Chapter 22 : Tabular Representation of Statistical Data - Rd Sharma Solutions for Class 9 Maths CBSE

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Chapter 22 - Tabular Representation of Statistical Data Excercise Ex. 22.1

Question 1
Solution 1
Question 2
Solution 2
Question 3
Solution 3

Question 4
Why do we group data?
Solution 4
The data obtained in original form are called raw data. Raw data does not give any useful information and is rather confusing to mind. Data is grouped so that it becomes understandable and can be interpreted. We form groups according to various characteristics. After grouping the data, we are in a position to make calculations of certain values which will help us in describing and analysing the data.
Question 5
Solution 5
Question 6
Solution 6
Question 7
Solution 7
Question 8
Solution 8
Question 9
Solution 9
The final marks in mathematics of 30 students are as follows:

53,61,48,60,78,68,55,100,67,90,75,88,77,37,84,58,60,48,62,56,44,58,52,64,98,59,70,39,50,60

(i)


Group I(30-39) II(40-49) III(50-59) IV(60-69) V(70-79) VI(80-89) VII(90-99) VIII(100-109)
Observations 37, 39 44, 48, 48 50, 52, 53, 55, 56, 58, 58, 59 60, 60, 60, 61, 62, 64, 67, 68 70, 75, 77, 78 84, 88 90, 98 100



(ii) Highest score = 100

(iii) Lowest score = 37

(iv) Range = 100 - 37 = 63

(v) If 40 is the pass mark, 2 students have failed.

(vi) 8 students have scored 75 or more.

(vii) Observations 51, 54, 57 between 50 and 60 have not actually appeared.

(viii) 5 students have scored less than 50.


Question 10
Solution 10
Question 11
The number of runs scored by a cricket player in 25 innings are as follows:

26,35,94,48,82,105,53,0,39,42,71,0,64,15,34,67,0,42,124,84,54,48,139,64,47.

(i) Rearrange these runs in ascending order.

(ii) Determine the player's highest score.

(iii) How many times did the player not score a run?

(iv) How many centuries did he score?

(v) How many times did he score more than 50 runs?
Solution 11
The numbers of runs scored by a player in 25 innings:

26, 35, 94, 48, 82, 105, 53, 0, 39, 42, 71, 0, 64, 15, 34, 67, 0, 42, 124, 84, 54, 48, 139, 64, 47.

(i) Runs in ascending order:- 0,0,0,15,26,34,35,39,42,42,47,48,48,53,54,64,64,67,71,82,84,94,105,124,139

(ii) The highest score = 139

(iii) The player did not score any run 3 times.

(iv) He scored 3 centuries.

(v) He scored more than 50 runs 12 times.
Question 12
Solution 12
Question 13
Write the class size and class limits in each of the following

(i) 104, 114, 124, 134, 144, 154, and 164

(ii) 47, 52, 57, 62, 67, 72, 82, 87, 92, 97 and 102

(iii) 12.5, 17.5, 22.5, 27.5, 32.5, 37.5, 42.5, 47.5
Solution 13
(i)



(ii)



(iii)


Question 14
Solution 14
Number of children Tally marks Number of families
0 5
1 ll 7
2 ll 12
3 5
4 l 6
5 lll 3
6 lll 3
Question 15
Solution 15
Marks Tally marks Frequency
20 - 30 l 1
30 - 40 lll 3
40 - 50 5
50 - 60 lll 8
60 - 70 lll 8
70 - 80 llll 9
80 - 90 llll 4
90 - 100 ll 2
Total = 40


Question 16

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

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

Prepare a frequency distribution table with 160-164 as one of the class intervals.

Solution 16
Heights (in cm)
Tally marks Frequency
145 - 149 llll 4
150 - 154 llll 9
155 - 159 ll 12
160 - 164 5
    Total = 30



Question 17

Solution 17
Height (in cm) Tally marks Frequency
800 - 810 lll 3
810 - 820 ll 2
820 - 830 l 1
830 - 840 lll 8
840 - 850 5
850 - 860 l 1
860 - 870 lll 3
870 - 880 l 1
880 - 890 l 1
890 - 900 5
Total = 30
Question 18
Solution 18
Maximum temperature (in degree Celsius) Tally marks Frequency
20.0 - 21.0 l 6
21.0 - 22.0 5
22.0 - 23.0 llll 9
23.0 - 24.0 5
24.0 - 25.0 lll 3
25.0 - 26.0 ll 2
Total = 30
Question 19
Solution 19
Monthly wages (in rupees) Tally marks Frequency
210 - 230 llll 4
230 - 250 llll 4
250 - 270 5
270 - 290 lll 3
290 - 310 ll 7
310 - 330 5
Total = 28
Question 20

The blood groups of 30 student of Class VIII are recoded as follows:

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

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

Represent this data in the form of a frequency distribution table. Which is the most common, and which is the rarest, blood group among these students?

