Quantitative Methods in Social Science 360-300 • CEGEP, Montreal

Revision sheet: Z scores, percentile ranks and the R score (360-300)

This sheet covers the chapter that decides admission to university: the Z score, the percentile rank and the CEGEP R score. It does not repeat the formulas, it gives what to do with them, and above all the six or seven confusions that cost marks on the exam and that feed every hallway myth about the R score.

It is written for Social Science students at CEGEP taking the compulsory Quantitative Methods course, 360-300, and it is also useful to anyone who simply wants to understand a transcript. The corrected exercise set on the same chapter goes through every calculation, up to the complete record of a semester.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

A Z score says nothing about your mark, it says your POSITION in your group. The whole chapter, R score included, consists of correcting for the fact that the same $82$ % is not worth the same thing depending on the group where it was obtained.

This chapter is part of 360-300, Quantitative Methods in Social Science

What is on this sheet

1. The essentials
2. The mistakes that cost marks
3. Which method to choose
4. How the answer is expected to be written
5. Check before you hand in
6. The typical problem, taken apart
7. Learn by heart
8. Frequently asked questions

Topics covered

Z score
Standardizing a data set
Affine transformation
Derived scales
T score
Percentile rank
Normal distribution
Empirical rule
College R score (cote R)
Group strength indicator (IFG)
Group dispersion indicator (IDGS)
Mean weighted by units

The essentials

What the Z score keeps and what it destroys

• $z=\dfrac{x-\overline{x}}{s}$ : it is a number of standard deviations, with no unit. A standardized data set ALWAYS has mean $0$ and standard deviation $1$ .
• These two numbers therefore carry no information: announcing them as a result earns nothing.
• The Z score is INVARIANT under any increasing affine transformation of the data: adding five points for everyone, or marking out of $50$ instead of $100$ , does not change it.
• It is an INTERVAL scale: the DIFFERENCES between Z scores can be compared, ratios are meaningless.

Standardizing means regraduating: in a group with mean 72 and standard deviation 8, every mark keeps its place, 72 becomes 0 and each step of 8 points becomes a step of 1.

What survives standardization is the SHAPE of the distribution and the order of the students; what disappears is the marking scale and the absolute level.

The five landmarks of the normal distribution, to know by heart

• $z=-2$ is about rank $2$ , $z=-1$ about $16$ , $z=0$ about $50$ , $z=+1$ about $84$ , $z=+2$ about $98$ .
• The PERCENTILE RANK is the percentage of the group that lies BELOW. It is never $0$ or $100$ .
• From $z=0$ to $z=1$ you gain $34$ ranks; from $z=1$ to $z=2$ you gain only $14$ . The same effort does not earn the same thing everywhere.
• Derived scales, all of the form $y=az+b$ : intelligence quotient $15z+100$ , T score $10z+50$ , R score $\left(z\times IDGS+IFG\right)\times 5+25$ .

The bell is symmetric and the percentile-rank scale is not: the

34

ranks gained between

z=0

and

z=+1

fall to

14

between

z=+1

and

z=+2

These landmarks are the empirical rule in disguise: 68% within one standard deviation, 95% within two, which leaves $2.5$ % on each side beyond $z=2$ .

The R score in three successive corrections

• We start from the Z score in the course, which says the position in HIS OR HER group.
• We multiply it by the IDGS, the group dispersion indicator, which COMPRESSES the Z score when the group was very homogeneous in secondary school.
• We add the IFG, the group strength indicator, built on secondary school records: it is used to compare GROUPS with each other.
• We bring everything back to the usual scale with $\left(\ldots\right)\times 5+25$ , then average the courses, weighting by UNITS.

Each box corrects a flaw of the previous one: the Z score ignores the group, the IDGS corrects its dispersion, the IFG its strength, and the last step is only a change of scale.

No R score is produced if the standard deviation of the group is zero or if the group is too small: the whole construction assumes variability.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Announcing the mean and standard deviation of the standardized data set

1 mark, and the concluding question misses the point

What not to write

"After standardization, the mean is $0$ and the standard deviation $1$ : the group is therefore homogeneous."

