# Ex.14.2 Q5 Statistics Solution - NCERT Maths Class 9

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## 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:

\begin{align}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{align}

1. Make a grouped frequency distribution table for this data with class intervals as 0.00 - 0.04, 0.04 - 0.08, and so on.
2. For how many days, was the concentration of sulphur dioxide more than 0.11 parts per million?

Video Solution
Statistics
Ex exercise-14-2 | Question 5

## Text Solution

What is known?

• Concentration of sulphur dioxide in the air (in ppm) of a certain city observed over $$30$$ days
• Class intervals of ‘$$0.00 – 0.04’, ‘0.04 - 0.08$$’, and so on. Hence class size is $$0.04.$$

What is unknown?

Constructing a grouped frequency table for the given data which will help us to find out for how many days was the concentration of sulphur dioxide more than $$0.11$$ ppm.

Reasoning:

According to class interval we can check how many numbers lie in between, like this we can draw frequency distribution table and can be conclude about no. of days in which Sulphur dioxide more than $$0.11$$ parts per million from table.

Steps:

A grouped frequency distribution table with a class size of $$‘0.04$$’ needs to be constructed for the given data.

 CONCENTRATION OF SULPHUR DI OXIDE (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 $$30$$

From this table we can see that the number of days during which the concentration of Sulphur dioxide is more than $$0.11$$ppm, falls over three class intervals, $$0.12-0.16, 0.16-0.20$$ and $$0.20-0.24.$$

So,$$2+4+2\text{ }\Rightarrow \text{ }8$$

$$8$$ days had a concentration of Sulphur dioxide more than $$0.11$$ ppm.

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