The Last Scholarship · Teacher Answers

🔐 Student Activity Unlock Codes (case-insensitive)

Phase 1 → 2

RULER01

Phase 2 → 3

ZSCORE02

Phase 3 → 4

TRUST03

Phase 4 → Done

MISSING04

Reference Data: The Three Finalists

Finalist	Exam	Score (x)	Mean (μ)	SD (σ)
Amara	Exam A	132	100	15
Diego	Exam B	78	65	8
Priya	Exam C	610	500	100

🎨 Oral Launch: Read Aloud to Class

"A university has one scholarship spot left. Three finalists have made it to the final round, each holding the top score from a completely different national exam. Different questions, different scales, different maximum marks. The committee has one job today: figure out, mathematically, who actually deserves that spot. All work goes on your whiteboard first."

Give code RULER01 only once a group has clearly attempted Task 1c on the board, not just discussed it. The struggle in 1b and 1c is the point.

Same Number, Different Meaning

SL & HL 4.12 · Why raw scores can't be compared directly

Task 1aYour first vote

No single correct answer. Most groups initially pick Priya (610 looks like the "biggest" number) or occasionally Amara. The point is only to get a committed, unexamined guess on the board before Task 1b unsettles it.

Task 1bAttack your own vote

Comparing 610, 132, and 78 directly is a category error: the exams have different means and different spreads, so the same raw score means something completely different on each one. A good analogy: comparing a temperature of 30 (Celsius) to a temperature of 30 (Fahrenheit), or comparing a height in centimetres to a weight in kilograms.

▸ Push groups to be specific about why it fails: it's not just "different exams," it's specifically that μ and σ differ, so the same raw gap from the mean represents a different amount of "unusualness."

Task 1cFind a fairer number

Expected reasoning: take (score − mean), then divide by the SD to express the gap "in SD units" rather than raw points.

\text{Amara: } \frac{132-100}{15} = \frac{32}{15} \approx 2.13 \qquad \text{Diego: } \frac{78-65}{8} = \frac{13}{8} = 1.625 \qquad \text{Priya: } \frac{610-500}{100} = \frac{110}{100} = 1.1

Amara ≈ 2.13 SDs above average · Diego = 1.625 SDs above average · Priya = 1.1 SDs above average

Expected sketch

Priya and Diego marked in gray, Amara (the standout) highlighted in red, all in SD units from the mean. A rough freehand curve is fine; the relative order is what matters.

Gate check before giving RULER01: all three values found correctly (subtract mean, divide by SD), even if a group hasn't yet written it as a general formula. The naive Phase 1a vote should now look shaky: Priya, the "biggest number," is actually the least exceptional of the three relative to her own exam.

Building the Universal Ruler

SL & HL 4.12 · The z-score formula · the standard normal distribution

Task 2aGeneralize your pattern

Students should arrive at (in their own notation, before it's named):

z = \frac{x-\mu}{\sigma}

▸ This is the first point the term z-score is introduced to the class, once a group has generalized it. Confirm it reproduces all three Phase 1 answers before moving on.

Task 2bQuick-fire check

x	μ	σ	z
54	50	2	2
42	50	4	−2
50	50	8	0
127	100	18	1.5
88	100	16	−0.75

▸ Row 2 confirms below-mean scores give negative z. Row 3 confirms a score exactly at the mean gives z = 0. Row 5 confirms z doesn't have to be a "nice" number.

Task 2cThe whoa question

Dataset 4, 6, 7, 9: mean = 6.5, standard deviation (population) ≈ 1.803 (3 s.f.).

z_1=\frac{4-6.5}{1.803}\approx -1.387 \quad z_2=\frac{6-6.5}{1.803}\approx -0.277 \quad z_3=\frac{7-6.5}{1.803}\approx 0.277 \quad z_4=\frac{9-6.5}{1.803}\approx 1.387

Mean of the four z-values = 0 · Standard deviation of the four z-values = 1

▸ This holds for any dataset, not just this one; it's an algebraic consequence of subtracting the mean and dividing by the SD, not a coincidence. This is exactly what Task 2d (HL) asks students to prove in general.

HL Extension · Task 2d · Full worked solution

Write \(Z = \dfrac{X-\mu}{\sigma} = \dfrac{1}{\sigma}X - \dfrac{\mu}{\sigma}\), which is of the form \(aX+b\) with \(a=\dfrac{1}{\sigma}\) and \(b=-\dfrac{\mu}{\sigma}\).

E(Z) = a·E(X) + b = (1/σ)·μ − μ/σ = μ/σ − μ/σ = 0

Var(Z) = a²·Var(X) = (1/σ)²·σ² = σ²/σ² = 1

So E(Z) = 0 and Var(Z) = 1 for any normal (or indeed any) random variable X with mean μ and variance σ², not just for a specific dataset.

✓ Gate check before giving ZSCORE02

SL minimum: Tasks 2a to 2c complete, z-score formula correctly generalized, and the mean-0/SD-1 pattern found numerically in Task 2c.

HL: additionally, Task 2d's algebraic proof completed using the E(aX+b) and Var(aX+b) rules.

Trust the Ruler

SL & HL 4.12 · Probability equivalence · verifying with the GDC

Task 3aSame z, same rareness?

Yes. Standardizing only relabels the horizontal axis; it doesn't stretch or squeeze the area under the curve. Two values with the same z-score sit in exactly the same relative position in their own distribution, so they have the same probability of being exceeded.

