Corrected exercises: Z scores, percentile ranks and the R score (360-300)

$z=\frac{81-72}{6}=\frac{9}{6}=1.5$: the mark is one and a half standard deviations above the mean.

$z=\frac{58-66}{5}=\frac{-8}{5}=-1.6$. Pitfall: writing $\frac{66-58}{5}=1.6$, with the numerator reversed, which puts a mark that is below the mean above it.

$x=\overline{x}+z\,s=64+(-0.75)\times 12=64-9=55$. Pitfall: dropping the sign of $z$, which gives $73$, a mark above the mean for a negative Z score.

From $z=\frac{x-\overline{x}}{s}$ we get $s=\frac{x-\overline{x}}{z}=\frac{85-75}{1.25}=8$. Pitfall: multiplying, $10\times 1.25=12.5$, when the standard deviation is in the DENOMINATOR.

$IQ=15z+100=15\times(-0.6)+100=-9+100=91$.

Go back through the Z score: $z=\frac{112-100}{15}=\frac{12}{15}=0.8$, then $T=10z+50=8+50=58$. Pitfall: converting directly, $112-100+50=62$, as if the two scales had the same standard deviation.

$z=\frac{T-50}{10}=\frac{36-50}{10}=-1.4$.

$z=\frac{53-62}{9}=-1$. The empirical rule puts $68\%$ of the observations between $z=-1$ and $z=1$, so $\frac{100-68}{2}=16\%$ lie below $z=-1$: percentile rank $16$. Pitfall: answering $32$, the total share outside the interval, instead of its lower half.

Four marks are below $67$, namely $47$, $54$, $59$ and $62$, and one other mark is equal to it: $PR=\frac{4+0.5\times 1}{10}\times 100=45$. Pitfall: counting the tie in full, which gives $50$, or forgetting it, which gives $40$.

Position $\frac{70}{100}\times 10=7$: the seventh value of the sorted list, that is $72$. Pitfall: reading the percentile rank $70$ as $70\%$ of the top mark, when it designates a POSITION in the group.

Mean: $\dfrac{70+74+78+82+86}{5}=\dfrac{390}{5}=78$. Deviations from the mean: $-8$, $-4$, $0$, $4$, $8$; their squares: 64, 16, 0, 16, 64, with sum 160. Population variance: $\dfrac{160}{5}=32$, so $\sigma=\sqrt{32}\approx 5.657$. Z scores: $-1.414$, $-0.707$, $0$, $0.707$ and $1.414$.

The Z scores add up to $-1.414-0.707+0+0.707+1.414=0$. This is always true because $\sum z_{i}=\dfrac{\sum \left(x_{i}-\overline{x}\right)}{s}$ and the sum of the deviations from the mean is zero by the very definition of the mean. As for the standard deviation of the Z scores, $\sigma_{z}=\dfrac{\sigma_{x}}{s}=1$, since dividing a whole data set by a constant divides its standard deviation by that same constant. A standardized data set therefore ALWAYS has mean 0 and standard deviation 1: these two numbers carry no information, they are imposed by the construction.

Adding 5 points to everyone shifts the mean by 5 and leaves the standard deviation unchanged: each numerator $x_{i}+5-\left(\overline{x}+5\right)$ gives back $x_{i}-\overline{x}$, so the Z scores are UNCHANGED. Multiplying by 1.2 multiplies both the mean and the standard deviation by 1.2: the numerator and the denominator are both multiplied by 1.2, and the Z scores are still UNCHANGED. This is why standardization is the right tool for comparing teachers who do not mark on the same scale: it is insensitive to any increasing affine transformation.

We can compare RELATIVE POSITIONS: the student is well below the group in the first course and slightly above it in the second. We can NOT conclude that the student has a better command of the second subject, or that the two groups are at a comparable level: a Z score of $+0.4$ in a weak group may correspond to a lower command of the material than a score of $-1.5$ in a very strong group. The Z score compares a student to HIS OR HER group, never two groups to each other. This is precisely the gap that the R score comes to fill.

