Chapter 22 – Tabular Representation of Statistical Data Exercise Ex. 22.1
Question 1
Solution 1
Question 2
Solution 2
Question 3
Solution 3
Question 4Why do we group data?Solution 4The 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 9The 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 11The 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 11The 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 13Write 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, 47.5Solution 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 20Here 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 21Three 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 21By 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 22Thirty 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 Exercise Ex. 22.2
Question 1
Solution 1
Question 2
Solution 2
Question 3
Solution 3
Question 4Following 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 8The 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
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