Rs Aggarwal 2019 Solutions for Class 9 Math Chapter 16 Presentation Of Data In Tabular Form are provided here with simple step-by-step explanations. These solutions for Presentation Of Data In Tabular Form are extremely popular among Class 9 students for Math Presentation Of Data In Tabular Form Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2019 Book of Class 9 Math Chapter 16 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2019 Solutions. All Rs Aggarwal 2019 Solutions for class Class 9 Math are prepared by experts and are 100% accurate.

Page No 623:

Question 1:

Define statistics as a subject.

Answer:

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

Page No 623:

Question 2:

Define some fundamental characteristics of statistics.

Answer:

The fundamental characteristics of data (statistics) are as follows:
(i) Numerical facts alone constitute data.
(ii) Qualitative characteristics like intelligence and poverty, which cannot be measured numerically, do not form data.
(iii) Data are aggregate of facts. A single observation does not form data.
(iv) Data collected for a definite purpose may not be suited for another purpose.
(v) Data in different experiments are comparable.

Page No 623:

Question 3:

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

Answer:

Primary data: The data collected by the investigator himself with a definite plan in mind are known as primary data.
Secondary data: The data collected by someone other than the investigator are known as secondary data.

Primary data are highly reliable and relevant because they are collected by the investigator himself with a definite plan in mind,  whereas secondary data are collected with a purpose different from that of the investigator and may not be fully relevant to the investigation.

Page No 623:

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

Answer:

(i) Variate : Any character which is capable of taking several different values is called a variant or a variable.
(ii) Class interval : Each group into which the raw data is condensed is called class interval .
(iii) Class size: The difference between the true upper limit and the true lower limit of a class is called its class size.
(iv) Class mark of a class: The class mark is given by Upper limit+Lower limit2.
(v) Class limit: Each class is bounded by two figures, which are called class limits.
(vi) True class limits: In the exclusive form, the upper and lower limits of a class are respectively known as true upper limit and true lower limit.
In the inclusive form of frequency distribution, the true lower limit of a class is obtained by subtracting 0.5 from the lower limit and the true upper limit of the class is obtained by adding 0.5 to the upper limit.
(vii) Frequency of a class: Frequency of a class is the number of times an observation occurs in that class.
(viii) Cumulative frequency of a class: Cummulative frequency of a class is the sum total of all the frequencies up to and including that class.



Page No 624:

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.

Answer:

(i) 

Blood group tally marks Number of students
A 9
B 6
O 12
AB 3

(ii) AB is rarest and O is most common. 

Page No 624:

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.

Answer:

Number of heads tally marks Frequency
0 6
1 10
2 9
3 5

Page No 624:

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 it in the form of a frequency distribution, taking classes 0−2, 1−4, etc.

Answer:

The minimum observation is 0 and the maximum observation is 8.
Therefore, classes of  the same size covering the given data are 0-2, 2-4, 4-6 and 6-8.          .


Frequency distribution table:
        Class               Tally mark          Frequency
        0-2            11
        2-4            17
        4-6             9
        6-8             3

Page No 624:

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?

Answer:

(i) 

Class interval tally marks Frequency
0-5 10
5-10 13
10-15 5
15-20 2

(ii) As we can see from the table, there are 2 children who watched tv for 15 hours or more. 

Page No 624:

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).

Answer:

The minimum observation is 0 and the maximum observation is 25.
Therefore, classes of the same size covering the given data are 0-5, 5-10, 10-15, 15-20 and 20-25.
Frequency distribution table:

                 Class Tally mark Frequency
                  0-5         6
                5-10       10
               10-15         8
               15-20         8
               20-25         8

Page No 624:

Question 10:

Construct a frequency table for the following ages (in years) of 30 students using equal class intervals, one of them being 912, 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.

