Chapter 16 : Presentation of Data in Tabular Form - R S Aggarwal And V Aggarwal Solutions for Class 9 Maths CBSE

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Chapter 16 - Presentation of Data in Tabular Form Excercise MCQ

Question 1

The range of the data

12,25,15,18,17,20,22,6,16,11,8,19,10,30,20,32 is

  1. 10
  2. 15
  3. 18
  4. 26
Solution 1

Correct option: (d)

Range = maximum value - minimum value

 = 32 - 6

 = 26

Question 2

The class mark of the class 100-120 is

  1. 100
  2. 110
  3. 115
  4. 120
Solution 2

Question 3

In the class intervals 10-20, 20-30, the number 20 is included in

  1. 10-20
  2. 20-30
  3. In each of 10-20 and 20-30
  4. In none of 10-20 and 20-30
Solution 3

Question 4

The class marks of a frequency distribution are 15, 20, 25, 30………. The class corresponding to the mark 20 is

  1. 12.5-17.5
  2. 17.5-22.5
  3. 18.5-21.5
  4. 19.5-20.5
Solution 4

Question 5

In a frequency distribution, the mid-value of a class is 10 and width of each class is 6. The lower limit of the class is

  1. 6
  2. 7
  3. 8
  4. 12
Solution 5

Question 6

The mid - value of a class interval is 42 and the class size is 10. The lower and upper limits are

  1. 37-47
  2. 37.5-47.5
  3. 36.5-47.5
  4. 36.5-46.5
Solution 6

Question 7

Let m be in the midpoint and u be the upper class limit of a class in a continuous frequency distribution. The lower class limit of the class is

  1. 2m - u
  2. 2m + u
  3. m - u
  4. m + u
Solution 7

Question 8

The width of each of the five continuous classes in a frequency distribution is 5 and the lower class limit of the class is

  1. 45
  2. 25
  3. 35
  4. 40
Solution 8

Question 9

Let L be the lower class boundary of a class in a frequency distribution and m be the midpoint of the class. Which one of the following is the upper class boundary of the class?

Solution 9

Chapter 16 - Presentation of Data in Tabular Form Excercise Ex. 16

Question 1

Define statistics as a subject.

Solution 1

Statistics is a branch of science which deals with the collection, presentation, analysis and interpretation of numerical data.

Question 2

Define some fundamental characteristics of statistics.

Solution 2

Fundamental characteristics of statistics :

(i) It deals only with the numerical data.

(ii) Qualitative characteristic such as illiteracy, intelligence, poverty etc cannot be measured numerically

(iii) Statistical inferences are not exact.

Question 3

What are the primary data and secondary data? Which of the two is more reliable and why?

Solution 3

Primary data: Primary data is the data collected by the investigator himself with a definite plan in his mind. These data are very accurate and reliable as these being collected by the investigator himself.

Secondary Data: Secondary data is the data collected by a person other than the investigator.

Secondary Data is not very reliable as these are collected by others with purpose other than the investigator and may not be fully relevant to the investigation.

Question 4

Explain the meaning of each of the following terms.

(i)Variate(ii) Class interval(iii)Class size

(iv)Class mark (v)Class limit(vi)True class limits

(vii)Frequency of a class(viii) Cumulative frequency of a class

Solution 4

(i)Variate : Any character which can assume many different values is called a variate.

(ii)Class Interval :Each group or class in which data is condensed is calleda class interval.

(iii)Class-Size :The difference between the true upper limitand the true lower limit of a class is called class size.

(iv)Classmark : The average of upper and lower limit of a class interval is called its class mark.

i.e Class mark=

(v) Class limit: Class limits are the two figures by which a class is bounded . The figure on the left side of a class is called lower lower limit and on the right side is called itsupper limit.

(vi)True class limits: In the case of exclusive form of frequency distribution, the upper class limits and lower classlimits are the true upper limits and thetrue lower limits. But in the case of inclusive form of frequency distribution , the true lower limit of a class is obtained by subtracting 0.5 from the lower limit of the class. And the true upper limit of the class is obtained by adding 0.5 to the upper limit.

(vii)Frequency of a class : The number of observations falling in aclass determines its frequency.

(viii)Cumulative frequency of a class: The sum of all frequenciesup to and including that class is called , the cumulative frequency of that class.

Question 5

The blood groups of 30 students of a class are recorded as under:

A, B, O, O, AB, O, A, O, A, B, O, B, A, O, O, A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O.