Task 3bVerify it, don't just guess

Both routes give the same probability, confirming Task 3a:

Expected sketch: same shape, relabeled axis

Amara's exam, N(100, 15²)

Standard normal, N(0, 1)

Identical shape and shaded area, only the axis label changes. This is the picture worth pausing on.

GDC keystrokes · Casio fx-CG50

P(X > 132), N(100, 15²): Ncd, Lower=132, Upper=1E99, σ=15, μ=100 → 0.01645
P(Z > 2.133): Ncd, Lower=2.133, Upper=1E99, σ=1, μ=0 → 0.01645

GDC keystrokes · TI-Nspire CX

P(X > 132): Normal Cdf, Lower=132, Upper=1E99, σ=15, μ=100 → 0.01645
P(Z > 2.133): Normal Cdf, Lower=2.133, Upper=1E99, σ=1, μ=0 → 0.01645

▸ 0.01645 (5 s.f.), i.e. about 1.64%. Small rounding differences are fine if students used their rounded z ≈ 2.13 instead of the stored value.

Task 3cSettle the vote, with numbers

Finalist	z	P(Z > z)
Amara	2.133	1.64%
Diego	1.625	5.21%
Priya	1.100	13.6%

Ranking (most to least exceptional): Amara, Diego, Priya. Amara has by far the smallest probability of scoring that well or better in her own cohort, so on this evidence, she is the strongest candidate, even though her raw score (132) was the smallest of the three raw scores. This directly overturns most groups' Phase 1a gut vote.

Expected sketch: all three tails on one standard normal curve

Each finalist's tail is a subset of the one to its left; Amara's is visibly the smallest and darkest region, the smallest area, the most exceptional.

✓ Gate check before giving TRUST03

Both probabilities in Task 3b confirmed equal (within rounding), all three Task 3c probabilities found, and a committee decision reached (Amara) with reasoning based on probability, not raw score.

The Missing Spread

SL & HL 4.12 · Inverse normal calculations with an unknown parameter

🗣 Phase 4 Prompt: Read Aloud

"One more file just came in. Jordan scored 150 on Exam D. We know the mean, 110, but nobody recorded the standard deviation. All the exam board can tell us is that only 2% of this year's candidates scored above 145. Can you still work out where Jordan really stands?"

Task 4aFind the missing z first

Need \(z\) such that \(P(Z>z)=0.02\), i.e. the 98th percentile of the standard normal.

GDC keystrokes · Casio fx-CG50

InvN: Tail=Right, Area=0.02, σ=1, μ=0 → z ≈ 2.0537

GDC keystrokes · TI-Nspire CX

Inverse Normal: Area=0.98, σ=1, μ=0 → z ≈ 2.0537

▸ z ≈ 2.0537 (stored to full precision; 2.05 to 3 s.f.). Watch for students entering Area = 0.02 on the TI by mistake, forgetting it wants the left-tail area; that gives z ≈ −2.05, the wrong sign.

Expected sketch

The unknown standard deviation doesn't stop us drawing this. The shape and the 0.02 area are known before σ is; the sketch comes first, the algebra in Task 4b fills in the number.

Task 4bSolve for the missing standard deviation

Rearranging \(z = \dfrac{x-\mu}{\sigma}\) for \(\sigma\):

\sigma = \frac{x-\mu}{z} = \frac{145-110}{2.0537} = \frac{35}{2.0537} \approx 17.0

σ ≈ 17.0 (3 s.f.)

Precision note: using the rounded z ≈ 2.05 instead of the stored value gives σ ≈ 17.07, close but not identical. Worth a quick word here on why carrying full precision through a chain of calculations matters, this is a common source of "wrong" answers that are actually a rounding artefact.

Task 4cRe-rank the finalists, one more time

Jordan's z-score:

z_{\text{Jordan}} = \frac{150-110}{17.042} \approx 2.347

Finalist	z	P(Z > z)
Jordan	2.347	0.946%
Amara	2.133	1.64%
Diego	1.625	5.21%
Priya	1.100	13.6%

Jordan wins the scholarship. Her reconstructed z-score (≈2.347) narrowly beats Amara's (≈2.133), even though her exam's spread was never directly recorded. This is the payoff of the whole activity: the same tool that let the committee compare three totally different exams in Phase 3 also lets them recover a missing parameter and slot a fourth, late candidate straight into the ranking.

Expected sketch: the final ranking, all four finalists

Jordan's region (red) sits inside Amara's (amber), which sits inside Diego's, which sits inside Priya's. The innermost, darkest sliver wins.

✓ Gate check before giving MISSING04

Correct z from InvNorm (≈2.05), σ correctly isolated and solved (≈17.0), Jordan's z and probability found, and the final ranking stated explicitly (Jordan first, Amara second).

✦

Consolidation from the Boards

Select 2–3 groups · 10–12 min debrief

Suggested discussion sequence

One board's Phase 1a vote next to their Phase 1c working, side by side. Ask the group to narrate what changed their mind, and why.
One board's Task 2a generalization, written as a formula. Name it as the z-score formula here if the class hasn't already landed on the name themselves.
The Task 2c "whoa" result (mean 0, SD 1). Ask: "would this work if I standardized the heights of every student in this school? Why?" Connect explicitly to the standard normal distribution, N(0, 1).
A board's Task 3b GDC screens (photograph or hold up) showing the two matching probabilities. This is the moment to state plainly: standardizing never changes a probability, it only changes the ruler.
The Phase 4 twist. Ask the class to explain, in their own words, why a missing standard deviation can still be recovered from a single percentile fact. This is genuinely the most conceptually demanding step; expect to lead it.