- 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:
We can observe that, 9 students has A blood group, 6 students has B blood group, 3 students has AB blood group, 12 students has O blood group.
Thus, let us represent the blood groups of 30 students in class as follows:
Blood Group | Number of Students |
A | 9 |
B | 6 |
AB | 3 |
O | 12 |
Total | 30 |
Here, the most common blood group is O which is 12 in number, and AB is the rarest blood group which 3 in number among given 30 students.
- 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:
It is given that, class size is 5. Therefore, the class intervals will be 0 – 5 , 5 – 10, 10 – 15… . Thus, a grouped frequency distribution table can be constructed as follows:
Distance(in km) | Tally mark | Number of Engineers |
0 – 5 | ||||| | 5 |
5 – 10 | ||||| ||||| | | 11 |
10 – 15 | ||||| ||||| | | 11 |
15 – 20 | ||||| |||| | 9 |
20 – 25 | | | 1 |
25 – 30 | | | 1 |
30 – 35 | || | 2 |
Total | 40 |
It is observed that, very few engineers whose homes are at more than or equal to 20 km distance from their work. Most of the engineers among 40 engineers are up to 15 km distance from their residences.
- 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) it given that, a grouped frequency distribution table of class size 2 has to be constructed . The class intervals will be 84 – 86, 86 – 88, 88 – 90, … Then, by given data, required table can be constructed as follows:
Relative humidity(in %) | Number of days(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) It can be observed that the relative humidity is high. Therefore, the data is abot a month of rainy season.
(iii) Range of data = Maximum value – Minimum value = 99.2 – 84.9 = 14.3
- 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) Given, the data given above by a grouped frequency distribution table, taking the class intervals as 160 – 165, 165 – 170 i.e., class interval will be 5. Then the required table by taking class interval 2 will be as follows:
Height (in cm) | Number of students (frequency) |
150 – 155 | 12 |
155 – 160 | 9 |
160 – 165 | 14 |
165 – 170 | 10 |
170 – 175 | 5 |
Total | 50 |
(ii) It can be concluded that more than 50% of the students are shorter than 165 cm.
- A study was conducted to find out the concentration of sulphur dioxide in the air in 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) It is given that, a grouped frequency distribution table for this data with class intervals as 0.00 – 0.04, 0.04 – 0.08, and so on i.e., the class interval will be 0.04. Then the required table is as follows:
Concentration of SO_{2}(in ppm) | Number of days (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 |
0.29 – 0.24 | 2 |
Total | 30 |
(ii)The number of days for which concentration of sulphur dioxide more than 0.11 parts per million in between 0.12 – 0.16, 0.16 – 0.20, 0.20 – 0.24
Required number of days = 2+ 4 + 2 = 8
Therefore, for 8 days, the concentration of SO_{2} is more than 0.11 ppm.
- 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:
The frequency distribution table for the given will be as follows:
Number of heads | Number of times(frequency) |
0 | 6 |
1 | 10 |
2 | 9 |
3 | 5 |
Total | 30 |
- The value of π up-to 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) Given, make a frequency distribution of the digits from 0 to 9 after the decimal point . The required table will be constructed as follows:
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)From the table, 3 and 9 are the most occurring digits and 0 is the least frequently occurring digits among the value of π up-to 50 decimal places.
- 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)Given, make a grouped frequency distribution table for this data, taking class width 5 and one of the class intervals as 5 – 10. i.e., class interval is 5.
Hours | Number of children |
0 – 5 | 10 |
5 – 10 | 13 |
10 – 15 | 5 |
15 – 20 | 2 |
Total | 30 |
(ii) 2 children watched television for 15 or more hours a week.
- 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:
A grouped frequency table of class size 0.5 has to be constructed starting fro class interval 2 – 2.5. Then the required table will be as follows:
Lives of batteries (in hours) | Number of batteries |
2 – 2.5 | 2 |
2.5 – 3.0 | 6 |
3.0 – 3.5 | 14 |
3.5 – 4.0 | 11 |
4.0 – 4.5 | 4 |
4.5 – 5.0 | 3 |
Total | 40 |
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