Finding Probabilities from Z-Scores
A z-score measures how many standard deviations a value sits from the mean, and the standard normal table translates that distance into a cumulative probability.
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How do you convert a raw score into a probability using a z-score and the standard normal distribution?
A class of students has a mean score of 70 with a standard deviation of 8. A student scored 86. What fraction of the class scored above this student?
Most learners assume you must use a formula and integrate the bell curve by hand. They feel stuck between knowing the answer exists somewhere under the curve and not knowing how to look it up.
A standard normal curve diagram marking the z-score on the horizontal axis, with the shaded area above it giving the probability. A side comparison shows the same z-value reading from a z-table.
Convert any raw score to a z-score, look up the cumulative probability in a standard normal table, and subtract from 1 if you need the upper tail — the bell curve becomes a measurable tool instead of a mystery.
- Computing z-scores themselves
- Inverse normal problems
- Non-normal distributions
- Manual integration of the normal curve
- 01The Mystery ScoreslideSlot 1Hook
Present the class scenario with mean 70, standard deviation 8, and a student who scored 86. Ask learners to estimate what percent of classmates scored higher, without giving them any tool yet.
- Mean = 70, Standard deviation = 8
- One student scored 86
- Question: What proportion of the class scored above 86?
PhenomenonAn exam score of 86 in a class with mean 70 and standard deviation 8 looks 'high,' but how high in probability terms?
QuestionWithout doing any math, roughly what fraction of students scored above 86 — most of them, about half, or just a few?
- 02Guess the Tail AreainteractiveSlot 2Tension
Show a standard normal curve with the region right of a shaded z-value hidden. Learners slide the z-value and try to match a guessed upper-tail probability (between 0 and 1). They quickly see that guessing is unreliable.
- Drag the z-value on the horizontal axis
- Guess the upper-tail probability
- Reveal: the true area depends precisely on the z-score
PredictionEstimate the area to the right of any z-score you select.
Tempting intuitionLearners assume the area changes linearly with z, so they guess rough percentages and get them wrong.
- 03From Raw Score to ProbabilityslideSlot 3Reveal
Show the three-step mechanism: (1) standardize using z = (x − μ) / σ, (2) look up the cumulative area to the left of z in a standard normal table, (3) subtract from 1 to get the upper-tail area. Use the class example: z = (86 − 70) / 8 = 2.00, cumulative = 0.9772, so P(X > 86) = 1 − 0.9772 = 0.0228.
- Step 1: Compute z = (x − μ) / σ
- Step 2: Read cumulative probability from the z-table
- Step 3: Subtract from 1 for the upper tail
Evidencez = 2.00 corresponds to a cumulative probability of 0.9772, so 97.72% of the class scored at or below 86 and only 2.28% scored above.
ConclusionThe proportion of students who scored above 86 is 0.0228, or about 2.3%.
Mechanism- 1Standardizing maps any normal distribution onto the standard normal curve, so one table works for all.
- 2The z-table reports the cumulative area to the LEFT of z, so the upper-tail area = 1 − table value.
- 04Any Score, Any ProbabilityslideSlot 4Takeaway
Transfer the same three-step procedure to a new situation: heights with mean 170 cm and standard deviation 6 cm, find P(X > 182). z = (182 − 170) / 6 = 2.00, cumulative = 0.9772, upper tail = 0.0228.
- Same three steps apply to any normally distributed variable
- Standardize, then read the table
- Subtract from 1 for 'greater than' questions
TransferGiven adult heights normally distributed with mean 170 cm and standard deviation 6 cm, what is the probability a randomly chosen adult is taller than 182 cm?
Expected inferenceLearners should compute z = 2.00, read 0.9772 from the table, and conclude P = 0.0228 without re-deriving the curve.
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