 Hello and welcome to the session. In this session, we are going to discuss the following question and the question says that, locate the median, upper and lower quartile, sixth desile, and 85th percentile for the following data. Here, the distribution is given such that the number of seats in a row is given from 10 to 18 with the corresponding number of rows given as 2, 5, 6, 11, 9, 8, 4, 3, 1. For discrete series, median md is given by the size of n plus 1 by 2th item, lower quartile, q1 is given by size of n plus 1 by fourth item, upper quartile is given by size of 3 into n plus 1 by fourth item, sixth desile, d6 is given by size of 6 into n plus 1 by tenth item, fifth percentile, 85 is given by size of 85 into n plus 1 by hundredth item, where n is the sum of the frequencies given by submission of f. Key idea, we shall proceed with the solution. The given distribution is as follows. Now, we shall find cumulative frequency and the first entry in the cumulative frequency column will be same as that of frequency, that is 2. Next will be 2 plus 5, that is 7, 7 plus 6, 13 and so on. Here, n is equal to sum of the frequencies, that is 49. From the key idea, we know that for discrete series, median md is equal to size of n plus 1 by 2th item, that is median md is equal to size of n plus 1 by 2th item, that is 49 plus 1 by 2th item, which is equal to size of 50 by 2th item, that is given by size of 25th item. Now, cumulative frequency just greater than 25 is 33, whose corresponding x value is 14. Therefore, median is equal to 14. Next, we shall find lower quartile, which is given by the size of n plus 1 by 4th item, that is lower quartile q1 is given by size of n plus 1 by 4th item, that is 49 plus 1 by 4th item equal to size of 50 by 4th item, that is the size of 12.5th item. Now, cumulative frequency just greater than 12.5 is 13, whose corresponding x value is 12. Therefore, lower quartile q1, which is equal to size of 12.5th item is given by 12. Next, we will find upper quartile and we know that upper quartile is given by the size of 3 into n plus 1 by 4th item, that is upper quartile q3 is equal to size of 3 into n plus 1 by 4th item, that is 49 plus 1 by 4th item, that is the size of 3 into 50 by 4th item, which is equal to the size of 3 into 12.5th item, that is the size of 37.5th item. Now, cumulative frequency just greater than 37.5 is 41, whose corresponding x value is 15. Therefore, upper quartile q3, which is equal to size of 37.5th item is given by 15. Now, we will find 6th item, that is 6th design given by the size of 6 into n plus 1 by 10th item, that is 6th design d6 is equal to size of 6 into n plus 1 by 10th item, that is 49 plus 1 by 10th item, that is the size of 6 into 50 by 10th item equal to the size of 30th item. Now, cumulative frequency just greater than 30 is 33, whose corresponding x value is 14. Therefore, 6th design, which is equal to the size of 30th item is given by 14. Now, we shall find out 85th percentile given by the size of 85 into n plus 1 by 100th item, that is 85th percentile p85 is given by size of 85 into n plus 1 by 100th item, that is 49 plus 1 by 100th item, which is equal to the size of 85 into 50 by 100th item, that is the size of 85 into 1 by 200th item equal to the size of 42.5th item. Now, cumulative frequency just greater than 42.5 is 45, whose corresponding x value is 16. Therefore, 85th percentile p85 which is equal to the size of 42.5th item is given by 16. Hence, we can say median is equal to 14, lower quartile q1 is given by 12th, upper quartile q3 is given by 15, 6th design d6 is given by 14 and 85th percentile p85 is equal to 16, which is the required answer. In this completed session, hope you enjoyed this session.