Fractiles-141012210550-conversion-gate01.pdf

FRACTILES GROUP 4

Fractiles are measures of location or position which include not only central location but also any position based on the number of equal divisions in a given distribution. If we divide the distribution into four equal divisions, then we have quartiles denoted by Q1, Q2, Q3, and Q4. The most commonly used fractiles are the quartiles, deciles, and percentiles.

Fractiles for Ungrouped Data QUARTILES divide a distribution into four equal parts. For example, Q1, or the first quartile, locates the point which is greater than 25% of the items in distribution. Q3 is the 3rd quartile  Q3 = 3N th item 4 This means that 75% of the observations lie below this value.

Q2 is the 2nd quartile  Q2 = 2N th item or the median 4 Q1 is the 1st quartile  Q1 = N th item 4

DECILES are values that divide a distribution into 10 equal parts. D1 is the 1st decile  D1 = N th item 10 D3 is the 3rd decile  D3 = 3N th item 10 D5 is the 5th decile  D5 = 5N th item or the median . 10

PERCENTILES are values that divide the distribution into 100 equal parts. P10 or the tenth percentile means the 10th item in the distribution which is 10% higher than the rest of the items. P1 is the 1st percentile  P1 = Nth item 100 P25 is the 25th percentile  P25 = 25Nth item or Q1 100

P50 is the 50th percentile  P50 = 50Nth item or the median 100 P67 is the 67th percentile  P67 = 67Nth item 100

Example 1: Calculate Q1, Q2, Q3, D1, D4, D5, D7, P10, P25, P50 and P70 for the following IQ scores: 87 90 95 96 97 98 98 99 100 100 100 100 100 101 101 102 102 102 103 104 105 107 110 Q1 = N th = 23 th = 5.75th item, which is 98. 4 4

Note: for odd number of observations, when the item number is fractional, take the next higher item. Since the 5th item 97, the 5.75th item is the next value, which is 98. This means that the score of 98 is higher than 25% of the items in the distribution. If the number of cases is even, take the point midway between the two items located at the middle of the distribution.

Q2 = 2N th = 2(23) th = 23 th = 11.5th item, 4 4 2 which is 100. This means that the score of 100 is higher than 50% of the items in distribution. Q3 = 3N th = 3(23) th = 69 th = 17.25 th item, 4 4 4 which is 102. D1 = N th = 23 th = 2.3 th item, which is 95. 10 10

D4 = 4N th= 4(23)th= 92 th=9.2th item, which is 100. 10 10 10 D5= 5N th= 5(23)th= 115 th= 11.5th item, which is 100. 10 10 10 D7= 7N th=7(23)th= 161 th=16.1th item, which is 102. 10 10 10 P10= 10N th=10(23)th= 23 th=2.3th item, which is 95. 100 100 10 P25= 25N th=25(23)th= 23 th=5.7th item, which is 98. 100 100 4

P50= 50Nth=50(23)th= 23th=11.5th item, which is 100. 100 100 2 P70= 70Nth=70(23)th= 1,610th=16.1th item, 100 100 100 which is 102. NOTE: that median is equal to Q2, D5, and P50

Grouped data • Estimate the cumulative frequencies on the table • Find N/4 or one-fourth of the number of the cases in the distribution. • Determine the class limit in which N/4 case falls • Compute Q1 by using the formula; Q1 = L+C (N/4-ΣCf<) fc

Where ; Q1 = is the first quartile L = lower real limit of the first quartile class N = total number of cases ΣCf< = sum of the cumulative frequencies “ lesser than” up to but below the first quartile class C = class interval fc = frequency of the Q1 class

CLASS LIMIT

FREQUENCY

CUMULATIVE FREQ. Cf<

46-48

1

35

43-45

1

34

40-42

2

33

37-39

3

31

34-36

3

28

31-33

4

25

28-30

7

21

25-27

5

14

22-24

3

9

19-21

2

6

16-18

2 1 1

4

13-15 10-12

2 1

Q1 = L+C (N/4- ΣCf<) fc = 21.5 + 3 (8.75 – 6) 3 = 21.5 + 3 ( 2.75) 3 = 21.5 + 2.75 Q1= 24.25

N/4 = 8.75 L = 21.5 ΣCf< = 6 fc = 3 C=3

Q3 = L+C (3N/4- ΣCf<) fc = 33.5 + 3 (26.5 – 25) 3 = 33.5 + 1.5 = 35

N/4 = 26.25 L = 33.5 ΣCf< = 25 fc = 3 C=3

Decile • Estimate the cumulative frequency “lesser than” • Look for 7N/10 or seven- tenth of the number of cases in distribution • Find the class limit in which the 7N/10 case falls • Compute D7 by using the formula; D7 = L+C (7N/10-ΣCf<) fc

Where ; D7= is the seventh L = lower real limit of the seventh decile class N = total number of cases ΣCf< = sum of the cumulative frequencies “ lesser than” up to but below the first quartile class C = class interval fc = frequency of the D7 class

D7 = L+C (7N/10-ΣCf<) fc = 30.5 + 3 (24.5 – 21) 4 = 30.5 + 3 (3.5) 4 = 30.5 + 10.5 4 = 30.5 + 2.625 D7= 33.125 or 33.13

7N/10 = 24.5 L = 30.5 ΣCf< = 21 fc = 4 C=3 N= 35

Percentile Formula; P90 = L+C (90N/100-ΣCf<) fc Where ; P90= is the ninetieth percentile L = lower real limit of the ninetieth percentile class N = total number of cases ΣCf< = sum of the cumulative frequencies “ lesser than” up to but below the ninetieth percentile class C = class interval fc = frequency of the P90 class

P90 = L+C (90N/100-ΣCf<) fc solution: = 39.5 + 3 ( 31.5 – 31) 2 = 39.5 + 3 (0.5) 2 = 39.5 + 1.5 2 = 39.5 + 0.75 P90 = 40.25

90N/100 = 31.5

L = 39.5 ΣCf< = 31 fc = 2 C=3 N= 35

On the other hand, special cases are just written as follows

P40 = 27.5 special case P60 = 30.5 special cases P80 = 36.5 special cases P40 = L+C (40N/100-ΣCf<) fc = 24.5 + 3 (14 – 9) 5 = 24.5 + 3 (5) 5 = 24. 5 + 3 P40 = 27.5

40N/100 =14

L = 24.5 ΣCf< = 9 fc = 5 C=3 N= 35

 If it is a special case, the percentile value of an upper real limit (2.75) can be determined by dividing the ΣCf< by N times 100 or ΣCf< * 100. N For instance, the ΣCf< is 14 and N is 35. 14/35 * 100 equals to 40. hence, 27.5 is P40.

Another case is percentile 60 where cumulative frequency “lesser than” is exactly 21. P60 = L+C (60N/100-ΣCf<) fc 60N/100 = 21 = 27.5 + 3 (21 -14) L = 27.5 7 ΣCf< = 14 = 27.5 +3 (7) fc = 7 7 C=3 P60= 30.5

N= 35

P80 = L+C (80N/100-ΣCf<) fc P80 = 33.5 + 3 (28 – 25) 3 80N/100 = 28 = 33.5 + 3 (3) L = 33.5 ΣCf< = 25 3 fc = 3 P80 = 36.5 C=3 N= 35