  • 8

  • 36

Q 29 | Page 68

If the arithmetic mean, 7, 8, x, 11, 14 is x, then x =

  • 9

  • 9.5

  • 10

  • 10.5

Q 29 | Page 68

If the arithmetic mean, 7, 8, x, 11, 14 is x, then x =

  • 9

  • 9.5

  • 10

  • 10.5

Q 30 | Page 68

If mode of a series exceeds its mean by 12, then mode exceeds the median by

  • 4

  • 8

  • 6

  • 10

Q 31 | Page 68

If the mean of first n natural number is 15, then n =

  • 15

  • 30

  • 14

  • 29

Q 31 | Page 68

If the mean of first n natural number is 15, then n =

  • 15

  • 30

  • 14

  • 29

Q 32 | Page 68

If the mean of observation \[x_1 , x_2 , . . . . , x_n is x\]  then the mean of x1 + a, x2 + a, ....., xn + a is 

  • a`overlineX`

  • `overlineX -a`

  • `overlineX +a`

  • `overlineX/a`

Q 32 | Page 68

If the mean of observation \[x_1 , x_2 , . . . . , x_n is x\]  then the mean of x1 + a, x2 + a, ....., xn + a is 

  • a`overlineX`

  • `overlineX -a`

  • `overlineX +a`

  • `overlineX/a`

Q 33 | Page 68

Mean of a certain number of observation is `overlineX`.  If each observation is divided by m(m ≠ 0) and increased by n, then the mean of new observation is

  • `overlineX/m +n`

  • `overlineX/n+m`

     

  • `overlineX +n/m`

  • `overlineX +m/n`

Q 33 | Page 68

Mean of a certain number of observation is `overlineX`.  If each observation is divided by m(m ≠ 0) and increased by n, then the mean of new observation is

  • `overlineX/m +n`

  • `overlineX/n+m`

     

  • `overlineX +n/m`

  • `overlineX +m/n`

Q 34 | Page 68

If \[u_i = \frac{x_i - 25}{10}, \Sigma f_i u_i = 20, \Sigma f_i = 100, \text { then }\]`overlineX`

  • 23

  • 24

  • 27

  • 25

Q 35 | Page 68

If 35 is removed from the data: 30, 34, 35, 36, 37, 38, 39, 40, then the median increased by

  • 2

  • 1.5

  • 1

  • 0.5

Q 35 | Page 68

If 35 is removed from the data: 30, 34, 35, 36, 37, 38, 39, 40, then the median increased by

  • 2

  • 1.5

  • 1

  • 0.5

Q 36 | Page 68

 While computing mean of grouped data , we assume that the frequencies are 

  • evenly distributed over all the classes.

  •  centred at the class marks of the classes .

  • centred at the upper limit of the classes.

  •  centred at the lower limit of the classes.

Q 36 | Page 68

 While computing mean of grouped data , we assume that the frequencies are 

  • evenly distributed over all the classes.

  •  centred at the class marks of the classes .

  • centred at the upper limit of the classes.

  •  centred at the lower limit of the classes.

Q 37 | Page 68

In the formula  \[\bar{X} = a + h \left( \frac{1}{N}\Sigma f_i u_i \right)\], for finding the mean of grouped frequency distribution  \[u_i\]

  • \[\frac{x_i + a}{h}\]

  • \[h\left( x_i - a \right)\]

  • \[\frac{x_i - a}{h}\]

  • \[\frac{a - x_i}{h}\]

Q 38 | Page 69

For the following distribution :

Class:               0-5         5-10        10-15        15-20        20-25

Frequency:        10            15            12            20              9

the sum of the lower limits  of the median and modal class is 

  • 15

  • 25

  • 30

  • 35

Q 38 | Page 69

For the following distribution :

Class:               0-5         5-10        10-15        15-20        20-25

Frequency:        10            15            12            20              9

the sum of the lower limits  of the median and modal class is 

  • 15

  • 25

  • 30

  • 35

Q 39 | Page 69

 For the following distribution: 
   Below:                       10         20       30      40      50     60

Number of students:      3          12        27      57      75      80

the modal class is 

  • 10-20

  • 20-30

  • 30-40

  • 50-60

Q 40 | Page 69

Consider the following frequency distributions

Class:   65-85    85-105   105-125 125-145 145-165  165-185  - 185-205
Frequency: 4 5 13 20 14 7 4

The difference of the upper limit of the median class and the lower limit of the modal class is 

  • 0

  • 19

  • 20

  • 38

Q 41 | Page 69

 In the formula 

`overlineX``\[= a + \frac{\Sigma f_i d_i}{\Sigma f_i}\], for finding the mean of grouped data \[{d_i}^{'^s}\]   are derivations from \[a\] of
  • lower limits of classes

  • upper limits ofclasses   

  •  mid-points of classes    

  •  frequency of the class marks 

Q 41 | Page 69

 In the formula 

`overlineX``\[= a + \frac{\Sigma f_i d_i}{\Sigma f_i}\], for finding the mean of grouped data \[{d_i}^{'^s}\]   are derivations from \[a\] of
  • lower limits of classes

  • upper limits ofclasses   

  •  mid-points of classes    

  •  frequency of the class marks 

Q 42 | Page 69

The abscissa of the point of intersection of less than type and of the more than types cumulative frequency curves of a grouped data gives its 

  •  mean   

  • median

  • mode

  • all the three above

Q 42 | Page 69

The abscissa of the point of intersection of less than type and of the more than types cumulative frequency curves of a grouped data gives its 

  •  mean   

  • median

  • mode

  • all the three above

Q 43 | Page 69

Consider the following frequency distribution :

Class: 0-5      6-11   12-17  18-23   24-29
Frequency:   13 10 15 8 11

The upper limit of the median class is 

  • 17

  •   17.5   

  • 18

  • 18.5

Chapter 15: Statistics

Ex. 15.1Ex. 15.2Ex. 15.3Ex. 15.4Ex. 15.5Ex. 115.5Ex. 15.6Others

RD Sharma 10 Mathematics

10 Mathematics

RD Sharma solutions for Class 10 Mathematics chapter 15 - Statistics

RD Sharma solutions for Class 10 Maths chapter 15 (Statistics) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CBSE 10 Mathematics solutions in a manner that help students grasp basic concepts better and faster.

Further, we at shaalaa.com are providing such solutions so that students can prepare for written exams. RD Sharma textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.

Concepts covered in Class 10 Mathematics chapter 15 Statistics are Reading of Pie Diagram, Frequency Polygon, Tabulation of Data, Inclusive and Exclusive Type of Tables, Median of Grouped Data, Mean of Grouped Data, Histograms, Pie Diagram, Ogives (Cumulative Frequency Graphs), Applications of Ogives in Determination of Median, Relation Between Measures of Central Tendency, Introduction to Normal Distribution, Properties of Normal Distribution, Graphical Representation of Histograms, Graphical Representation of Cumulative Frequency Distribution, Mode of Grouped Data, Ogives (Cumulative Frequency Graphs), Mean of Grouped Data, Median of Grouped Data, Introduction of Statistics, Statistics Examples and Solutions.

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