Statistics for Dummies (25 page)
Percentiles are used in a variety of ways for comparison purposes and to determine relative standing. Babies' weights, lengths, and head circumferences are reported and interpreted in terms of percentiles, for example. Percentiles are also used by companies to get a handle on where they stand in terms of the competition on sales, profits, customer satisfaction, and so on. (For the details on percentiles, see
Chapter 5
.) And the relationship between percentiles and standard scores is an important one, as the next example shows.
Suppose Rhodie (a hypothetical physical therapy student) takes a physical therapy certification test and gets a score of 235. She's trying to figure out what that score really means. She is told that the test scores have a normal distribution with mean 250 and standard deviation 15, which means her standard score on the test is
−
1.0 (one standard deviation below the mean). Her score is below average, but is it still good enough to pass the test? Each year, the certification test is different, so the cutoff score for passing and failing changes, too. However, the test administrators always pass the top 60% of the scores and fail the bottom 40%. Knowing this information, does Rhodie pass the test? And what's the cutoff score for this year's test? Percentiles help Rhodie unravel the mystery of her standard score and relieve her anxiety.
If the test administrators always pass the top 60% and fail the bottom 40%, that means the cutoff score for pass/fail is at the 40th percentile. (Remember that percentile means percentage
below.
) At what percentile is Rhodie's score? Referring to
Figure 8-7
and using the empirical rule, you know that her score is 1 standard deviation below the mean. So, about 68% of the scores lie between 235 and 265 (within 1 standard deviation of the mean) and the rest (100%
−
68% = 32%) lie outside that range. Half of those outside the ±2 standard deviation range (or 32% ÷ 2 = 16%) lie below 235. This means that Rhodie's score is at the 16th percentile (approximately), so she doesn't pass the test. To pass, she needs to at least be at the 40th percentile or better.
The empirical rule can take you only so far in determining percentiles; if you notice, Rhodie's score was conveniently chosen by me so that it fell right on one of the tick marks. But suppose she scored somewhere in between the tick marks? Never fear, the standard normal table is here.
Table 8-1
shows the corresponding
percentile
(percentage below) for any standard score between
−
3.4 and +3.4. This covers well over 99.7% of the situations you'll ever come across. Notice that as the standard scores get larger, the percentile also gets larger. Also note that a standard score of 0 is at the 50th percentile point of the data, which is the same as the median. (See
Chapter 5
for more on means and medians.)
 |
Standard Score | Percentile | Standard Score | Percentile | Standard Score | Percentile |
|---|---|---|---|---|---|
− | 0.03% | − | 13.57% | +1.2 | 88.49% |
− | 0.05% | − | 15.87% | +1.3 | 90.32% |
− | 0.07% | − | 18.41% | +1.4 | 91.92% |
− | 0.10% | − | 21.19% | +1.5 | 93.32% |
− | 0.13% | − | 24.20% | +1.6 | 94.52% |
− | 0.19% | − | 27.42% | +1.7 | 95.54% |
− | 0.26% | − | 30.85% | +1.8 | 96.41% |
− | 0.35% | − | 34.46% | +1.9 | 97.13% |
− | 0.47% | − | 38.21% | +2.0 | 97.73% |
− | 0.62% | − | 42.07% | +2.1 | 98.21% |
− | 0.82% | − | 46.02% | +2.2 | 98.61% |
− | 1.07% | 0.0 | 50.00% | +2.3 | 98.93% |
− | 1.39% | +0.1 | 53.98% | +2.4 | 99.18% |
− | 1.79% | +0.2 | 57.93% | +2.5 | 99.38% |
− | 2.27% | +0.3 | 61.79% | +2.6 | 99.53% |
− | 2.87% | +0.4 | 65.54% | +2.7 | 99.65% |
− | 3.59% | +0.5 | 69.15% | +2.8 | 99.74% |
− | 4.46% | +0.6 | 72.58% | +2.9 | 99.81% |
− | 5.48% | +0.7 | 75.80% | +3.0 | 99.87% |
− | 6.68% | +0.8 | 78.81% | +3.1 | 99.90% |
− | 8.08% | +0.9 | 81.59% | +3.2 | 99.93% |
− | 9.68% | +1.0 | 84.13% | +3.3 | 99.95% |
− | 11.51% | +1.1 | 86.43% | +3.4 | 99.97% |
|
| HEADS UP | In order to use |
To calculate a percentile when the data have a normal distribution:
Convert the original score to a standard score by taking the original score minus the mean, and then dividing that difference by the standard deviation. (The notation for this is
Use
Table 8-1
and find the corresponding percentile for the standard score.
You already know that Rhodie scored at the 16th percentile with her score of 235. Suppose Clint also takes the same test, and he scores 260. Does he pass? To find out, you can convert his score to a standard score and find the corresponding percentile. His standard score is (260
−
250) ÷ 15 = 10 ÷ 15 = 0.67. Using
Table 8-1
, Clint's score is somewhere between the 72.58th percentile and the 75.80th percentile. (To be conservative, go with the lower percentile.) In any case, Clint does pass, because he performs better than at least 72% of the test takers (including Rhodie). And his percentile is higher than 40, beating the cutoff percentile.
Wouldn't it be nice not to have to convert every person's score to a standard score just to determine whether he or she passed? Why not just find the cutoff score, in terms of original units, and compare everyone's result to that? You know that the cutoff score is at the 40th percentile. Using
Table 8-1
, you can see that the corresponding standard score closest to the 40th percentile is
−
0.3 (use the table backward). This means that the cutoff score is 0.3 standard deviations below the mean. What's this score in standard units? You can use the standard score formula and solve it backwards to find the original score.
Recall that the formula for converting an original score (
x
) to a standard score (call it Z) is
. With a little algebra, you can rewrite this equation so that you can convert from a standard score (Z) back to the original score (
x
). That formula looks like this:
x
= Z
σ
+
μ
.
To convert a standard score to a score in original units (original score):
Find the mean and the standard deviation of the population that you're working with.
Take the standard score (Z) and multiply by the standard deviation.
Add the mean to your result from Step 2.
