 Hi and welcome to the session. Let us discuss the following questions that questions face a study was conducted to find out the concentration of sulphur dioxide in the air in parts per million of a certain city. The data obtained for 30 days is as follows. This is the data given to us. First part is make a look frequency distribution table for this data with class intervals as 0.000 to 0.04, 0.04 to 0.08 and so on. Second part is for how many days was the concentration of sulphur dioxide more than 0.11 parts per million. Let us now begin with the solution. In the first part we have to make a frequency distribution table for the given data. Let us make the table now. We have made two columns one for the concentration of sulphur dioxide and another for the frequency. Now here the minimum value is 0.01 and maximum value is 0.22. It is given in the question that we have to take the class intervals as 0.00 to 0.04, 0.04 to 0.08 and so on. So the class intervals are 0.00 to 0.04, 0.04 to 0.08, 0.08 to 0.12, 0.12 to 0.16, 0.16 to 0.20, 0.20 to 0.24. Remember that 0.04 is not included in this interval, it is included in the next interval. Similarly 0.08 is not included in this interval, it is included in next interval. So keeping this in mind we will now fill the frequency column. We will first find the number of values which are covered in the interval 0.00 to 0.04. Now the values which are covered in the first interval are 0.03, 0.02, 0.01 and 0.01. As there are four values which are covered in the first interval, therefore frequency of the first interval is 4. Now we will find the number of values which are covered in the next interval that is 0.04 to 0.08. Look at the data here. Values covered in the interval 0.04 to 0.08 are 0.04, 0.05, 0.06, 0.07, 0.06, 0.07, 0.05, 0.07 and 0.04. So there are nine values which are covered in the second interval. So this means frequency of the second interval is 9. In the same pattern we will calculate the frequency of other intervals. Now on adding all these frequencies we get 30 past number of days. This is the required frequency distribution table. This completes the first part. Let's now move on to the second part. In second part we have to find for how many days was the conservation of sulfur dioxide more than 0.11 parts per million. Now the number of days for which the concentration of sulfur dioxide was more than 0.11 parts per million, we add the frequencies of class 0.12 to 0.16, 0.16 to 0.20, 0.20 to 0.24. Now the frequencies of these classes are 2, 4, 2. On adding all this we get 8. For 8 days the concentration of sulfur dioxide was more than 0.11 parts per million. Hence our required answer is the concentration of sulfur dioxide was more than 0.11 parts per million for 8 days. This is our required answer. So this completes the session. Bye and take care.