Standardization destroys the ABSOLUTE LEVEL and the UNIT. After the transformation, we no longer know whether the mean of the group was 45 or 85, or whether the exam was out of 20 or out of 100. The serious case is a group in which everyone failed: a student who gets 48 out of 100 in a group with mean 40 shows a flattering Z score of $+1$ without having reached the passing mark. The Z score never says whether a goal was reached, only where one stands relative to the others. Any decision about a competency, as opposed to a ranking, must therefore go back to the raw mark.

$IQ=15z+100$. For $z=1.2$: $IQ=15\times 1.2+100=18+100=118$. The correct reading is that the person stands 1.2 standard deviations above the mean of the age group, which, under a normal distribution, corresponds to about the 88th percentile rank. The number 118 has no existence of its own: it is a position disguised as a quantity.

$T=10z+50$, so $z=\dfrac{T-50}{10}=\dfrac{63-50}{10}=1.3$. As an IQ: $15\times 1.3+100=19.5+100=119.5$, or about 120. Note that the two numbers, 63 and 119.5, describe exactly the same position: going from one to the other only changes the unit, like going from centimetres to inches.

In any group, the mean of the Z scores is 0 by construction. The mean R score is therefore $\left(0+0.8\right)\times 5+25=4+25=29$. This is the exact effect of the group strength indicator: it SHIFTS the whole distribution of the group's R scores up or down, without changing the gaps between students of the same group. A strong group, with a high IFG, sees all its R scores rise together.

$z=\dfrac{128-100}{15}=\dfrac{28}{15}\approx 1.867$. Corresponding T score: $10\times 1.867+50\approx 68.7$. Consistency can be checked another way: 128 is a little under two standard deviations above 100, and 68.7 is a little under two standard deviations above 50. The two scales tell the same story.

For three practical reasons. First, a standard deviation of 1 would give scores between $-3$ and $3$, with negatives and decimals, which users reject. Second, a larger standard deviation makes it possible to write INTEGERS that are fine enough: with the IQ, a difference of one point represents $\dfrac{1}{15}$ of a standard deviation, a convenient resolution. Finally, the chosen scale hides the relative nature of the score, which is a commercial advantage and a scientific flaw: nobody asks "118 compared to whom?", when that is the only question worth asking.

The percentile rank of a value is the percentage of the observations that are BELOW it, ties counting for half. The mark 71 is in 11th place; ten marks are below it and there are no ties, so $PR=\dfrac{10+0}{20}\times 100=50$. With the convention that counts the value itself for half, we would get $\dfrac{10+0.5}{20}\times 100=52.5$: both conventions are in circulation, and a serious problem statement says which one it uses, since the gap can reach several ranks on a small sample.

For 94, nineteen marks are below: $PR=\dfrac{19}{20}\times 100=95$, or 97.5 with the second convention. The percentile rank can never be 100, since an observation cannot be strictly above itself: there is always at least one value, its own, that is not below. Symmetrically, rank 0 is impossible. Software that displays "percentile rank 100" has rounded.

The mark of 71 out of 100 measures a PERFORMANCE: the share of the exam that was answered correctly, judged against the content. The percentile rank of 52.5 measures a POSITION: about half of the group did worse. The two can diverge completely. On a very hard exam, a mark of 45 may correspond to rank 95; on an easy exam, a mark of 88 may correspond to rank 30. What to tell the parent: the mark answers "what did the student learn?", the percentile rank answers "where does the student stand?", and a limited-enrolment admission file only cares about the second question.

Position 15 corresponds to the 15th value of the sorted list, that is 78. We could refine this by interpolating between the 15th and 16th values, but with 20 observations the precision displayed would be illusory. Above all, remember that percentile rank 75 is not "75% of the top mark": here it is worth 78 points, and it could have been worth 40 in a weaker group.

