Quartiles, Deciles, and Percentiles

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Grant Owens
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1 Quartiles, Deciles, and Percentiles From the definition of median that it s the middle point in the axis frequency distribution curve, and it is divided the area under the curve for two areas have the same area in the left, and in the right From this may be divided the area under the curve for four equally area and this called quartiles, in the same procedure divided the area for ten equally pieces of area is called deciles, finally where divided the area for hundred equally pieces of area is called percentiles A2 A Q1 = first quartile Q2 = second quartile Q = third quartile Where: A1 = A2 = A = A A1 A Q1 Q2 Q Data

2 A A2 A1 D1 D2 D A A5 A6 Data A A8 A A10 D D5 D6 D D8 D D1 = first decile D2 = second decile D = third decile D = ninthe decile Where: A1 = A2 = A = A= =A10

3 The same procedure for division is done for finding percentiles for any frequency distributed curve P1 = first percentile P2 = second percentile P = third percentile P = ninety ninth percentile Where: A1 = A2 = A = A= =A100 To find the quartiles, or desiles, or percentiles we follow the same procedure to fined the median Arrangement the data in ascending form only If numbering arrangement of quartiles, desiles, and percentiles is fraction then its value is for the number greater than it, if true number the value is the mean of its and the greater numbers

4 Example: find the quartiles Q1, Q2, and Q of the following data 20, 0, 25, 2, 22, 2, 6 Solution: Arrange data in ascending form, and n = odd number Ascending arrangement q1 = (1/) x n = (1/) x = 15 q1 = 2 Q1 = 22 q2 = (2/) x n = (2/) x = 5 q2 = Q2 = 25 q = (/) x n = (/) x = 525 q = 6 Q = 2

5 Example: find the quartiles Q1, Q2, and Q of the following data 20, 0, 25, 2, 22, 2, 6, 18 Solution: Arrange data in ascending form, and n = 8 even number Ascending arrangement q1 = (1/) x n = (1/) x 8 = 2 q1 = mean of (2), and () Q1 = (20+22)x(1/2) = 21 q2 = (2/) x n = (2/) x 8 = q2 = mean of (), and (5) Q2 = (2+25)x(1/2) = 2 q = (/) x n = (/) x 8 = 6 q = mean of (6), and () Q = (0+2)x(1/2) = 1

6 Example: find the the quartiles Q1, Q2, and Q of the following data Columns Load (fi) Solution: Columns Load ) find the cumulative frequency and the summation of frequencies and real interval limit (fi) Cumulative frequency Real interval ) find the arrangement number of quartiles to find quartile interval 1 q1 = (1/) x n = (1/) x 2 = 65 The interval of quartile number 1 is have the cumulative frequency = a = the real lower limit of quartiles interval = 65 n 1 = the cumulative frequency of the previous interval of the quartiles interval = f q = the frequency of quartiles interval = C = the length of quartiles interval = 20

7 ) find the arrangement number of quartiles to find quartile interval 2 q2 = (2/) x n = (2/) x 2 = 15 The interval of quartile number 2 is have the cumulative frequency = ) find the arrangement number of quartiles to find quartile interval q = (/) x n = (/) x 2 = 2025 The interval of quartile number is have the cumulative frequency =

8 Columns Load (fi) Cumulative frequency Real interval

9 Example: find the desiles D1, D5, and D8 of the following data 20, 0, 25, 2, 22, 2, 6 Solution: Arrange data in ascending form, and n = odd number Ascending arrangement d1 = (1/10) x n = (1/10) x = 0 d1 = 1 D1 = 20 d5 = (5/10) x n = (5/10) x = 5 d5 = D5 = 25 d8 = (8/10) x n = (8/10) x = 56 d8 = 6 D8 = 2

10 Example: find the desiles D1, D5, and D8 of the following data 20, 0, 25, 2, 22, 2, 6, 18 Solution: Arrange data in ascending form, and n = 8 even number Ascending arrangement d1 = (1/10) x n = (1/10) x 8 = 08 d1 = 1 D1 = 18 d5 = (5/10) x n = (5/10) x 8 = d5 = mean of (), and (5) D5 = (2+25)x(1/2) = 2 d8 = (8/10) x n = (8/10) x 8 = 6 d8 = D8 = 2

11 Example: find the desiles D1, D5, and D of the following data Columns Load (fi) Solution: Columns Load ) find the cumulative frequency and the summation of frequencies and real interval limit (fi) Cumulative frequency Real interval ) find the arrangement number of desiles to find desiles interval D1 d1 = (1/10) x n = (1/10) x 2 = 2 The interval of desiles number 1 is have the cumulative frequency = a = the real lower limit of desiles interval = 5 n 1 = the cumulative frequency of the previous interval of the desiles interval = 0 f d = the frequency of desiles interval = C = the length of desiles interval = 20

12 ) find the arrangement number of desiles to find desiles interval D5 d5 = (5/10) x n = (5/10) x 2 = 15 The interval of desiles number D5 is have the cumulative frequency = ) find the arrangement number of desiles to find desiles interval D d = (/10) x n = (/10) x 2 = 2 The interval of desiles number D5 is have the cumulative frequency =

13 Columns Load 50 6 (fi) Cumulative frequency Real interval

14 Example: find the percentiles P8, P50, and P85 of the following data 20, 0, 25, 2, 22, 2, 6 Solution: Arrange data in ascending form, and n = odd number Ascending arrangement p8 = (8/100) x n = (8/100) x = 056 p8 = 1 P8 = 20 p50 = (50/100) x n = (50/100) x = 5 p50 = P50 = 25 p85 = (85/100) x n = (85/100) x = 55 p85 = 6 P85 = 2

15 Example: find the percentiles P8, P50, and P85 of the following data 20, 0, 25, 2, 22, 2, 6, 18 Solution: Arrange data in ascending form, and n = 8 even number Ascending arrangement p8 = (8/100) x n = (8/100) x 8 = 06 p8 = 1 P8 = 18 p50 = (50/100) x n = (50/100) x 8 = p50 = mean of (), and (5) P50 = (2+25)x(1/2) = 2 p85 = (85/100) x n = (85/100) x 8 = 68 p85 = P85 = 2

16 Example: find the percentiles P8, P50, and P85 of the following data Columns Load (fi) Solution: Columns Load ) find the cumulative frequency and the summation of frequencies and real interval limit (fi) Cumulative frequency Real interval ) find the arrangement number of percentiles to find percentiles interval P8 p8 = (8/100) x n = (8/100) x 2 = 216 The interval of percentiles number 8 is have the cumulative frequency = a = the real lower limit of percentiles interval = 5 n 1 = the cumulative frequency of the previous interval of the percentiles interval = 0 f p = the frequency of percentiles interval = C = the length of percentiles interval = 20

17 ) find the arrangement number of percentiles to find percentiles interval P50 p50 = (50/100) x n = (50/100) x 2 = 15 The interval of percentiles number P50 is have the cumulative frequency = ) find the arrangement number of percentiles to find percentiles interval P85 p85 = (85/100) x n = (85/100) x 2 = 225 The interval of percentiles number P85 is have the cumulative frequency =

18 Columns Load 50 6 (fi) Cumulative frequency Real interval Example: find the percentiles arrangement of 115 in the above data a = the real lower limit of percentiles interval = 105 n = the cumulative frequency of the previous interval of the percentiles interval = 1 1 f p = the frequency of percentiles interval = X C = the length of percentiles interval = , 151, % 56% 2

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