The Normal Curve and Transformations
5.1 Percentile Rank
5
Formula for finding the Percentile Rank ofX,PR PR of X= B+ 1 2E
N 100 (Formula 5.1)
where
B =the number of scoresbelowthe given scoreX
E =the number of scoresexactlythe same as X(if there is only oneXscore, thenE= 1)
N= the total number of scores in the distribution
■QuestionFor the following distribution, what is the percentile rank of 16?
12, 13, 13, 14, 16, 18, 22
SolutionN= 7,B= 4,E= 1. Using Formula 5.1, we find that PR of 16 = 4 + 1 2 1
7 100
PR of 16 = 4 + 0 50
7 100
PR of 16 = 0 64 100 PR of 16 =64%
This means we can say that 64% of the scores of this distribution fall at or below a score of 16. The score 16 is at the 64th percentile.■
For small samples like in the example above, the percentile rank concept becomes a bit fuzzy. On the face of it, we see that the score of 16 is the fifth number up from the bottom. Since we are looking to find the percent of scores at or below 16 and there are seven numbers, we should be diving 5 by 7 to get 71%, correct? Well, we need to remember that the score of 16, if on a continuous scale, is actually at the midpoint of the interval between 15.5 and 16.5. In a larger sample (with, say, 1000 values), we might have multiple 16’s. These“16’s”are assumed to be evenly distributed. That is, we would assume that about half of them, if measured more precisely, would have values between 15.5 and 16, therefore at or below 16. The other half, if measured more precisely, would be assumed to have values between 16 and 16.5, therefore above 16. This is why we use half of this frequency number when calculating percentile rank.
Finding a Score Value Given the Percentile Rank
Suppose we administer an achievement test to a group of students. Formula 5.1 can be used to transform each student’s score to a percentile rank; this, in turn, allows us to determine how any one student scored on the test with respect to
the group. However, what if we wanted to work things the other way? Instead of using a score to find a percentile rank, we would use a percentile rank to find a score. Formula 5.2 is used to identify the score that is a given percentile rank.
Formula for findingXgiven a Percentile Rank,Xp
Xp=L+ N P −F
f h (Formula 5.2)
where
Xp=the score at a given percentile
N =the total number of scores in the distribution P =the desired percentile, expressed as a proportion L =the exact lower limit of the class interval F =the sum of all frequencies belowL
f =the number of scores in the critical interval h =the width of the critical interval
Although Formula 5.2 requires us to determine six values to find the score at a given percentile rank, determining these values is really quite easy. The compu- tational steps of Formula 5.2 are specified in the context of the next worked example.
■QuestionTwo hundred twenty-four students are administered an achieve- ment test. The grouped frequency distribution of the obtained scores is presented in the following table. What score is at the 85th percentile?
Class Interval Frequency Cum f
450–499 15 224
400–449 29 209
350–399 46 180
300–349 65 134
250–299 32 69
200–249 20 37
150–199 9 17
100–149 8 8
Solution
Step 1. Determine N. The highest number in the cum fcolumn is the total number of scores in the distribution.N= 224.
Step 2. Identify the critical interval within which lies the score at the 85th percentile. We want to find the score below which falls 85% of the total
5.1 Percentile Rank 129
number of scores. Eighty-five percent of 224 = (0.85)(224) = 190.40, or rounded, 190. The interval 400–449 contains the 181st through the 209th scores. The 190th score is somewhere in this interval.
Step 3.DetermineL. The exact lower limit of the critical interval is 399.5.
Step 4.Determine F. F is the sum of the scores below the critical interval.
Simply look at the cumulative frequency just below the critical interval:
F =180.
Step 5.Determinef. The number of scores in the critical interval is 29. It is assumed that the scores within the interval are evenly distributed.
Step 6.Determineh. The real limits of the critical interval are 399.5–449.5. The interval width,h, is 449.5−399.5 = 50.
Step 7.DetermineP.P is the desired percentile rank, stated as a proportion.
