# R S AGGARWAL AND V AGGARWAL Solutions for Class 9 Maths Chapter 16 – Presentation of Data in Tabular Form

## Chapter 16 – Presentation of Data in Tabular Form Exercise 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

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 2

Grouped Frequency Distribution Table:

Question 3

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 3

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 4

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 4

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 5

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 5

Grouped Frequency Distribution Table and
Cumulative Frequency Table:

Question 6

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 6
 Age (in years) (age) No of patients (Frequency) Cumulative Frequency 10-2020-3030-4040-5050-6060-70 905060805030 90140200280330360 Total 360

Question 7

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 7
 Marks (below) No of students(Cumulative Frequency.) Class Intervals Frequency 102030405060 51232404548 0-1010-2020-3030-4040-5050-60 512 – 5 = 732 – 12 = 2040 – 32 = 845 – 40 = 548 – 45 = 3 Total 48

Question 8

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 8
 Marks (below) No of students(Cumulative Frequency) Class Intervals Frequency 102030405060 172229375060 0-1010-2020-3030-4040-5050-60 1722 – 17 = 529 – 22 = 737 – 29 = 850 – 37 = 1360 – 50 = 10 Total 60

Question 9

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 9
 Marks (below) No of student s(C.F.) Class Intervals Frequency More than 60More than 50More than 40More than 30More than 20More than 10More than 0 01640758792100 More than 6050-6040-5030-4020-3010-200-10 016-0=1640-16=2475-40=3587-75=1292-87=5100-92=8 Total 100

Question 10

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 10

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 11

(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 11

Question 12

Define some fundamental characteristics of statistics.

Solution 12

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 13

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 Cumulativefrequency 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 13

 Height (in cm) Frequency Cumulativefrequency 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

Question 14

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

Solution 14

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 15

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 15

(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 16

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 16

(i) Frequency Distribution Table:

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

Question 17

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 17

Frequency Distribution Table:

Question 18

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 18

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 19

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 19

(i) Grouped Frequency Distribution
Table:

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

Question 20

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 20

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 :

## Chapter 16 – Presentation of Data in Tabular Form Exercise 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

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