A study was conducted to find the concentration of Sulfur dioxide (SO2) in a city
atmosphere in ppm. The information received in 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
Based on this information,
(i) Make a grouped frequency distribution table for this class intervals as 0.00-0.04,
0.04-0.08… and so on.
(ii) For how many days will the concentration of SO2be more than 0.11 ppm?
Answers
Given : 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
concentration of Sulfur dioxide (SO2) in a city atmosphere in ppm for 30 day
To Find : Make a grouped frequency distribution table for this class intervals as 0.00-0.04,
For how many days will the concentration of SO2be more than 0.11 ppm?
Solution:
Class Interval 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
more than 0.11 ppm = 0.12 and above
= 2 + 4 + 2
= 8
8 days the concentration of SO2be more than 0.11 ppm
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Answer:
Given : 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
concentration of Sulfur dioxide (SO2) in a city atmosphere in ppm for 30 day
To Find : Make a grouped frequency distribution table for this class intervals as 0.00-0.04,
For how many days will the concentration of SO2be more than 0.11 ppm?
Solution:
Class Interval 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
more than 0.11 ppm = 0.12 and above
= 2 + 4 + 2
= 8
8 days the concentration of SO2be more than 0.11 ppm
Learn More:
The tally marks |||| represents the number count (a) 5 (b)
brainly.in/question/4227546
The representation of 'one picture to many objects' in a Pictograph is ...
brainly.in/question/11758889