Advertisements
Advertisements
प्रश्न
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?
उत्तर
Taking class intervals as 0.00, −0.04, 0.04, −0.08, and so on, a grouped frequency table can be constructed as follows.
|
Concentration of SO2 (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 |
| 0.20 − 0.24 | 2 |
| Total | 30 |
The number of days for which the concentration of SO2 is more than 0.11 is the number of days for which the concentration is 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 SO2 is more than 0.11 ppm.
APPEARS IN
संबंधित प्रश्न
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.
The monthly wages of 30 workers in a factory are given below:
83.0, 835, 890, 810, 835, 836, 869, 845, 898, 890, 820, 860, 832, 833, 855, 845, 804, 808,
812, 840, 885, 835, 836, 878, 840, 868, 890, 806, 840, 890.
Represent the data in the form of a frequency distribution with class size 10.
The daily minimum temperatures in degrees Ce1siu& recorded in a certain Arctic region are
as follows:
−12.5, −10.8, −18.6, −8.4, −10.8, −4.2, −4.8, −6.7, −13.2, −11.8, −2.3, 1.2, 2.6, 0, 2.4,
0, 3.2, 2.7, 3.4, 0, − 2.4, − 2.4, 0, 3.2, 2.7, 3.4, 0, − 2.4, − 5.8, -8.9, 14.6, 12.3, 11.5, 7.8,2.9.
Represent them as frequency distribution table taking − 19.9 to − 15 as the first class
interval.
Thirty children were asked about the number of hours they watched T.V. programmers 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?
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 |
In the class intervals 10-20, 20-30, 20 is taken in
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
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.
Prepare a continuous grouped frequency distribution from the following data:
| Mid-point | Frequency |
| 5 | 4 |
| 15 | 8 |
| 25 | 13 |
| 35 | 12 |
| 45 | 6 |
Also find the size of class intervals.
