 Hello and welcome to the session. In this session, we are going to discuss the following question and the question says that calculate the values of Q1, Q3, D8, P56 and median for the following data. Here, the distribution is given such that the wages are given as 0 to 10, 10 to 20, 20 to 30, 30 to 40, 40 to 50, 50 to 60 with the corresponding number of workers as 20, 39, 32, 29, 25 and 15. For continuous series, lower quartiles can be calculated by taking the size of n by 4th item and is given by the formula Q1 is equal to L plus I upon F into n by 4 minus C. Upper quartile can be calculated by taking the size of 3n by 4th item and is given by the formula Q3 is equal to L plus I upon F into 3n by 4 minus C. Eighth decile can be calculated by taking the size of 8n by 10th item and is given by the formula D8 is equal to L plus I upon F into 8n by 10 minus C. 56 percentile can be calculated by taking the size of 56n by 100 and is given by the formula P56 is equal to L plus I upon F into 56n by 100 minus C and medium can be calculated by taking the size of n by 2th item and is given by the formula MD is equal to L plus I upon F into n by 2 minus C where in each of the above equations n is equal to sum of the frequencies and is given by submission of F. With this key idea, let us proceed with the solution. The given distribution is as follows. Next we shall find cumulative frequency and the first entry in the cumulative frequency column will be same as that of frequency that is 20. Next will be 20 plus 39 that is 59, 59 plus 32, 91 and so on. Here n is equal to the sum of the frequencies that is 160. Let us first find the lower quartile. Lower quartile is given by the size of n by 4th item and n is equal to 160 therefore 160 by 4th item which is equal to size of 40th item. Now cumulative frequency just greater than 40 is 59 whose corresponding class interval is 10 to 20 therefore Q1 lies in the class interval 10 to 20 and from the key idea we know that lower quartile Q1 is equal to L plus I upon F into n by 4 minus C where L is the lower limit of the class interval that is 10, I is the width of the class interval given by 20 minus 10 that is 10, F is the frequency of the class interval that is 39, C is the cumulative frequency of the class just lower than the quartile class and is given by 20 therefore C is equal to 20. Now putting all these values in the above formula we get Q1 is equal to L that is 10 plus I upon F that is 10 upon 39 into n by 4 minus C that is 160 by 4 minus 20 that is 10 plus 10 by 39 into 40 minus 20 which is equal to 10 plus 10 by 39 into 20 that is 10 plus 200 by 39 which is equal to 10 plus 5.12 and thus we get lower quartile is equal to 15.12 upper quartile is given by the size of 3n by 4th item that is 3 into 160 by 4th item as the value of n is 160 which is equal to the size of 120th item now cumulative frequency equal to 120 lies in the class interval 30 to 40 therefore upper quartile Q3 lies in the class interval 30 to 40. Now from the key idea upper quartile is given by the formula Q3 is equal to L plus I upon F into 3n by 4 minus C I is the width of the class interval given by 40 minus 30 that is 10 F is the frequency of the class interval that is 29 C is the cumulative frequency of the class just lower than the quartile class and is given by 91 therefore C is equal to 91 putting all these values in the above formula we get Q3 is equal to L that is 30 plus I upon F that is 10 upon 29 into 3n by 4 that is 3 into 160 by 4 minus C that is 91 which is equal to 30 plus 10 upon 29 into 120 minus 91 that is 30 plus 10 upon 29 into 29 which is equal to 30 plus 10 therefore we get upper quartile Q3 is equal to 40 now we will calculate 8th decil which is given by the size of 810 by 10th item that is the size of 8 into 160 by 10th item which is given by size of 128th item now cumulative frequency just greater than 128 is 145 whose corresponding interval is 40 to 50 therefore d8 lies in the class interval 40 to 50 and we know that 8th decil d8 is given by the formula d8 is equal to L plus I upon F into 810 by 10 minus C where L is equal to the lower limit of the class interval that is 40 I is the width of the class interval given by 50 minus 40 that is 10 F is the frequency of the class interval that is 25 and C is the cumulative frequency of the class just lower than the decil class and is given by 120 therefore C is equal to 120 now substituting all these values in the above formula we get 8th decil d8 is equal to L that is 40 plus I upon F that is 10 upon 25 into 8n by 10 that is 8 into 160 by 10 minus C that is 120 which is equal to 40 plus 10 by 25 into 128 minus 120 that is 40 plus 10 by 25 into 8 which is equal to 40 plus 10 by 5 that is 40 plus 3.2 which gives the value of 8 decil as 43.2 next we shall find 56 percentile which is given by the size of 56n by 100 item that is the size of 56 into 160 by 100 item that is the size of 448 by 5th item which is equal to the size of 89.6 item now cumulative frequency just greater than 89.6 is 91 whose class interval is 20 to 30 therefore 56 percentile p 56 lies in the class interval 20 to 30 and we know that p 56 is given by L plus I upon F into 56n by 100 minus C where L is the lower limit of the class interval that is 20 I is the width of the class interval given by 30 minus 20 that is 10 F is the frequency of the class interval that is 32 and C is the cumulative frequency of the class just lower than the percentile class and is given by 59 therefore C is equal to 59 now putting all these values in the above formula we get 56 percentile p 56 is equal to L that is 20 plus I upon F that is 10 upon 32 into 56n by 100 that is 56 into 160 by 100 minus C that is 59 which is equal to 20 plus 10 by 32 into 89.6 minus 59 which is equal to 20 plus 10 by 32 into 30.6 which gives 20 plus 306 upon 32 that is 20 plus 153 by 16 which is equal to 20 plus 9.56 that is 29.56 therefore p 56 is equal to 29.56 now we will calculate median which is given by the size of n by 2th item that is the size of 160 by 2th item equal to the size of 8th item now cumulative frequency just greater than 80 is 91 whose corresponding class interval is 20 to 30 therefore median lies in the class interval 20 to 30 from the key idea we know that median is given by md is equal to L plus I upon F into n by 2 minus C where L is the lower limit of the class interval that is 20 I is the width of the class interval given by 30 minus 20 that is 10 F is the frequency of the class interval that is 32 C is the cumulative frequency of the class just lower than the median class and is given by 59 therefore C is equal to 59 now substituting all these values in the formula we get median md is equal to L that is 20 plus I upon F that is 10 upon 32 into n by 2 that is 160 by 2 minus C that is 59 which is equal to 20 plus 10 by 32 into 80 minus 59 which gives 20 plus 10 by 32 into 21 which is equal to 20 plus 105 upon 16 that is 20 plus 6.56 which is equal to 26.56 therefore median is equal to 26.56 thus we can say lower quartile q1 is equal to 15.12 upper quartile q3 is equal to 40 eighth this is d8 is given by 43.256 percentile p56 is equal to 29.56 and median md is given by 26.56 which is the required answer this completes our session hope you enjoyed this session