Because the percentile rank depends on the DENSITY of the observations in the region concerned. At the ends of the distribution, where observations are sparse, it takes many points to gain a few ranks: from 83 to 94, eleven points, only three observations are passed (86, 90 and 94). In the centre, where everyone is crowded together, the same distance goes much further: from 60 to 66, six points, four observations are passed (61, 63, 65 and 66). On this list of twenty, each observation passed is worth 5 ranks, so the three observations passed at the top gain 15 ranks over eleven points, and the four passed in the middle gain 20 ranks over six points. This is the direct translation of the bell shape: the same distance in marks is not worth the same number of ranks depending on where you are, and this is why the percentile rank is not an interval scale.

Empirical rule: about 68% of the observations fall within $\left[\mu-\sigma, \mu+\sigma\right]$, about 95% within $\left[\mu-2\sigma, \mu+2\sigma\right]$ and about 99.7% within three standard deviations. By symmetry, below $z=-1$ there remains $\dfrac{100-68}{2}=16\%$, so the percentile rank of $z=-1$ is 16, and that of $z=1$ is $16+68=84$. Below $z=-2$: $\dfrac{100-95}{2}=2.5\%$, rounded to 2 with the exact values of the table, and $z=2$ corresponds to rank 98. Finally $z=0$ is the median, rank 50.

Percentile rank 84 is $z=1$, so the score is $500+1\times 100=600$. Percentile rank 16 is $z=-1$, so $500-100=400$. These two values enclose the central two thirds of the candidates: a score of 600 places you above about five candidates out of six.

We need $z=1.28$, so a score of $500+1.28\times 100=628$. Note in passing that gaining the six percentile ranks between 84 and 90 costs 28 score points, while the 34 ranks between 50 and 84 cost 100: the progression is not linear, and it gets more and more expensive as you climb.

No, the gap is not the same. Between $z=0$ and $z=1$: from 50 to 84, that is 34 ranks. Between $z=1$ and $z=2$: from 84 to 97.5 (98 with the rounded landmark), that is about 13.5 ranks, or 14 rounded to the unit. The same one-standard-deviation gap is therefore worth more than twice as many ranks at the centre as at the edge, because the density of the curve is much higher there. This is the numerical demonstration of the point made in the previous exercise: the percentile rank is ORDINAL, it cannot be subtracted.

The conversion stops being valid. The correspondences $z=1$ to rank 84 and $z=2$ to rank 98 are properties of the NORMAL distribution, not properties of the Z score. On a distribution stretched toward the right, such as incomes or response times, a Z score of 1 may correspond to rank 90 or more. The Z score remains computable and interpretable as a distance in standard deviations, but any conversion into a percentile rank requires either normality or the empirical distribution itself. A useful reminder of Chebyshev's inequality: with no assumption about shape, all we know is that at most $\dfrac{1}{4}$ of the observations lie beyond two standard deviations of the mean, which is much looser than the 5% of the normal distribution.

The Z score answers the objection of TEACHERS WHO DO NOT MARK ALIKE. If applications were ranked on the raw average, one student would land with a generous marker and gain points without learning anything more, while another, in a course known for low averages, would be penalized for the whole university career. By standardizing within each course group, we erase the severity of the marker and the scale of the exam, since the Z score is invariant under any increasing affine transformation of the marks.

The group strength indicator answers the SYMMETRIC objection: after standardization, all groups have the same mean, 0, whereas the groups are not at the same level. Without the IFG, an average student in a very strong group and an average student in a very weak group would have exactly the same score, which would encourage people to pick the weakest groups possible. The IFG is estimated from the SECONDARY school records of the group's members, that is from information that comes before and from outside the course, and it shifts all the scores of the group up or down.

For purely cosmetic and practical reasons, exactly as for the IQ. Multiplying by 5 spreads the scale so that one standard deviation is worth 5 R score points, which gives a convenient resolution. Adding 25 eliminates negative values: nearly all scores then fall between 15 and 35. It must be stressed that these two operations change nothing about the RANKING: they are affine and increasing, so the order of the students is strictly the same as with the raw Z scores.