Answer:

The minimum observation is 6 and the maximum observation is 24.
Therefore, classes of the same size covering the given data are 6-9, 9-12, 12-15, 15-18, 18-21 and 21-24.
Frequency distribution table:
         Class            Tally mark          Frequency
          6-9                             5
         9-12                              4
       12-15                              4
       15-18                              7
       18-21                              3
       21-24                              7

Page No 624:

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.

Answer:

The minimum observation is 210 and the maximum observation is 330.
Therefore, classes of the same size covering the given data are 210-230, 230-250,250-270,270-290,290-310 and 310-330.
Frequency distribution table:
 

            Class        Tally mark           Frequency
           210-230                            4
           230-250                            4
          250-270                           5
          270-290                             3
          290-310                           7
          310-330                           5



Page No 625:

Question 12:

The weights (in grams) of 40 oranges picked at random from a basket are as follows:
40,50,60,65,45,55,30,90,75,85,70,85,75,80,100,110,70,55,30,35,45,70,80,85,95,70,60,70,75,100,65,60,40,100,75,110,30,45,84.
Construct a frequency table as well as a cumulative frequency table.

Answer:

The minimum observation is 30 and the maximum observation is 120.
                                                
Frequency distribution table:

           Class                 Tally mark            Frequency
30-40                                     4
40-50                                    6
50-60                                     3
60-70                                    5
70-80                                    9
80-90                                     6
90-100                                       2
100-110                                     3
110-120                                      2

                                     
Cumulative frequency table:

   Class    Tally mark    Frequency      Cumulative frequency
30-40                     4                   4
40-50                    6                  10
50-60                     3                  13
60-70                    5                  18
70-80                9                  27
80-90                  6                  33
90-100                     2                  35
100-110                    3                  38
110-120                     2                  40

Page No 625:

Question 13:

The heights (inAns 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.

Answer:

Class tally marks Frequency Cumulative frequency
145-150 4 4
150-155 9 4 + 9 = 13
155-160 12 13 + 12 = 25
160-165 5 25 + 5 = 30

Page No 625:

Question 14:

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

Ages (in years) 1020 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.

Answer:

The cumulative frequency table can be presented as given below:
 

      Age (in years ) No. of patients      Cumulative frequency
          10-20 90                 90         
          20-30 50        140
          30-40 60         200
          40-50 80         280
          50-60 50         330
          60-70 30         360

Page No 625:

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

Answer:

The grouped frequency table can be presented as given below:
 

            Marks         No. of students
              0-10                      5
            10-20                      7
            20-30                    20
            30-40                     8
            40-50                     5
            50-60                     3

Page No 625:

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.

Answer:

The frequency table can be presented as given below:
 

             Marks   Number of students
              0-10                   17
             10-20                    5
             20-30                    7
             30-40                    8
             40-50                  13
             50-60                  10

Page No 625:

Question 17:

Make a frequency table from the following:

Marks obtained Number of students
More than 60 0
More than 50 16
More than 40 40
More than 30 75
More than 20 87
More than 10 92
More than 0 100

Answer:

The frequency table can be presented as below:
 

            Class     Frequency
               0-10            8
              10-20            5
               20-30          12
               30-40          35
               40-50          24
               50-60          16



Page No 626:

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.

Answer:

Range = Maximum value - minimu value
= 100 - 46 = 54
Thus, the range is 54.

Page No 626:

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.

Answer:

(i) class mark=upper limit+lower limit2=120+902=2102=105
(ii) mid-value = 10
width = 6
Let the lower limit of the class be x
upper limit = x + 6
class mark/mid-value=upper limit+lower limit2
x+x+62=10x=7
(iii) width = 5
lower class limit of lowest class = 10
The classes will be 10-15, 15-20, 20-25, 25-30, 30-35.
Upper class limit of the highest class = 35.
(iv) Class marks = 15, 20, 25, ...
class size = 20 - 15 = 5
Let lower limit of class be x.
x+x+52=20x=17.5
Thus, the class is 17.5-22.5. 
(v) 20 will be included in the class interval 20-30. 

