$A$ 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:
$\begin{array}{llllll}0.03 & 0.08 & 0.08 & 0.09 & 0.04 & 0.17 \\ 0.16 & 0.05 & 0.02 & 0.06 & 0.18 & 0.20 \\ 0.11 & 0.08 & 0.12 & 0.13 & 0.22 & 0.07 \\ 0.08 & 0.01 & 0.10 & 0.06 & 0.09 & 0.18 \\ 0.11 & 0.07 & 0.05 & 0.07 & 0.01 & 0.04\end{array}$
$(i)$ 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?
$\begin{array}{llllll}0.03 & 0.08 & 0.08 & 0.09 & 0.04 & 0.17 \\ 0.16 & 0.05 & 0.02 & 0.06 & 0.18 & 0.20 \\ 0.11 & 0.08 & 0.12 & 0.13 & 0.22 & 0.07 \\ 0.08 & 0.01 & 0.10 & 0.06 & 0.09 & 0.18 \\ 0.11 & 0.07 & 0.05 & 0.07 & 0.01 & 0.04\end{array}$
$(i)$ 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?
(8) To construct the grouped frequency distribution table,we count the number of observations falling into each class interval. Note that the upper limit of each class is excluded from that class.
$(ii)$ The number of days for which the concentration of $SO_2$ is more than $0.11$ $ppm$ corresponds to the sum of frequencies for the intervals $0.12 - 0.16, 0.16 - 0.20$,and $0.20 - 0.24$.
Required number of days $= 2 + 4 + 2 = 8$.
Therefore,for $8$ days,the concentration of $SO_2$ was more than $0.11$ $ppm$.
| Concentration of $SO_2$ (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 | $30$ |
$(ii)$ The number of days for which the concentration of $SO_2$ is more than $0.11$ $ppm$ corresponds to the sum of frequencies for the intervals $0.12 - 0.16, 0.16 - 0.20$,and $0.20 - 0.24$.
Required number of days $= 2 + 4 + 2 = 8$.
Therefore,for $8$ days,the concentration of $SO_2$ was more than $0.11$ $ppm$.
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