What to write

"The mean $0$ and standard deviation $1$ are AUTOMATIC: every standardized data set has them. Homogeneity is read from the standard deviation of the RAW data."

The tight data set of the sheet (blue) and the same data set stretched out (orange) give exactly the same Z scores: mean 0 and standard deviation 1 say nothing about homogeneity.

Why: Standardization is a construction: it manufactures these two numbers by definition. A result that you would get on any data whatsoever is not a result.

2. Comparing two Z scores by a ratio

1 mark, and the mistake is spotted at once

What not to write

"His Z score is $1.6$ and mine is $0.8$ : he is twice as good."

What to write

"He stands $0.8$ standard deviation above me. The Z score is an INTERVAL scale: differences can be compared, ratios cannot."

The zero of the Z scores is the mean of the group, a choice: the gap of 0.8 standard deviation has a meaning (green), the ratio of 2 has none (red).

Why: The zero of a Z score is the mean of the group, so a convention. A Z score of $0$ is not the absence of performance, and a ratio is meaningless without an absolute zero.

3. Reading the percentile rank as a percentage of correct answers

2 marks, and the interpretation of the whole table

What not to write

"My percentile rank is $84$ : so I got 84% of the exam right."

What to write

"My percentile rank is $84$ : 84% of the group did WORSE than I did. That says nothing about my mark, which could be $62$ % as easily as $95$ %."

Two illustrative groups of 20: 62% in the weak group and 95% in the strong group give the same percentile rank 95. The rank is a position, not a mark.

Why: A percentile rank is a POSITION, a percentage of correct answers is a PERFORMANCE. In a very weak group, a rank of $95$ can go with a mediocre mark.

4. Believing that ten percentile ranks are always the same effort

1 mark, and the question on the shape of the curve

What not to write

"Going from rank $50$ to rank $60$ or from rank $85$ to rank $95$ is the same progress."

What to write

"From rank $50$ to rank $60$ you need to gain about $0.25$ standard deviation; from rank $85$ to rank $95$ you need $0.61$ , two and a half times as much."

The two areas are worth 10%: near the centre, 0.25 standard deviation is enough (blue); in the tail, you need 0.61 (orange), because the bell is low there.

Why: The density of the normal distribution collapses in the tails: there are very few people between $z=1.5$ and $z=2$ , so passing them costs many standard deviations.

5. Believing that a general rise in marks raises the Z score

1.5 marks, and the question on affine invariance

What not to write

"The teacher added five points for everyone: my Z score goes up."

What to write

"Adding the same amount for everyone shifts the mean by the same amount and leaves the standard deviation unchanged: the Z score does not move by a thousandth."

Five more points for everyone: the mean also moves up by five, the student's gap to the mean stays at 8 points, one standard deviation, and the Z score stays 1.

Why: The Z score measures a relative position. Only a change that alters the GAPS between students, such as bonus marks for some, can move it.

6. Comparing two students from different groups by their marks

the whole question, which is exactly about this

What not to write

"We both got $82$ %: we are at the same level."

What to write

"In a group with mean $74$ and standard deviation $4$ , $82$ gives $z=2.0$ , that is rank $98$ . In a group with the same mean but standard deviation $12$ , the same $82$ gives $z\approx 0.67$ , that is rank $75$ ."

Both groups have the same mean of

74

. In the tight group, the mark

82

is almost at the end of the curve; in the spread-out group, it is still in the thick of the pack.

Why: This is the reason the Z score exists: the raw mark mixes the level of the student and the severity of the marking. The dispersion of the group decides everything.

7. Averaging course R scores instead of weighting by units

1.5 marks, and a real gap of a few tenths on an admission file

What not to write

"My three courses give $33$ , $29$ and $29.13$ : my R score is $30.38$ ."

What to write

"The R score of the record is $\dfrac{\sum u_{i}CRC_{i}}{\sum u_{i}}$ : weighted by UNITS, it is $30.19$ and not $30.38$ ."

Each course is a rectangle whose width is its units: the weighted mean (green) takes into account the 2.66 units of course 2, the simple mean (red) does not. Axis cut at 28.