(i) Represent this data in the form of a frequency distribution table.

(ii) Find out which is the most common and which is the rarest blood group among these students.

Solution 5

(i) Frequency Distribution Table:

 

 

(ii) The most common blood group is 'O' and the rarest blood group is 'AB'.

Question 6

Three coins are tossed 30 times. Each time the number of heads occurring was noted down as follows:

0, 1, 2, 2, 1, 2, 3, 1, 3, 0, 1, 3, 1, 1, 2, 2, 0, 1, 2, 1, 0, 3, 0, 2, 1, 1, 3, 2, 0, 2.

Prepare a frequency distribution table.

Solution 6

Frequency Distribution Table:

Question 7

Following data gives the number of children in 40 families :

1, 2, 6, 5, 1, 5, 1, 3, 2, 6, 2, 3, 4, 2, 0, 4, 4, 3, 2, 2, 0, 0, 1,2, 2, 4, 3, 2, 1, 0, 5, 1, 2, 4, 3, 4, 1, 6, 2, 2.

Represent in the form of a frequency distribution, taking classes 0-2, 2-4, etc.

Solution 7
nimum observation is 0 and maximum observation is 6. The classes of equal size covering the given data are : (0-2), (2-4), (4-6) and (6-8).

Thus , the frequency distribution may be given as under:

Question 8

Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as under:

8, 4, 8, 5, 1, 6, 2, 5, 3, 12, 3, 10, 4, 12, 2, 8, 15, 1, 6, 17, 5, 8, 2, 3, 9, 6, 7, 8, 14, 12.

(i) Make a grouped frequency distribution table for this data, taking class width 5 and one of the class interval as 5 - 10.

(ii) How many children watched television for 15 or more hours a week?

Solution 8

(i) Grouped Frequency Distribution Table:

 

 

(ii) 2 children watch television for 15 or more hours a week. 

Question 9

The marks obtained by 40 students of a class in an examination are given below .

3, 20, 13, 1, 21, 13, 3, 23, 16, 13, 18, 12, 5, 12, 5, 24, 9, 2, 7, 18, 20, 3, 10, 12, 7, 18, 2, 5, 7, 10, 16, 8, 16, 17, 8, 23, 24, 6, 23, 15.

Present the data in the form of a frequency distribution using equal class size, one such class being 10-15(15 not included).

Solution 9

Minimum observation is 1 and minimum observation is 24. The classes of equal size converging the given data are : (0-5), (5-10), (10-15), (15-20), (20-25)

Thus, the frequency distribution may be given as under :

 

Question 10

Construct a frequency table for the following ages (in years) of 30 students using equal class intervals, one of them being 9-12, where 12 is not included.

18, 12, 7, 6, 11, 15, 21, 9, 8, 13, 15, 17, 22, 19, 14, 21, 23, 8, 12, 17, 15, 6, 18, 23, 22, 16, 9, 21, 11, 16.

Solution 10

Grouped Frequency Distribution Table:

 

Question 11

Construct a frequency table with equal class intervals from the following data on the monthly wages (in rupees) of 28 labourers working in a factory, taking one of the class intervals as 210-230 (230 not included).

220, 268, 258, 242, 210, 268, 272, 242, 311, 290, 300, 320, 319, 304, 302, 318, 306, 292, 254, 278, 210, 240, 280, 316, 306, 215, 256, 236.

Solution 11

Minimum observation is 210 and maximum observation =320

So the range is (320-210)=110

The classes of equal size covering the given data are :

(210-230), (230-250), (250-270) , (270-290), (290-310), (310-330)

Thus the frequency distribution may be given as under :


Question 12

The weights (in grams ) of 40 oranges picked at random from a basket are as follow :

40, 50, 60, 65, 45, 55, 30, 90, 75, 85, 70,85, 75, 80, 100, 110, 70, 55, 30, 35, 45, 70, 80, 85, 95, 70, 60, 70, 75, 40, 100, 65, 60, 40, 100, 75, 110, 30, 45, 84.

Construct a frequency table as well as a cumulative frequency table.

Solution 12

 

Minimum observation is 30 and maximum observation is 110

So, range is 100-30=80

The classes of equal size covering the given data are :

(30-40) ,(40-50) , (50-60) ,(60-70) , (70-80), (80-90),(90-100),(100-110), (110-120)

Thus , the frequency and cumulative frequency table may be given as under :

 

Question 13

The heights (in cm) of 30 students of a class are given below:

161, 155, 159, 153, 150, 158, 154, 158, 160, 148, 149, 162, 163, 159, 148, 153, 157, 151, 154, 157, 153, 156, 152, 156, 160, 152, 147, 155, 155, 157.