Page No 626:

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  

Answer:

The  complete table will be

Height (in cm) Frequency Cumulative frequency
160 – 165 15 a = 15
165 – 170 b = 35 – 15 = 20  35
170 – 175 12 = 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 = 15 + 20 + 12
+ 3 + 5 + 5 = 60
 

 



Page No 628:

Question 1:

The range of the data
12, 25, 15, 18, 17, 20, 22, 6, 16, 11, 8, 19, 10, 30, 20, 32 is
(a) 10
(b) 15
(c) 18
(d) 26

Answer:

(d) 26

We have:
Maximum value = 32
Minimum value = 6
We know:
Range = Maximum value - Minimum value 
          =32 - 6
          =26

Page No 628:

Question 2:

The class mark of the class 100−120 is
(a) 100
(b) 110
(c) 115
(d) 120

Answer:

(b) 110

Class mark = Upper limit+Lower limit2=120+1002=110



Page No 629:

Question 3:

In the class intervals 1020, 20−30, the number 20 is included in
(a) 10−20
(b) 20−30
(c) in each of 10−20 and 20−30
(d) in none of 10−20 and 20−30

Answer:

(b) 2030
This is the continuous form of frequency distribution. Here, the upper limit of each class is excluded, while the lower limit is included. So, the number 20 is included in the class interval 2030.

Page No 629:

Question 4:

The class marks of a frequency distribution are 15, 20, 25, 30, .. . The class corresponding to the class marks 20 is
(a) 12.5−17.5
(b) 17.5−22.5
(c) 18.5−21.5
(d) 19.5−20.5

Answer:

(b) 17.5-22.5

We are given frequency distribution 15, 20, 25, 30,...
Class size = 20 - 15 = 5
Class marks = 20
Now,
Lower limit=20-52=352=17.5Upper limit =20+52=452=22.5

Thus, the required class is 17.5-22.5.

Page No 629:

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
(a) 6
(b) 7
(c) 8
(d) 12

Answer:

(b) 7

Given:
Mid value of the class = 10
Width of each class = 6
Now,
Let the lower limit be x.
We know:
Upper limit = Lower limit + Class size
                  = x + 6
Also,
Mid value=x+x+62=2x+62=x+3x+3=10x=7

Thus, the lower limit is 7.

Page No 629:

Question 6:

The mid-value of a class interval is 42 and the class size is 10. The lower and upper limits are
(a) 37−47
(b) 37.5−47.5
(c) 36.5−47.5
(d) 36.5−46.5

Answer:

(a) 37–47

Let the lower limit be x.
Here,
Class size = 10
∴ Upper limit = Class size + Lower limit
Upper limit = (x + 10)
Mid value of the class interval = 42

x+x+102=422x+102=422x+10=842x=74x=37Thus, we have:Lower limit=37 Upper limit=37+10=47

Page No 629:

Question 7:

Let m be 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
(a) 2m u
(b) 2m + u
(c) mu
(d) m + u

Answer:

(a) 2m - u

Given:
Mid value = m
Upper limit = u

We know:
Lower limit+Upper limit2=Mid value Lower limit +u2=mLower limit+u=2mLower limit=2m-u

Page No 629:

Question 8:

The width of each of the five continuous classes in a frequency distribution is 5 and the lower class limit of the lowest class is 10. The upper class limit of the highest class is
(a) 45
(b) 25
(c) 35
(d) 40

Answer:

(c) 35

We have:
Class width = 5
Lower class limit of the lowest class = 10
Now,
Upper class limit of the highest class = 10 + 5 × 5 = 35

Page No 629:

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?
(a) m+(m + L)2
(b) L + m + L2
(c) 2m - L
(d) m - 2L

Answer:

(c) 2m-L

Mid value=Lower limit+Upper limit2m=L+U2U=2m-LUpper class boundary of the class=2m-L



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