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प्रश्न
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?
उत्तर
(i) Our class intervals will be 0 − 5, 5 − 10, 10 −15…..
The grouped frequency distribution table can be constructed 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., the number of children in class interval 15 − 20).
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संबंधित प्रश्न
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
(ii) Which month or season do you think this data is about?
(iii) What is the range of this data?
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
Represent them as a frequency distribution table with class size 1°C.
The marks scored by 55 students in a test are given below:
| Marks | 0-5 | 5-10 | 10-15 | 15-20 | 20-25 | 25-30 | 30-35 |
| No. of students | 2 | 6 | 13 | 17 | 11 | 4 | 2 |
Prepare a cumulative frequency table:
The number of books in different shelves of a library are as follows:
30, 32, 28, 24, 20, 25, 38, 37, 40, 45, 16, 20
19, 24, 27, 30, 32, 34, 35, 42, 27, 28, 19, 34,
38, 39, 42, 29, 24, 27, 22, 29, 31, 19, 27, 25
28, 23, 24, 32, 34, 18, 27, 25, 37, 31, 24, 23,
43, 32, 28, 31, 24, 23, 26, 36, 32, 29, 28, 21.
Prepare a cumulative frequency distribution table using 45-49 as the last class interval.
Given below are the cumulative frequencies showing the weights of 685 students of a school. Prepare a frequency distribution table.
| Weight (in kg) | No. of students |
| Below 25 | 0 |
| Below 30 | 24 |
| Below 35 | 78 |
| Below 40 | 183 |
| Below 45 | 294 |
| Below 50 | 408 |
| Below 55 | 543 |
| Below 60 | 621 |
| Below 65 | 674 |
| Below 70 | 685 |
The difference between the highest and lowest values of the observations is called
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
30 children were asked about the number of hours they watched TV programmes last week. The results are recorded as under:
| Number of hours | 0 – 5 | 5 – 10 | 10 – 15 | 15 – 20 |
| Frequency | 8 | 16 | 4 | 2 |
Can we say that the number of children who watched TV for 10 or more hours a week is 22? Justify your answer.
The blood groups of 30 students are recorded as follows:
A, B, O, A, AB, O, A, O, B, A, O, B, A, AB, B, A, AB, B, A, A, O, A, AB, B, A, O, B, A, B, A
Prepare a frequency distribution table for the data.
The value of π upto 35 decimal places is given below:
3.14159265358979323846264338327950288
Make a frequency distribution of the digits 0 to 9 after the decimal point.
