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Class 9 RD Sharma- Chapter 22 Tabular Representation of Statistical Data – Exercise 22.1 | Set 2

  • Last Updated : 30 Apr, 2021
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Question 11. The number of runs scored by a cricket player in 25 innings is as follows:

26, 35, 94, 48, 82, 105, 53, 0, 39, 42, 71, 0, 64, 15, 34, 15, 34, 6, 71, 0, 64, 15, 34, 15, 34, 67, 0, 42, 124, 84, 54, 48, 139, 64, 47

(i) Rearrange these runs in ascending order.

(ii) Determine the player, is the 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:

(i) Arranging the runs in ascending order as:

0, 0, 0, 0, 6, 15, 15, 15, 15, 26, 34, 34, 34, 34, 35, 39, 42, 42, 47, 48, 48, 53, 54, 64, 64, 64, 67, 71, 71, 82, 84, 90, 94, 124, 139.

(ii) From the above arrangement, it is observed that,

The Highest Score out of all the runs is 139.

(iii) 3 times, the player was not successful in scoring any run.

(iv) He scored 3 centuries at all.



(v) 12 times, he scored more than 50 runs.

Question 12. The class size of distribution is 25 and the first class interval is 200-224. There are seven class-interval.

(i) Write the class intervals.

(ii) Write the class marks of each interval.

Solution:

Given that,

The Class size is equal to 25,

The first class interval is 200-224

There are seven class intervals in all. 

(i) These seven class interval are as follows:

200-224, 225-249, 250-274, 275-299, 300-324, 325-349, 350-374.

(ii) Class marks of the following class intervals are:

  • For class interval, 200-224:

 \begin{aligned}\frac{200+224}{2}&=\frac{424}{2}\\&=212\end{aligned}

  • For class interval, 225-249:

\begin{aligned}\frac{225+249}{2}&=\frac{479}{2}\\&=237\end{aligned}

  • For class interval, 250-274:

\begin{aligned}\frac{250+274}{2}&=\frac{524}{2}\\&=287\end{aligned}

  • For class interval, 300-324:

\begin{aligned}\frac{300+324}{2}&=\frac{674}{2}\\&=312\end{aligned}

  • For class interval, 325-349:

\begin{aligned}\frac{325+349}{2}&=\frac{674}{2}\\&=337\end{aligned}

  • For class interval, 350-374:

\begin{aligned}\frac{350+374}{2}&=\frac{724}{2}\\&=362\end{aligned}

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, 78, 82, 87, 92, 97, 102

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



Solution:

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

The class size for the given data is:

114 – 104 = 10

The class limits are obtained as: 

Class mark

Lower class limit

Upper class limit

Class limit

104

104 – 10/2 = 99

104 + 10/2 = 109

99 – 109

114

114 – 10/2 = 109

114 + 10/2 = 119

109 – 119

124

119

129

119 – 129

134

129

139

129 – 139

114

139

149

139 – 149

154



149

159

149 – 159

164

159

169

159 – 169

(ii) 47, 52, 57, 62, 67, 72, 78, 82, 87, 92, 97, 102

The class size for the given data is:

52 – 47 = 5

The class limits are obtained as:  

Class mark

Lower class limit

Upper class limit

Class limit

47

47 – 5/2 = 44.5

47 + 5/2 = 49.5

44.5 + 49.5

52

49.5

54.5

49.5 + 54.5

57

54.5

59.5

54.5 – 59.5

62

59.5

64.5



59.5 – 64.5

67

64.5

69.5

64.5 – 69.5

72

69.5

74.5

69.5 – 74.5

77

74.5

79.5

74.5 – 79.5

82

79.5

84.5

79.5 – 84.5

87

84.5

89.5

84.5 – 89.5

92

89.5

94.5

89.5 – 94.5

97

94.5

99.5

94.5 – 99.5

102

99.5

104.5

99.5 – 104.5

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

The class size for the given data is:

17.5 – 12.5 = 5

The class limits are obtained as: 

Class mark

Lower-class limit

Upper-class limit



Class limit

12.5

12.5 – 2.5 = 10

12.5 + 2.5 = 15

10 – 15

17.5

17.5 – 2.5 = 15

17.5 + 2.5 = 20

15 – 20

22.5

22.5 – 2.5 = 20

22.5 + 2.5 = 25

20 – 25

27.5

27.5 – 2.5 = 25

27.5 + 2.5 = 30

25 – 30

32.5

32.5 – 2.5 = 30

32.5 + 2.5 = 35

30 – 35

37.5

37.5 – 2.5 = 35

37.5 + 2.5 = 40

35 – 40

42.5

42.5 – 2.5 = 40

42.5 + 2.5 = 45

40 – 45

47.5

47.5 – 2.5 = 45

47.5 + 2.5 = 50

45 – 50

Question 14. 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, 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.

Solution:

The frequency distribution of the following sequence of data : 

Number of children

Tally marks

Number of families



0

\bcancel{||||}

5

1

\bcancel{||||}\,||

7

2

\bcancel{||||}\,\bcancel{||||}\,||

12

3

\bcancel{||||}

5

4

\bcancel{||||}\,|

6

5

|||

3

6

|||

3

Question 15. Marks scored by 40 students of class IX in mathematics are given below:

81, 55, 68, 79, 85, 43, 29, 68, 54, 73, 47, 35, 72, 64, 95, 44, 50, 77, 64, 35, 79, 52, 45, 54, 70, 83, 62, 64, 72, 92, 84, 76, 63, 43, 54, 38, 73, 68, 52, 54.