P =0.85.
Step 8.Plug the preceding values into Formula 5.2.
Xp=L+ N P −F
f h
X85= 399 50 + 224 0 85 −180
29 50
X85= 399 50 + 190 40−180
29 50
X85= 399 50 + 0 36 50 X85= 399 50 + 18 X85= 417 50or418
The score that is at the 85th percentile is 418. Alternatively, 85% of the achievement scores fall below a score of 418.■
Notice that inner workings of Formula 5.2 bear a striking resemblance to the intuitive calculation of the median. This should come as no surprise since the median is the 50th percentile. If we wanted to find the median of the foregoing distribution, we would begin by multiplying 224 by 0.50 instead of 0.85. It is important to note that every percentile rank should be considered an approx- imation due to the assumption that scores are evenly distributed across an interval.
Some Characteristics of Percentile Ranks Consider the following worked example.
■QuestionLaMarr and Margaret are in different statistics classes. They each scored 42 on the midterm. Who did better?
LaMarr’s Class
50 49 49 47 44 42 42 42 42 41 39 37 37 36
Margaret’s Class
44 44 43 42 41 40 40 39 39 35 32 30
SolutionUsing Formula 5.1, PR of X= B+ 1 2E
N 100
LaMarr's PR= 5 + 1 2 4
14 100 =50 Margaret's PR= 8 + 1 2 1
12 100 =71
Even though both students received identical scores, Margaret’s performance could be considered superior since her percentile rank was higher, that is, if we are primarily interested in performance as performance relative to others in a distribution. Of course, we are not always interested in measuring perfor- mance in this manner. For example, if the tests taken in the two courses were identical, it might be reasonable to conclude that neither student outperformed the other.■
Suppose we had the task of admitting students into our university. It would probably be a mistake to base our decision for admittance on only students’ percentile ranks taken from high school grade point averages. Surely, some high schools have different degrees of rigor associated with their classes.
Some data sets, even if they are ostensibly measuring the same thing, may not be equivalent. This is one reason why undergraduate admission commit- tees also consider students’ performances on nationally standardized tests (e.g. the Scholastic Aptitude Test [SAT]). The SAT is the same measure throughout the country. By using the percentile ranks of SAT scores, a student from Redding, California, can be compared with a student from Ypsilanti, Michigan.
However, the manner in which the percentile rank locates scores has an important weakness. Percentile ranks are based solely on the rank ordering of scores. (Recall the limitations associated with ordinal scales discussed in Chapter 2.) Similar to the median, a percentile rank is determined merely by the relative position of the scores. The magnitude of the difference, if it can even be determined, is not taken into account. The next worked example highlights this difficulty.
5.1 Percentile Rank 131
■QuestionWhat are the percentile ranks for the score of 80 in both distributions A and B?
Solution
DistributionA DistributionB
82 90
81 89
81 87
81 82
80 80
60 79
60 79
59 79
58 77
56 76
40 76
Distribution A PR= 6 + 1 2 1
11 100 =59 Distribution B PR= 6 + 1 2 1
11 100 =59
Even though the percentile rank of 59% is the same, the distributions are clearly different. The score of 80 is 2 away from the top and 40 away from the bottom in DistributionA, whereas it is 10 away from the top and 4 away from the bottom in DistributionB. If the underlying scale is ordinal, then this is an unresolvable problem. The percentile rank does accurately communicate the relative position of the value 80 in both distributions, and no improved descriptor can be calculated. Percentile ranks only consider the number of scores below a given value, not the magnitude of differences between those scores.■
One clear advantage of percentile ranks is that they are versatile; they can be used with any shaped distribution, skewed, or otherwise. However, just as an ordinal scale can communicate relative position but not the magnitude of dif- ferences between values, so also are the limitations of percentile ranks.
A different system, one that measureshow muchmore or less one value is from another, will need to be employed to locate more precisely a given score within a distribution of scores.