RD SHARMA Solutions for Class 9 Maths Chapter 22 - Tabular Representation of Statistical Data
Chapter 22 - Tabular Representation of Statistical Data Exercise Ex. 22.1
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.
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?
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.
(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
(ii)
(iii)
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 |
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 |
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.
Heights (in cm) |
Tally marks | Frequency |
145 - 149 | llll | 4 |
150 - 154 | llll | 9 |
155 - 159 | ll | 12 |
160 - 164 | 5 | |
Total = 30 |
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 |
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 |
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 |
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?
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.
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.
Number of heads |
Number of times (frequency) |
0 |
6 |
1 |
10 |
2 |
9 |
3 |
5 |
Total |
30 |
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?
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).
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 Exercise Ex. 22.2
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.
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 |
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.
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 | |
Chapter 22 - Tabular Representation of Statistical Data Exercise 22.26
Tally marks are used to find
(a) class intervals
(b) range
(c) frequency
(d) upper limits
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).
The difference between the highest and lowest values of the observations is called
(a) frequesncy
(b) mean
(c) range
(d) class-intervals
The difference between the highest and lowest value of observations is called 'Range' of observations.
Hence, correct option is (c).
The difference between the upper and the lower class limits is called
(a) mid-points
(b) class size
(c) frequency
(d) mean
The difference between the upper class limit and the lower class limit is called class size.
Hence, correct option is (b).
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
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).
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
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
Chapter 22 - Tabular Representation of Statistical Data Exercise 22.27
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
The number of times a particular item occurs in a given data is called its Frequency.
Hence, correct option is (b).
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
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 (9^{th}) class
Upper limit of the highest (9^{th}) class = 13.1 + 8 × 2.5 = 33.1
Hence, correct option is (b).
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
Range of observations = Highest observation - Lowest observation
= 95 - 62
= 33
Hence, correct option is (d).
Tally are usually marked in a bunch of
(a) 3
(b) 4
(c) 5
(d) 6
Tally are usually marked in a bunch of 5: 4 in a vertical line and one is placed diagonally.
Hence, correct option is (c).
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
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