One curve, and a way of putting any measurement on it. The z-score is what lets a height and an exam mark be compared at all.
Which score is better?
Maths: 78 where μ = 70, σ = 8.
English: 85 where μ = 75, σ = 20.
z = (x − μ)/σ. Maths gives 78 − 708 = 1.0; English gives 85 − 7520 = 0.5. The maths mark stands a full standard deviation above its class, the English mark only half of one, so maths is the stronger performance. A compares the raw marks, which come from different tests with different spreads — the whole reason z-scores exist.
z = (x − μ) / σ
z = 174 − 1637
subtract first, then divide
z = 117 ≈ 1.57
just over 1½ SDs above
A z-score has no units. That is the point: once a height and a mark are both expressed in standard deviations, they sit on the same scale and can be compared directly.
Which is better: 70 in Algebra (μ = 62, σ = 4) or 88 in History (μ = 79, σ = 6)?
Algebra: z = 70 − 624 = 2
two SDs above its mean
History: z = 88 − 796 = 1.5
one and a half SDs above
the larger z wins: Algebra
even though 88 > 70
A value below the mean gives a negative z, and that is simply the left half of the curve. Dropping the minus sign turns “shorter than average” into “taller than average” and reverses every area that follows. Keep the sign all the way to the table.
z = 1.57
from the last step
table: area to the left = 0.9418
a standard z-table
P(X < 174) ≈ 94.2%
and 5.8% are taller
Sketch the curve and shade what the question asks for before touching the table. Almost every mistake here is a correct lookup used the wrong way round, and the sketch is what catches it.
| Within | Holds | Outside |
|---|---|---|
| μ ± 1σ | 68% | 32% |
| μ ± 2σ | 95% | 5% |
| μ ± 3σ | 99.7% | 0.3% |
Use it as a check, not as the method. Our z = 1.57 sits between 1 and 2, so the area to its left must sit between 84% and 97.5% — and 94.2% does. A table answer outside that band means the lookup went astray.
top 10% → area to the left is 0.90
the table is a left-area table
z ≈ 1.28
read the table backwards
x = μ + zσ = 72 + 1.28(10)
rearranged from z = (x−μ)/σ
x ≈ 84.8
above the mean, as it must be
Any cut-off above the mean must have a positive z, and any cut-off below it a negative one. Checking that before the arithmetic catches the “top 10%” being looked up as 0.10 rather than 0.90.
The same z = (x − μ)/σ and the same left-area table, from three other angles.
Weights are normal with μ = 50 kg, σ = 4 kg. Find P(46 < X < 56).
z = 46 − 504 = −1, z = 56 − 504 = 1.5
one on each side of μ
table: 0.1587 and 0.9332
both left areas
P = 0.9332 − 0.1587 = 0.7745
bigger minus smaller
Adding the two table values gives more than 1, which no probability can be.
Scores are normal with μ = 60, σ = 5. Below what score are the bottom 5%?
left area 0.05 → z ≈ −1.645
below the mean, so z is negative
x = 60 + (−1.645)(5) = 60 − 8.225
x = μ + zσ
x ≈ 51.8
under 60 ✓
A bottom cut-off reads the table at the small area itself, and the z carries a minus sign.
On a test with mean 80, a score of 92 has z = 1.5. Find σ.
1.5 = (92 − 80)/σ
substitute what we know
1.5σ = 12
multiply both sides by σ
σ = 8
check: 12 ÷ 8 = 1.5 ✓
Any one of x, μ, σ and z can be the unknown; the formula is the equation every time.
Normal-curve items ask for a z-score, an area (below, above or between), a cut-off for a top or bottom percentage, or which of two scores is relatively better. Sketch and shade first, convert with z = (x − μ)/σ, then use the left-area table once or twice. The 10-second check: the empirical rule — a z between 1 and 2 must give a left area between 0.84 and 0.975.
Find the z-score
Daily sales at a sari-sari store are normal with mean ₱4 200 and SD ₱600. What is the z-score of a ₱3 000 day?
z = 3 000 − 4 200600 = −1200600 = −2.0. B drops the sign, turning a poor day into an excellent one and reversing every area after it. C stops before dividing by σ. D divides 3 000 by 4 200. Since ₱3 000 is below the mean, the answer has to be negative before any arithmetic is done.
Hint: subtract the mean first, then divide by the standard deviation, and keep the sign. A value below the mean always has a negative z.