Solution:

Obtaining the frequency distribution for the marks scored by 40 students of class IX are as follows: 

Marks

Tally marks

Frequency

20 – 30

|

1



30 – 40

|||

3

40 – 50

\bcancel{||||}

5

50 – 60

\bcancel{||||}\,|||

8

60 – 70

\bcancel{||||}\,|||

8

70 – 80

\bcancel{||||}\,||||

9

80 – 90

||||

4

90 – 100

||

2

Total = 40

Question 16. 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, 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:

The frequency distribution table with 160 – 164 as one of the class intervals is as follows : 

Height (in cm)

Tally marks

Frequency

145 – 149

||||

4

150 – 154

\bcancel{||||}\,||||

9

155 – 159

\bcancel{||||}\,\bcancel{||||}\,||

12

160 – 164

\bcancel{||||}\,|



6

Total = 30

Question 17. The monthly wages of 30 workers in a factory are given below:

830, 835, 890, 810, 835, 836, 869, 845, 898, 890, 820, 860, 832, 833, 855, 845, 804, 808, 812, 840, 885, 835, 836, 878, 840, 868, 890, 806, 840, 890.

Solution:

Obtaining the frequency distribution for the monthly wages of 30 workers in a factory are as follows: 

Height (in cm)

Tally marks

Frequency

800 – 810

|||

3

810 – 820

||

2

820 – 830

|

1

830 – 840

\bcancel{||||}\,|||

8

840 – 850

\bcancel{||||}

5

850 – 860

|

1

860 – 870

|||

3

870 – 880

|

1

880 – 890

|

1

890 – 900

\bcancel{||||}

5

Total = 30

Question 18. The daily maximum temperatures (in degree Celsius) recorded in a certain city during the month of November are as follows:

25.8, 24.5, 25.6, 20.7, 21.8, 20.5, 20.6, 20.9, 22.3, 22.7, 23.1, 22.8, 22.9, 21.7, 21.3, 20.5, 20.9, 23.1, 22.4, 21.5, 22.7, 22.8, 22.0, 23.9, 24.7, 22.8, 23.8, 24.6, 23.9, 21.1.

Solution:

Obtaining the frequency distribution for the daily maximum temperatures (in degree Celsius) recorded in a certain city during the month of November are as follows: 

Maximum temperature (in degree Celsius)

Tally marks

Frequency

20.0 – 21.0

\bcancel{||||}\,|

6

21.0 – 22.0

\bcancel{||||}



5

22.0 – 23.0

\bcancel{||||}\,||||

9

23.0 – 24.0

\bcancel{||||}

5

24.0 – 25.0

|||

3

25.0 – 26.0

||

2

Total = 30

Question 19. Construct a frequency table with equal class intervals from the following data on the monthly wages (in rupees) of 28 laborers 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, 218, 306, 292, 254, 278, 210, 240, 280, 316, 306, 215, 256, 236.

Solution:

The following table depicts the frequency distribution on the monthly wages (in rupees) of 28 laborers working in a factory : 

Monthly wages (in rupees)

Tally marks

Frequency

210 – 230 

||||

4

230 – 250

||||

4

250 – 270

\bcancel{||||}

5

270 – 290

|||

3

290 – 310

\bcancel{||||}\,||

7

310 – 330

\bcancel{||||}

5

Total = 28

Question 20. The daily minimum temperatures in degree Celsius recorded in a certain arctic region are as follows:

-12.5, -10.8, -18.6, -8.4, -10.8, -4.2, -4.8, -6.7, -13.2, -11.8, -2.3, -1.2, -2.6, 0, 2.4, 0, 3.2, 2.7, 3.4, 0, -2.4, 0, 3.2, 2.7, 3.4, 0, -2.4, -5.8, -8.9, -14.6, -12.3, -11.5, -7.8, -2.9.



Represent them as frequency distribution table taking -19.9 to  -15 as the first class interval.

Solution:

Frequency distribution with lower limit included and upper limit excluded is depicted taking the first class interval as -19.9 to  -15 :

Temperature

Tally marks

Frequency

-19.9 to -15

||

2

-15 to -10.1

\bcancel{||||}\,||

7

-10.1 to -5.2

\bcancel{||||}

5

-5.2 to -0.3

||||

4

-0.3 to -4.6

\bcancel{||||}\,\bcancel{||||}\,\bcancel{||||}\,||

17

Total = 35

Question 21. The blood groups of 30 students of class VIII are recorded 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. Find out which is the most common and which is the most rarest blood group among these students.

Solution:

Given that,

9 students out of the given blood groups have blood group A, 6 as B, 3 as AB and 12 as O

The frequency table representing the data along with the corresponding frequency is as follows:

Blood groupNumber of students

A

9

B

6

AB

3

O

12

Total

30

Since, 12 students have their blood group as O, which is the maximum, therefore the most common blood group is O. Also, 3 students have their blood group as AB, which implies that the rarest blood group is AB.

Question 22. Three coins were 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

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

Prepare a frequency distribution table for the data given above.

Solution:

The frequency distribution table for the data of the three coins being tossed 30 times can be computed as follows:

Number of heads

Frequency

0

6

1



10

2

9

3

5

Total=30

Question 23. Thirty children were asked about the number of hours they watched TV programs 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, 2.

(i) Make a 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:

(i) The following class intervals are 0 – 5, 5 – 10, 10 – 15 constructed.

The grouped frequency distribution table taking into account these class intervals 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 at least 15  hours a week is 2, which implies that the number of children in the class interval 15 – 20 is equivalent to 2.

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