Question

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,{\text{ }}0.08,{\text{ }}0.08,{\text{ }}0.09,{\text{ }}0.04,{\text{ }}0.17} \\\ {0.16,{\text{ }}0.05,{\text{ }}0.02,{\text{ }}0.06,{\text{ }}0.18,{\text{ }}0.20} \\\ {0.11,{\text{ }}0.08,{\text{ }}0.12,{\text{ }}0.13,{\text{ }}0.22,{\text{ }}0.07} \\\ {0.08,{\text{ }}0.01,{\text{ }}0.10,{\text{ }}0.06,{\text{ }}0.09,{\text{ }}0.18} \\\ {0.11,{\text{ }}0.07,{\text{ }}0.05,{\text{ }}0.07,{\text{ }}0.01,{\text{ }}0.04} \end{array}$$ Using this table, find the probability of the concentration of sulphur dioxide in the interval 0.12-0.16 on any of these days.

Answer 1

Find the number of days whose concentration lies in the given interval and find the probability for it.
First, we find what is given, we are given30 days of data, it means that there are 30 possibilities and then calculate on how many days the concentration of sulphur dioxide lies in the given interval. Then we use the probability formula and substitute according and get the required probability.

Complete step by step solution:
For probability to be calculated, we need to first find out the total possible events and also favourable outcomes out of the total outcomes.
First, we calculate the total outcomes, we are given a total of 30 days data, then it is clear that the total maximum outcomes are 30.
Now, coming to the favourable outcomes.
From the given 30 days, we can see that there are only two days which have concentration of the given range, which are as following
$0.12,0.13$
Hence, the favourable outcomes from the total outcomes are 2.
Now that we have favourable outcome and total outcomes, we can calculate the probability
The probability of an event = $\dfrac{{favourable\;outcomes}}{{total\;outcomes}}$
Probability of a days with concentration of interval 0.12-0.16
= $\dfrac{{number{\text{ }}of{\text{ days }}with{\text{ concentration}}}}{{totaldays}}$

Probability of a days with concentration of interval 0.12-0.16
=$\dfrac{2}{{30}} = \dfrac{1}{{15}}$

Note: We have to carefully notice what has been asked in the question, since we have been asked for the range with intervals, the lower limit of the given interval gets included and the lower limit doesn’t, that is why 0.12 is considered and 0.16 is not considered.