Why: A course of $2.66$ units weighs more than a course of $2.00$ . The gap looks tiny, but competitive programs are decided to the tenth of an R score point.

8. Believing that a weak group is a good deal

the whole question on the role of the two indicators

What not to write

"I enrol in the weakest group: my Z score will be higher there, so my R score too."

What to write

"The Z score will be higher there, but the group's IFG will be lower and the IDGS will compress the Z score if the group was homogeneous in secondary school: the two corrections are made to cancel that advantage."

Why: The IFG and the IDGS exist precisely to make scores comparable from one group to another. This is the most widespread hallway myth, and the exam often takes it as its subject.

Which method to choose

What does the problem ask for: a score, a rank or a mark?

Spot what is GIVEN and what is ASKED, then follow the path: raw mark, Z score, percentile rank.

The path of the chapter: from the mark to the Z score, from the Z score to the percentile rank through the table, and from the Z score to a derived scale by y = az + b.

If the mark, the mean and the standard deviation of the group are given → compute $z=\dfrac{x-\overline{x}}{s}$ , the compulsory step of the whole chapter

Example: $82$ in a group $(74, 8)$ gives $z=1.0$
If the Z score is given and a position in the group is asked → read the percentile rank from the landmarks, or from the normal table

Example: $z=+1$ gives rank $84$
If the percentile rank is given and a mark is asked → go from the rank back to the Z score by the table, THEN $x=\overline{x}+zs$

Example: rank $98$ in a group $(74, 4)$ gives $x=82$
If a derived score is asked: IQ, T score, R score → apply the affine form $y=az+b$ of that scale, after computing $z$

Example: $z=1.2$ gives a T score of $62$
If two students from different groups are compared → standardize EACH in his or her group, then compare the Z scores, never the marks

Example: $82$ in $(74, 4)$ beats $88$ in $(80, 12)$

The normal table is only useful if the problem says, or implies, that the distribution is normal. Otherwise, only the observed ranks can be used.

Which part of the R score corrects what?

Faced with a question on the R score, first look for WHICH flaw the question puts in doubt.

The two flaws that the R score corrects: two groups of different levels (the IFG), two groups of different dispersions (the IDGS).

If the problem says the groups are not at the same level → the IFG answers: it is built on secondary school records and compares the groups with each other

Example: an elite group and an ordinary group
If the problem says a group is very homogeneous → the IDGS answers: it compresses the Z score inflated by an artificially small standard deviation

Example: a group where everyone hovers around 84%
If the problem is about a course with more or fewer units → the final weighting answers: the mean is taken over units, not over the number of courses

Example: a course of 2.66 units against a course of 1.00
If the problem says the group has five students, or that everyone has the same mark → no R score is produced: the construction requires variability and a sufficient group size

Example: a make-up course with only five students

If the question is about fairness between programs rather than a calculation, the expected answer is that the R score compares positions, never course contents.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Comparing two students from different groups

When to use it: As soon as the problem gives two marks, each with the mean and the standard deviation of its group.

1 Write for each student the mark, the mean and the standard deviation of HIS OR HER group, in a three-column table.
2 Compute the two Z scores, with at least two decimals.
3 Translate each Z score into a percentile rank, stating that the distribution is assumed normal.
4 Compare the Z scores by their DIFFERENCE, never by their ratio.
5 Conclude with a sentence that names the position, not the mark.

Concluding sentence

"With $z_{A}=2.00$ against $z_{B}=0.67$ , student A stands $1.33$ standard deviations higher in his or her group than student B in his or hers, that is approximately rank $98$ against rank $75$ , assuming both distributions are normal."

The trap: Standardizing both marks with the same mean and the same standard deviation: each student is compared to HIS OR HER group, never to the other.

Marking: 2 marks for the two Z scores, 1 mark for the ranks, 1 mark for the written normality assumption, 1 mark for the conclusion.

Computing the R score of a semester

When to use it: When the problem gives, for each course, the mark, the statistics of the group, the IFG, the IDGS and the number of units.