With $IFG=0.9$: $CRC=\left(1.4\times 1.0+0.9\right)\times 5+25=2.3\times 5+25=11.5+25=36.5$. With $IFG=0.3$: $CRC=\left(1.4+0.3\right)\times 5+25=1.7\times 5+25=8.5+25=33.5$. Three R score points of difference for the same relative performance, only because of the level of the group. On a full record, such a gap moves a candidate by several hundred places in a competitive program, which explains why the construction of the IFG is closely monitored.

$CRC=\left(1.4\times 0.85+0.9\right)\times 5+25=\left(1.19+0.9\right)\times 5+25=2.09\times 5+25\approx 35.45$. The correction costs about one point. Its meaning: in a group that was very HOMOGENEOUS in secondary school, the standard deviation of the course marks is artificially small, so that an ordinary performance produces an inflated Z score. The IDGS, below 1 in this case, compresses the Z score to cancel that inflation. It is the answer to a third objection: without it, a student would gain by ending up in a group where everyone looks alike.

In group A: $z=\dfrac{78-70}{4}=2.0$. In group B: $z=\dfrac{78-82}{4}=-1.0$. The same mark of 78 is therefore worth two standard deviations above in one case and one standard deviation below in the other: this is the largest gap met in practice, and it shows why a raw mark means nothing without its group.

Group A: $CRC=\left(2.0+0.4\right)\times 5+25=2.4\times 5+25=37$. Group B: $CRC=\left(-1.0+1.1\right)\times 5+25=0.1\times 5+25=25.5$. The gap is 11.5 R score points, enormous. The group strength correction, which gives group B 0.7 more Z points, that is 3.5 R score points, is very far from making up for the three standard deviations that separate the two relative positions.

The two effects must be separated. Through the Z score, yes: in a weak group, the same absolute performance ranks higher, and the Z score rises. Through the IFG, no: a weak group has a lower IFG, and all its R scores are lowered by the same amount. The calculation shows that the compensation is PARTIAL, because the IFG is estimated from secondary school records and varies by only about one point from one extreme to the other, whereas the Z score can vary by three. In practice, the residual advantage of a weak group exists but remains modest, and it vanishes as soon as the weak group teaches less well, which is paid for in later courses. The honest answer to give a student is therefore: this parameter is marginal compared with the work put in, and it cannot be controlled at registration time.

$z=\dfrac{78-70}{8}=1.0$ instead of 2.0. The Z score is HALVED although neither the mark nor the mean has moved. The reason is that the Z score measures a distance in standard-deviation units: in a very spread-out group, eight points above the mean is commonplace, while in a homogeneous group it is exceptional. This is exactly the phenomenon that the IDGS comes to correct when the homogeneity comes from recruitment rather than from the course.

The standard deviation of the group is 0, and the Z score would be $\dfrac{0}{0}$, therefore undefined. The system then computes no Z score: the regulations provide that no R score is produced when the dispersion is zero or the group is too small, in practice fewer than six students with a usable CEGEP record. The course is then removed from the calculation of the record, without penalty. This is an important reminder: the whole construction rests on the existence of variability in the group, and it makes no sense without it.

With 8 students, the mean and especially the standard deviation of the group are estimated from very few observations, so they are very sensitive to each value. If the group included a student at 95 and that student drops out, the mean falls and the standard deviation DROPS: the other Z scores are recomputed on a tighter base, and those of the good students rise mechanically. A numerical example: on the marks 60, 64, 68, 70, 72, 76, 80 and 95, the mean is 73.125 and the population standard deviation about 10.17; the student at 80 has a Z score of 0.676. Without the 95, the mean falls to 70 and the standard deviation to 6.32, and the same 80 is now worth 1.58. The R score gains more than four points without the student having done anything. This is precisely why a minimum group size is imposed.

Both results count: the failure goes into the record with its score, and the retake with its own, each weighted by the units of the course. Nothing erases the failure. The criticism is twofold: it durably penalizes a bad start, often linked to circumstances outside school, and it discourages risk-taking, since a student has every interest in never enrolling in a difficult course outside his or her program. In practice, a whole semester ruined for health reasons can be removed from the calculation on request, but that is an administrative procedure, not an automatic mechanism.