Solution 20
Here 9 students  have blood groups  A, 6 as B, 3 as AB and 12 as O.
So, the table representing the data is as follows:  

Blood group

Number of students

A

9

B

6

AB

3

O

12

Total

30



As 12 students have the blood group O and 3 have their blood group as AB. Clearly, the most common blood group among these students is O and the rarest blood group among these students is AB.
Question 21
Three coins were tossed 30 times simultaneously. 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
              3    0    0    1    1    2    3    2    2    0

Prepare a frequency distribution table for the data given above.   

Solution 21
By observing the data given above following frequency distribution table can be constructed
 

Number of heads

Number of times (frequency)

0

 6

1

10

2

  9

3

  5

Total

30

Question 22
Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows:

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

    (i)    Make a grouped frequency distribution table for this data, taking class width 5 and one of the class intervals as 5 - 10.
    (ii)    How many children watched television for 15 or more hours a week? 
                                                           

Solution 22
(i) Class intervals will be 0 - 5, 5 - 10, 10 -15.....
    The grouped frequency distribution table is as follows:
    
                       

Hours

Number of children

0 - 5

10

5 - 10

13

10 - 15

 5

15 - 20

 2

Total

30

                                                                                         
(ii) The number of children, who watched TV for 15 or more hours a week
        is 2 (i.e. number of children in class interval 15 - 20).

Question 23

Solution 23

Since first class interval is -19.9 to -15

Frequency distribution with lower limit included and upper limit excluded is:

Temperature Tally marks Frequency
-19.9 to -15 ll 2
-15 to -10.1  ll  7
-10.1 to -5.2   5
-5.2 to -0.3 llll 4
-0.3 to 4.6 ll 17
Total   35



Chapter 22 - Tabular Representation of Statistical Data Excercise Ex. 22.2

Question 1

Solution 1
Question 2
Solution 2
Question 3

Solution 3

Question 4
Following are the the ages of 360 patients getting medical treatment in a hospital on a day:

Age (in years): 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70
No. of Patients: 90 50 60 80 50 30

Construct a cumulative frequency distribution.
Solution 4


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

Question 5
Solution 5

Question 6
Solution 6

Question 7

Solution 7

Question 8
The following cumulative frequency distribution table shows the daily electricity consumption (in kW) of 40 factories in an industrial state:
Consumption (in KW) No. of Factories
Below 240

Below 270

Below 300

Below 330

Below 360

Below 390

Below 420
1

4

8

24

33

38

40


(i) Represent this as a frequency distribution table.

(ii) Prepare a cumulative frequency table.
Solution 8
(i)

Consumption (in kW) No. of Factories Class interval Frequency
Below 240 1 0 - 240 1
Below 270 4 240 - 270 4 - 1 = 3
Below 300 8 270 - 300 8 - 4 = 4
Below 330 24 300 - 330 24 - 8 = 16
Below 360 33 330 - 360 33 - 24 = 9
Below 390 38 360 - 390 38 - 33 = 5
Below 420 40 390 - 420 40 - 38 = 2


(ii)

 

Class interval Frequency Consumption (in kW) No. of factories
0 - 240 1 More than 0 40
240 - 270 3 More than 270 40 - 1 = 39
270 - 300 4 More than 270 39 - 3 = 36
300 - 330 16 More than 300 36 - 4 = 32
330 - 360 9 More than 330 32 - 16 = 16
360 - 390 5 More than 360 16 - 9 = 7
390 - 420 2 More than 390 7 - 5 = 2
    More than 420
2 - 2 = 0
  N = 40  
 


Question 9

Solution 9

Chapter 22 - Tabular Representation of Statistical Data Excercise 22.26

Question 1

Tally marks are used to find

(a) class intervals

(b) range

(c) frequency

(d) upper limits

Solution 1

When observations are large, it may not be easy to find the frequencies by simple counting.

So, we make use of tally marks.

Thus, Tally marks are used to find frequency.

Hence, correct option is (c).

Question 2

The difference between the highest and lowest values of the observations is called

(a) frequesncy

(b) mean

(c) range

(d) class-intervals

Solution 2

The difference between the highest and lowest value of observations is called 'Range' of observations.

Hence, correct option is (c).

Question 3

The difference between the upper and the lower class limits is called

(a) mid-points

(b) class size

(c) frequency

(d) mean

Solution 3

The difference between the upper class limit and the lower class limit is called class size.

Hence, correct option is (b).

Question 4

In the class intervals 10 - 20, 20 - 30, 20 is taken in

(a) the interval 10 - 20

(b) the interval 20 - 30

(c) both intervals 10 - 20, 20 - 30

(d) none of the intervals

Solution 4

Since, 10 - 20, 20 - 30 are Exclusive Class Intervals, the upper limit of a class is not included in the class.