1 Compute the Z score of each course separately, in its own group.
2 Apply $CRC=\left(z\times IDGS+IFG\right)\times 5+25$ course by course, and keep two decimals.
3 Multiply each CRC by the units of the course, then add.
4 Divide by the sum of the units, not by the number of courses.
5 Check that the result falls between the smallest and the largest of the CRCs.

Concluding sentence

"The R score of the semester is $\dfrac{\sum u_{i}CRC_{i}}{\sum u_{i}}=\dfrac{211.00}{6.99}\approx 30.19$ ."

The trap: Rounding the CRCs to the unit before weighting: on an admission file, a few hundredths show.

Marking: 1 mark per Z score, 1 mark per CRC, 1 mark for the weighting by units.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The sign of the Z score

Compare the mark with the mean of the group: above it, the Z score is positive; below it, negative. No calculation is needed.

A mark of $68$ in a group with mean $74$ cannot give $z=+0.75$ .
The percentile rank agrees with the sign

A positive Z score gives a rank above $50$ , a negative Z score a rank below. Rank $50$ corresponds exactly to the mean.

$z=-0.4$ with a stated rank of $66$ : the two contradict each other, one of them is wrong.
The order of magnitude of a Z score

In an ordinary group, almost all Z scores lie between $-3$ and $+3$ . Beyond that, you divided by the variance instead of the standard deviation.

A $z=8.4$ almost always comes from a standard deviation confused with a variance of $s^{2}$ .
The R score stays between its components

The weighted mean of the CRCs cannot leave the interval of the CRCs: this is the immediate check of a weighting.

CRCs of $29$ , $29.1$ and $33$ cannot give an R score of $34$ .

The typical problem, taken apart

The complete record of a semester, from the transcript to the R score

Course 1: mark $82$ , group with mean $74$ and standard deviation $8$ , $IFG=0.60$ , $IDGS=1.00$ , $2.00$ units.

Course 2: mark $78$ , group with mean $70$ and standard deviation $12$ , $IFG=0.20$ , $IDGS=0.90$ , $2.66$ units.

Course 3: mark $90$ , group with mean $84$ and standard deviation $4$ , $IFG=-0.30$ , $IDGS=0.75$ , $2.33$ units.

The three course groups of the problem, each with its mean and its spread, and the student's mark (in blue) in each.

Step 1

$z_{1}=\dfrac{82-74}{8}=1.000$ ; $z_{2}=\dfrac{78-70}{12}\approx 0.667$ ; $z_{3}=\dfrac{90-84}{4}=1.500$ .

Why

Each course has its own group, so its own mean and its own standard deviation. Mixing the three groups is the mistake that costs the whole question.

Step 2

$CRC_{1}=\left(1.000\times 1.00+0.60\right)\times 5+25=33.00$ .

Why

The IDGS is $1.00$ here: the group had a normal dispersion in secondary school, so the Z score goes through uncompressed.

Step 3

$CRC_{2}=\left(0.667\times 0.90+0.20\right)\times 5+25=29.00$ and $CRC_{3}=\left(1.500\times 0.75-0.30\right)\times 5+25\approx 29.13$ .

Why

Course 3 has the best mark and the best Z score, and yet its CRC is barely above the lowest of the three and far below the $33.00$ of course 1: the group was homogeneous, so compressed by the IDGS, and weak, so penalized by a negative IFG.

Step 4

$\sum u_{i}CRC_{i}=2.00(33.00)+2.66(29.00)+2.33(29.13)\approx 211.00$ and $\sum u_{i}=6.99$ .

Why

We multiply BEFORE adding: this is the very definition of a weighted mean, and the reverse order gives nonsense.

Step 5

R score $=\dfrac{211.00}{6.99}\approx 30.19$ , against $\dfrac{33.00+29.00+29.13}{3}\approx 30.38$ as a simple mean.

Why

The gap of $0.19$ comes from the course with the highest CRC (course 1) carrying the fewest units ( $2.00$ ) while the lowest (course 2) carries the most ( $2.66$ ). On a competitive admission file, two tenths of an R score point separate hundreds of candidates.

Step 6

Check: $29.00\le 30.19\le 33.00$ .

Why

A weighted mean always stays between the smallest and the largest of its terms; a result outside that interval signals an inverted weighting.