Prepare a frequency table as well as a cumulative frequency table with 160-165 (165 not included) as one of the class intervals.

Solution 13

Grouped Frequency Distribution Table and Cumulative Frequency Table:

  

Question 14

Following are the ages (in years ) of 360 patients , getting medical treatment in a hospital:

Age (in years)

10-20

20-30

30-40

40-50

50-60

60-70

Number of patients

90

50

60

80

50

30

Construct the cumulative frequency table for the above data.

Solution 14

Age (in years)

(age)

No of patients (Frequency)

Cumulative Frequency

10-20

20-30

30-40

40-50

50-60

60-70

90

50

60

80

50

30

90

140

200

280

330

360

Total

360

Question 15

Present the following as an ordinary grouped frequency table :

Marks(below)

10

20

30

40

50

60

Number of students

5

12

32

40

45

48

Solution 15

Marks (below)

No of students(Cumulative Frequency.)

Class Intervals

Frequency

10

20

30

40

50

60

5

12

32

40

45

48

0-10

10-20

20-30

30-40

40-50

50-60

5

12 - 5 = 7

32 - 12 = 20

40 - 32 = 8

45 - 40 = 5

48 - 45 = 3

Total

48

 

Question 16

Given below is a cumulative frequency table ;

Marks

Number of students

Below 10

17

Below 20

22

Below 30

29

Below 40

37

Below 50

50

Below 60

60

Extract a frequency table from the above .

Solution 16

Marks (below)

No of students(Cumulative Frequency)

Class Intervals

Frequency

10

20

30

40

50

60

17

22

29

37

50

60

0-10

10-20

20-30

30-40

40-50

50-60

17

22 - 17 = 5

29 - 22 = 7

37 - 29 = 8

50 - 37 = 13

60 - 50 = 10

Total

60

 

Question 17

Make a frequency table from the following ;

Marks obtained

Number of students

More than 60

0

More than50

16

More than40

40

More than30

75

More than20

87

More than10

92

More than0

100

Solution 17

Marks (below)

No of student s(C.F.)

Class Intervals

Frequency

More than 60

More than 50

More than 40

More than 30

More than 20

More than 10

More than 0

0

16

40

75

87

92

100

More than 60

50-60

40-50

30-40

20-30

10-20

0-10

0

16-0=16

40-16=24

75-40=35

87-75=12

92-87=5

100-92=8

Total

100

 

Question 18

The marks obtained by 17 students in a mathematics test (out of 100) are given below:

90, 79, 76, 82, 65, 96, 100, 91, 82, 100, 49, 46, 64, 48, 72, 66, 68.

Find the range of the above data.

Solution 18

Arranging data in ascending order, we have

46, 48, 49, 64, 65, 66, 68, 72, 76, 79, 82, 82, 90, 91, 96, 100, 100

Minimum marks = 46

Maximum Marks = 100

Range of the above data = Maximum Marks - Minimum Marks

 = 100 - 46

= 54

Question 19

(i) Find the class mark of the class 90 - 120.

(ii) In a frequency distribution, the mid-value of the class is 10 and width of the class is 6. Find the lower limit of the class.

(iii) The width of each of five continuous classes in a frequency distribution is 5 and lower class limit of the lowest class is 10. What is the upper class limit of the highest class?

(iv) The class marks of a frequency distribution are 15, 20, 25, ... Find the class corresponding to the class mark 20.

(v) In the class intervals 10-20, 20-30, find the class in which 20 is included.

Solution 19

Question 20

Find the values of a, b, c, d, e, f, g from the following frequency distribution of the heights of 50 students in a class:

Height (in cm)

Frequency

Cumulative

frequency

160-165

15

a

165-170

b

35

170-175

12

c

175-180

d

50

180-185

e

55

185-190

5

f

 

g

 

 

Solution 20

 

Height

(in cm)

Frequency

Cumulative

frequency

160-165

15

a = 15

165-170

b = 35 - 15 = 20

35

170-175

12

c = 35 + 12 = 47

175-180

d = 50 - 47 = 3

50

180-185

e = 55 - 50 = 5

55

185-190

5

f = 55 + 5 = 60

 

g = 60

 

 

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