Because three of the four ingredients of the formula do not depend on the student. The composition of the group changes every semester, so the mean and the standard deviation on which the Z score is computed change too; the IFG changes with the cohort; the courses taken do not have the same weighting in units. A perfectly consistent student will therefore see the score fluctuate by a point or two from one semester to the next, which is the normal noise of the instrument.

The cutoff is administrative, not statistical. The gap of 0.4 R score points corresponds to less than a tenth of a standard deviation of performance, that is, to an indistinguishable difference, of the same order as the noise described in the previous question. Treating 29.8 and 30.2 as two distinct cases is a management choice that nothing in the measure justifies. What can correctly be said: the R score ranks large groups well, and it does not decide between two neighbouring files. Programs that know this complete the ranking with a letter, an interview or a test.

Yes, in principle, and that is the whole point of the construction. The first term, the Z score, is computed inside a course group of a given CEGEP, which allows no direct comparison between institutions. It is the IFG that makes comparison possible, because it is built on a COMMON reference outside the CEGEPs, the secondary school marks, themselves standardized across Quebec. Comparability therefore rests entirely on the quality of that reference: if secondary school marks were themselves incomparable from one school to another, the whole edifice would lose its foothold. This is the fundamental objection that regularly comes back in public debate.

FALSE. Percentile rank 80 means that 80% of the group did WORSE, and says nothing about the mark. On a hard exam, it can correspond to 58 out of 100; on an easy exam, to 92. Correct statement: the percentile rank is a position in a group, the mark is a performance on a body of content, and the two coincide only by chance.

FALSE. The Z score is an INTERVAL scale, with a conventional zero placed at the mean: ratios are meaningless on it, exactly as for degrees Celsius. Going from 1 to 2 also corresponds to gaining only 14 percentile ranks, against 34 for going from 0 to 1. Correct statement: a Z score of 2 is one standard deviation higher than a score of 1, which is a considerable gap, but the expression "twice as good" has no meaning.

FALSE, and in fact the opposite. If everyone got 85, the standard deviation of the group is zero, the Z score is undefined and no R score is produced for that course; it is removed from the record. More generally, in a very homogeneous group the standard deviation is small and the Z scores are extreme in both directions, which benefits some and heavily penalizes others. Correct statement: the R score exists only if there is variability in the group, and a high mark shared by everyone produces no advantage.

FALSE. Percentile ranks are not equally spaced in performance. Under a normal distribution, going from rank 50 to rank 60 takes about 0.25 standard deviation; going from rank 85 to rank 95 takes about 0.60, nearly two and a half times as much. Correct statement: ten percentile ranks cost more the higher you are, because the density of the distribution decreases toward the ends.

FALSE. The CEGEP record is a mean WEIGHTED by the number of units of each course: a 3-unit course weighs more than a 1-unit course, and physical education or make-up courses do not carry the same weight as a course in the specific education component. Correct statement: the R score of the record is $\dfrac{\sum u_{i}CRC_{i}}{\sum u_{i}}$, and two students with exactly the same course R scores can have different records if their distribution of units differs.

Quantitative Methods: $z=\dfrac{84-71}{8}=1.625$. Philosophy: $z=\dfrac{76-73}{6}=0.5$. Psychology: $z=\dfrac{81-72}{9}=1.0$. English: $z=\dfrac{70-74}{7}\approx -0.571$. Note that the best MARK, 84, also corresponds to the best position, but this is not automatic: 81 in psychology, in a very spread-out group, is worth less than 84 in a group that is less so.

Quantitative Methods: $\left(1.625+0.7\right)\times 5+25=2.325\times 5+25\approx 36.63$. Philosophy: $\left(0.5+0.5\right)\times 5+25=30.0$. Psychology: $\left(1.0+0.8\right)\times 5+25=34.0$. English: $\left(-0.571+0.9\right)\times 5+25\approx 26.64$. Note that English, the only course where the student is below the mean, stays above 25 thanks to a high IFG: the group was strong, and that partially protects the student.

