Question

Question 12
$\begin{array}{cc}{\text{Concentration of SO}}_2\text{(in ppm)}& \text{Number of days (frequency)}\\ 0.00-0.04& 4\\ 0.04-0.08& 9\\ 0.08-0.12& 9\\ 0.12-0.16& 2\\ 0.16-0.20& 4\\ 0.20-0.24& 2\\ \text{Total}& 30\end{array}$
The above frequency distribution table represents the concentration of Sulphur dioxide in the air in parts per million of a certain city for 30 days. 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

Total number of days =30
Number of days for which the concentration of Sulphur dioxide was in the interval of 0.12 −0.16= 2

$\text{P(concentration of Sulphur dioxide in the interval 0.12 - 0.16)}=\frac{\text{Number of days for which the concentration of Sulphur dioxide was in the interval of 0.12 - 0.16}}{\text{Total number of days}}$
Hence, required probability, $P=\frac{2}{30}=\frac{1}{15}$