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 0.08 0.08 0.09 0.04 0.170.16 0.05 0.02 0.06 0.18 0.200.11 0.08 0.12 0.13 0.22 0.070.08 0.01 0.10 0.06 0.09 0.180.11 0.07 0.05 0.07 0.01 0.04 (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?

Options

A. `6`
B. `7`
C. `9`
D. `8`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Organize the Data
We have the following data for the concentration of sulphur dioxide in parts per million (ppm) over 30 days:
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

### Step 2: Define Class Intervals
We will create class intervals as follows:
- 0.00 - 0.04
- 0.04 - 0.08
- 0.08 - 0.12
- 0.12 - 0.16
- 0.16 - 0.20
- 0.20 - 0.24

### Step 3: Count the Frequency for Each Interval
Now we will count how many days fall into each class interval.

- **0.00 - 0.04**: 0.03, 0.02, 0.01, 0.01, 0.04 (Total: 4 days)
- **0.04 - 0.08**: 0.04, 0.05, 0.06, 0.07, 0.07, 0.06, 0.08, 0.08, 0.08 (Total: 9 days)
- **0.08 - 0.12**: 0.09, 0.09, 0.10, 0.11, 0.08, 0.12 (Total: 6 days)
- **0.12 - 0.16**: 0.13, 0.12, 0.16 (Total: 3 days)
- **0.16 - 0.20**: 0.17, 0.18, 0.18, 0.20 (Total: 4 days)
- **0.20 - 0.24**: 0.22 (Total: 1 day)

### Step 4: Create the Frequency Distribution Table
Now we can summarize this information in a frequency distribution table:

| Class Interval | Frequency (Number of Days) |
|----------------|---------------------------|
| 0.00 - 0.04 | 4 |
| 0.04 - 0.08 | 9 |
| 0.08 - 0.12 | 6 |
| 0.12 - 0.16 | 3 |
| 0.16 - 0.20 | 4 |
| 0.20 - 0.24 | 1 |
| **Total** | **30** |

### Step 5: Answer the Second Question
Now we need to find out for how many days the concentration of sulphur dioxide was more than 0.11 ppm. This means we will count the days in the intervals:
- 0.12 - 0.16 (3 days)
- 0.16 - 0.20 (4 days)
- 0.20 - 0.24 (1 day)

Adding these gives us:
3 + 4 + 1 = 8 days

### Final Answers:
(i) The grouped frequency distribution table is as shown above.
(ii) The concentration of sulphur dioxide was more than 0.11 ppm for **8 days**.