Sum of the units: $2+2.33+2+2=8.33$. Weighted sum: $2\times 36.63+2.33\times 30.0+2\times 34.0+2\times 26.64=73.26+69.90+68.00+53.28=264.44$. Record: $\dfrac{264.44}{8.33}\approx 31.75$. The weighting by units has a real effect here: philosophy, a weaker score but with more units, pulls the record down by about a tenth of a point compared with a simple mean.

Without English, the sum of the units becomes 6.33 and the weighted sum $264.44-53.28=211.16$, so a record of $\dfrac{211.16}{6.33}\approx 33.36$, a gain of 1.6 points. Arithmetically, dropping the course therefore improves the record. Is it a good idea? Rarely: a withdrawal puts a notation on the transcript, delays graduation, may cost full-time status and therefore financial aid, and above all the course will have to be retaken, with the risk that the new score is worse. The prudent rule is to drop a course you are going to fail, not a course where you are slightly below the mean.

The record is already at 31.75, well above the threshold of 29: no improvement is needed. We can turn the question around and look for the minimum R score in English that would keep the record at 29: we would need $\dfrac{73.26+69.90+68.00+2x}{8.33}=29$, so $2x=29\times 8.33-211.16=241.57-211.16=30.41$, that is $x\approx 15.2$. An R score of 15 corresponds to about two standard deviations below the mean of an ordinary group, that is, to a clear failure. The threshold is therefore reached with a very comfortable margin, and this is exactly the kind of check a student should make before worrying.

Candidate A: the R score has by construction a mean of 25 and a standard deviation of 5, so $z=\dfrac{31-25}{5}=1.2$. Candidate B: $z=\dfrac{612-500}{100}=1.12$. Candidate C: $z=\dfrac{14.6-12.4}{1.8}\approx 1.222$. Candidate D: rank 88 corresponds, under the normal distribution, to $z\approx 1.175$, a value read from the table between 1.17 and 1.18.

Ranking by decreasing Z score: C ($1.222$), A ($1.200$), D ($1.175$), B ($1.120$). The gap between the first and the last is $1.222-1.120=0.102$ standard deviation, about a tenth of a standard deviation. This is a tiny gap: on a normal distribution, it separates percentile rank 89 from rank 87. In other words the four candidates are indistinguishable, and the order obtained comes down to rounding.

Converting a percentile rank into a Z score assumes the NORMALITY of the original distribution. Without it, rank 88 may correspond to a very different Z score: on a distribution stretched toward the right, it could be worth 0.9; on a tighter distribution with a few extreme values, 1.6. With no information about the shape, the only thing known for certain is the ORDER: candidate D is above 88% of the group, which is enough for a rough ranking and not for a comparison to the tenth.

With $y=8z+70$: candidate A, $8\times 1.2+70=79.6$; candidate B, $8\times 1.12+70=78.96$; candidate C, $8\times 1.222+70\approx 79.78$; candidate D, $8\times 1.175+70=79.4$. The four marks fit within eight tenths of a point, which makes visible at a glance what the raw numbers, 31, 612, 14.6 and 88, completely hid.

Three reservations. First reservation: the four Z scores are computed on different REFERENCE POPULATIONS, Quebec CEGEP students for A, the test takers for B, one lycée class for C; being above one's group says nothing as long as the groups are not comparable, and nothing here guarantees that they are. Second reservation: the gap measured, a tenth of a standard deviation, is smaller than the noise of each of the instruments, so the ranking obtained is not reproducible. Third reservation: for C, the standard deviation of 1.8 points comes from a single class, hence a small sample, which makes its Z score unstable exactly as in the case of the small CEGEP groups. The conclusion to write is that the four files are equivalent given the data available, and that the tie must be broken on other criteria, explicit and written in advance.