Thus, 20, will be taken in the class 20 - 30.

Hence, correct option is (b).

Question 5

In a frequency distribution, the mid-value of a class is 15 and the class intervals is 4. The lower limit of the class is

(a) 10

(b) 12

(c) 13

(d) 14

Solution 5

begin mathsize 12px style Let space the space lower space limit space be space straight x.
Class space interval space equals space 4
rightwards double arrow Upper space limit space equals space straight x space plus space 4
Now comma space mid minus value space of space straight a space class space equals space fraction numerator straight x space plus space 4 space plus space straight x over denominator 2 end fraction equals straight x space plus space 2 space equals space 15 space left parenthesis given right parenthesis
rightwards double arrow straight x space equals space 13 space equals space lower space limit
Hence comma space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 6

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

(a) 47 and 37

(b) 37 and 47

(c) 37.5 and 47.5

(d) 47.5 and 37.5

Solution 6

begin mathsize 12px style Let space the space lower space limit space of space straight a space class space equals space straight x
Class space size space equals space 10
rightwards double arrow Upper space limit space equals space straight x space plus space 10
Now comma space mid minus value space equals space fraction numerator straight x space plus space 10 space plus space straight x over denominator 2 end fraction equals straight x space plus space 5 space equals space 42 space left parenthesis given right parenthesis
rightwards double arrow straight x space equals space 37 space equals space lower space limit
rightwards double arrow straight x space plus space 10 space equals space 47 space equals space upper space limit
Thus comma space upper space and space lower space limits space are space 47 comma space 37.
Hence comma space correct space option space is space left parenthesis straight a right parenthesis. space end style

Chapter 22 - Tabular Representation of Statistical Data Excercise 22.27

Question 1

The number of times a particular item occurs in a given data is called its

(a) variation

(b) frequency

(c) cumulative frequency

(d) class-size

Solution 1

The number of times a particular item occurs in a given data is called its Frequency.

Hence, correct option is (b).

Question 2

The width of each of nine classes in a frequency distribution is 2.5 and the lower class boundary of the lowest class 10.6. Then the upper class boundary of the highest class is

(a) 35.6

(b) 33.1

(c) 30.6

(d) 28.1

Solution 2

Number of classes = 9

Lower limit of the lowest class = 10.6

Width of each class = 2.5

So, Upper limit of the lowest class = 10.6 + 2.5 = 13.1

Now, Upper limit of the lowest class + Width of each class = Upper limit of the next class

Thus, we have

Upper limit of the lowest class + 8 × width of each class = Upper limit of the highest (9th) class

Upper limit of the highest (9th) class = 13.1 + 8 × 2.5 = 33.1

Hence, correct option is (b).

  

                                 

Question 3

The following marks were obtained by the students in a test:

81, 72, 90, 90, 86, 85, 92, 70, 71, 83, 89, 95, 85, 79, 62

The range of the marks is

(a) 9

(b) 17

(c) 27

(d) 33 

Solution 3

Range of observations = Highest observation - Lowest observation

                             = 95 - 62

                             = 33

Hence, correct option is (d).

Question 4

Tally are usually marked in a bunch of

(a) 3

(b) 4

(c) 5

(d) 6

Solution 4

Tally are usually marked in a bunch of 5: 4 in a vertical line and one is placed diagonally.

Hence, correct option is (c).

Question 5

Let l be the lower class limit of a class-interval in a frequency distribution and m be the mid-point of the class. Then, the upper class limit of the class is

begin mathsize 12px style left parenthesis straight a right parenthesis space straight m plus fraction numerator straight l plus straight m over denominator 2 end fraction
left parenthesis straight b right parenthesis space straight l plus fraction numerator straight m plus straight l over denominator 2 end fraction
left parenthesis straight c right parenthesis space 2 straight m space minus space straight l
left parenthesis straight d right parenthesis space straight m space minus space 2 straight l end style

Solution 5

begin mathsize 12px style Let space the space class space size space equals space straight S
Lower space limit space space equals space straight l
rightwards double arrow Upper space limit space equals space straight l space plus space straight S
Now comma space mid minus point space equals space fraction numerator straight l space plus space straight l space plus space straight S over denominator 2 end fraction space equals space straight m space space left parenthesis given right parenthesis
rightwards double arrow straight S space equals space 2 straight m space minus space 2 straight l
Then comma space Upper space limit space equals space straight l space plus space straight S space equals space straight l space plus space 2 straight m space minus space 2 straight l space equals space 2 straight m space minus space straight l
Hence comma space correct space option space is space left parenthesis straight c right parenthesis. end style

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