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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?


Answer :

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 \(S\left(O_{2}\right)\) (in ppm)Number of days (frequency)
0.00 - 0.044
0.04 - 0.089
0.08 - 0.129
0.12 - 0.162
0.16 - 0.204
0.20 - 0.242
Total30


The number of days for which the concentration of is more than 0.11 is the number of days for 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 \(S\left(O_{2}\right)\) is more than 0.11 ppm.

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