Problem 1: 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. Which is the most common, and which is the rarest, blood group among these students?
Solution:
Frequency:- No. of students having the same blood group.
| Blood Group | No. of Students |
|---|
| A | 9 |
| B | 6 |
| AB | 3 |
| O | 12 |
| Total | 30 |
Most common blood group (highest frequency blood group) :- O
Rarest blood group (least frequency blood group) :- AB.
Problem 2: The distance (in km) of 40 engineers from their residence to their place of work were found as follows:
5 3 10 20 25 11 13 7 12 31
19 10 12 17 18 11 32 17 16 2
7 9 7 8 3 5 12 15 18 3
12 14 2 9 6 15 15 7 6 12
Construct a grouped frequency distribution table with class size 5 for the data given above taking the first interval as 0-5 (5 not included). What main features do you observe from this tabular representation?
Solution:
Grouped frequency distribution table of class size 5:
| Distance (in Kms) | Frequency |
|---|
| 0-5 | 5 |
| 5-10 | 11 |
| 10-15 | 11 |
| 15-20 | 9 |
| 20-25 | 1 |
| 25-30 | 1 |
| 30-35 | 2 |
| Total | 40 |
We observe that, most of the engineers live within 20 kms from their place of work. Only 4 employees out of 40 employees are living in between 20 and 35 kms from their place of work.
Problem 3: The relative humidity (in %) of a certain city for a month of 30 days was as follows:
98.1 98.6 99.2 90.3 86.5 95.3 92.9 96.3 94.2 95.1
89.2 92.3 97.1 93.5 92.7 95.1 97.2 93.3 95.2 97.3
96.2 92.1 84.9 90.2 95.7 98.3 97.3 96.1 92.1 89
(i) Construct a grouped frequency distribution table with classes 84 – 86, 86 – 88, etc.
(ii) Which month or season do you think this data is about?
(iii) What is the range of this data?
Solution:
(i) grouped frequency distribution table with classes 84 – 86, 86 – 88, etc….
| Relative humidity (in %) | Frequency |
|---|
| 84-86 | 1 |
| 86-88 | 1 |
| 88-90 | 2 |
| 90-92 | 2 |
| 92-94 | 7 |
| 94-96 | 6 |
| 96-98 | 7 |
| 98-100 | 4 |
| Total | 30 |
(ii) In the most of days the humidity is very high, so this data should be from rainy season.
(iii) range of the data = maximum value in the data – minimum value in the data
= 99.2 – 84.9
= 14.3
Problem 4: The heights of 50 students, measured to the nearest centimetres, have been found to be as follows:
161 150 154 165 168 161 154 162 150 151
162 164 171 165 158 154 156 172 160 170
153 159 161 170 162 165 166 168 165 164
154 152 153 156 158 162 160 161 173 166
161 159 162 167 168 159 158 153 154 159
(i) Represent the data given above by a grouped frequency distribution table, taking the class intervals as 160 – 165, 165 – 170, etc.
(ii) What can you conclude about their heights from the table?
Solution:
(i) grouped frequency distribution table with classes 160 – 165, 165 – 170, etc…
| Height (in cm) | Frequency |
|---|
| 150-155 | 12 |
| 155-160 | 9 |
| 160-165 | 14 |
| 165-170 | 10 |
| 170-175 | 5 |
| Total | 50 |
(ii) from the above table we can tell that height of most of the students (35 students) is less than 165 cms.
Problem 5: A study was conducted to find out the concentration of sulphur dioxide in the air in
2020-21 parts per million (ppm) of a certain city. The data obtained for 30 days is as follows:
0.03 0.08 0.08 0.09 0.04 0.17
0.16 0.05 0.02 0.06 0.18 0.20
0.11 0.08 0.12 0.13 0.22 0.07
0.08 0.01 0.10 0.06 0.09 0.18
0.11 0.07 0.05 0.07 0.01 0.04
(i) Make a grouped frequency distribution table for this data with class intervals as 0.00 – 0.04, 0.04 – 0.08, and so on.
(ii) For how many days, was the concentration of sulphur dioxide more than 0.11 parts per million?
Solution:
(i) grouped frequency distribution table with classes 0.00 – 0.04, 0.04 – 0.08, etc..
Concentration of sulphur dioxide in air (in ppm) | Frequency |
|---|
| 0.00-0.04 | 4 |
| 0.04-0.08 | 9 |
| 0.08-0.12 | 9 |
| 0.12-0.16 | 2 |
| 0.16-0.20 | 4 |
| 0.20-0.24 | 2 |
| Total | 30 |
(ii) concentration of sulphur dioxide was more than 0.11 parts per million for 8 days. ((2 + 4 + 2) days).
Problem 6: 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:
Frequency distribution table:
| No. of heads | Frequency |
|---|
| 0 | 6 |
| 1 | 10 |
| 2 | 9 |
| 3 | 5 |
| Total | 30 |
Problem 7: The value of π upto 50 decimal places is given below:
3.14159265358979323846264338327950288419716939937510
(i) Make a frequency distribution of the digits from 0 to 9 after the decimal point.
(ii) What are the most and the least frequently occurring digits?
Solution:
(i) frequency distribution table:-
| Digit | Frequency |
|---|
| 0 | 2 |
| 1 | 5 |
| 2 | 5 |
| 3 | 8 |
| 4 | 4 |
| 5 | 5 |
| 6 | 4 |
| 7 | 4 |
| 8 | 5 |
| 9 | 8 |
| Total | 50 |
(ii) most frequently occurring digits : 3 & 9 (both occurred 8 times)
least frequently occurring digits : 0 (occurred only twice).
Problem 8: 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:
(i) grouped frequency distribution table with class width 5 :
| No. of hours | frequency |
|---|
| 0-5 | 10 |
| 5-10 | 13 |
| 10-15 | 5 |
| 15-20 | 2 |
| Total | 30 |
(ii) from the above table, we can tell that only 2 children watched television for 15 or more hours a week.
Problem 9: A company manufactures car batteries of a particular type. The lives (in years) of 40 such batteries were recorded as follows:
2.6 3.0 3.7 3.2 2.2 4.1 3.5 4.5
3.5 2.3 3.2 3.4 3.8 3.2 4.6 3.7
2.5 4.4 3.4 3.3 2.9 3.0 4.3 2.8
3.5 3.2 3.9 3.2 3.2 3.1 3.7 3.4
4.6 3.8 3.2 2.6 3.5 4.2 2.9 3.6
Construct a grouped frequency distribution table for this data, using class intervals of size 0.5 starting from the interval 2 – 2.5.
Solution:
Grouped frequency distribution table with class intervals of size 0.5 starting from the interval 2 – 2.5:
| life of battery (in years) | Frequency |
|---|
| 2-2.5 | 2 |
| 2.5-3 | 6 |
| 3-3.5 | 14 |
| 3.5-4 | 11 |
| 4-4.5 | 4 |
| 4.5-5 | 3 |
| Total | 40 |