The conclusion, written out

"The R score of the semester is $30.19$ , obtained by weighting the three course performance scores by the units of the courses. The course with the highest raw mark ends up with an R score barely above the lowest, because its group was both homogeneous and weak."

The classic mistake on this problem: Taking the arithmetic mean of the three CRCs: you get $30.38$ and ignore that the course with the highest CRC carries only $2.00$ units out of $6.99$ , while the lowest carries $2.66$ .

Learn by heart

• $z=\dfrac{x-\overline{x}}{s}$ . A standardized data set ALWAYS has mean $0$ and standard deviation $1$ : these two numbers teach nothing.
• Landmarks of the normal distribution: $z=-2\to 2$ , $z=-1\to 16$ , $z=0\to 50$ , $z=+1\to 84$ , $z=+2\to 98$ .
• The PERCENTILE RANK is the percentage of the group BELOW. It is never $0$ or $100$ , and it is nothing like a percentage of correct answers.
• The Z score is INVARIANT under an increasing affine transformation: adding points for everyone does not change it.
• $CRC=\left(z\times IDGS+IFG\right)\times 5+25$ , and the R score of the record is $\dfrac{\sum u_{i}CRC_{i}}{\sum u_{i}}$ , weighted by UNITS.
• Differences between Z scores can be compared, ratios never: it is an interval scale.
• No R score without variability: zero standard deviation or too small a group, and no score is produced.

Frequently asked questions

How do I compute my Z score from my mark?

Subtract the mean of the group from your mark, then divide by the standard deviation of the group. A mark of 82 in a group with mean 74 and standard deviation 8 gives a Z score of 1.00, one standard deviation above the mean. The result has no unit: it is read as a number of standard deviations, never as points.

What does a percentile rank of 84 mean?

That 84 percent of the group is below you. It is not a percentage of correct answers: your mark could be 62 as easily as 95, it all depends on the group. A percentile rank is a position in a ranking, and it is never 0 or 100 since you cannot pass yourself.

How is the R score calculated at CEGEP?

For each course, take the student's Z score in his or her group, multiply it by the group dispersion indicator, add the group strength indicator, then multiply by five and add twenty-five. The R score of the record is then the mean of these course performance scores, weighted by the units of each course.

Is it an advantage to be in a weak group for the R score?

No, and that is exactly what the two indicators correct. In a weak group, the Z score goes up, but the group strength indicator is lower and offsets it. If the group is also very homogeneous, the dispersion indicator compresses the Z score. The whole construction aims to make scores comparable from one group to another.

Why doesn't my Z score go up when the teacher adds points?

Because a bonus given to everyone shifts the mean by exactly as much as your mark, and leaves the standard deviation unchanged. The Z score measures a relative position: it only moves if the gaps between students change. A general rise changes nobody's ranking.

Practise it

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

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

11 corrected exercises
110 points
150 minutes

Do the exercises

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-qm-z-scores-percentile-ranks. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: Z scores, percentile ranks and the R score (360-300)

The essentials

What the Z score keeps and what it destroys

The five landmarks of the normal distribution, to know by heart

The R score in three successive corrections

The mistakes that cost marks

1. Announcing the mean and standard deviation of the standardized data set

2. Comparing two Z scores by a ratio

3. Reading the percentile rank as a percentage of correct answers

4. Believing that ten percentile ranks are always the same effort

5. Believing that a general rise in marks raises the Z score

6. Comparing two students from different groups by their marks

7. Averaging course R scores instead of weighting by units

8. Believing that a weak group is a good deal

Which method to choose

What does the problem ask for: a score, a rank or a mark?

Which part of the R score corrects what?

How the answer is expected to be written

Comparing two students from different groups

Computing the R score of a semester

Check before you hand in

The typical problem, taken apart

The complete record of a semester, from the transcript to the R score

Learn by heart

Frequently asked questions

How do I compute my Z score from my mark?

What does a percentile rank of 84 mean?

How is the R score calculated at CEGEP?

Is it an advantage to be in a weak group for the R score?

Why doesn't my Z score go up when the teacher adds points?

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

See also

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