Z Score Table Examples
Ten fully worked problems, solved from the table step by step — left, right, between, both-negative, reverse lookups and raw-score conversions. Each one names the move it uses, because in real questions the hard part is never the arithmetic: it is recognising which of the few moves the question is asking for.
Example 1 — area to the left (the direct read)
Question: Find P(Z ≤ 0.56).
Move: read the table directly — row 0.5, column 0.06.
Answer: 0.7123. 71.23% of values fall below z = 0.56; it is the 71.23rd percentile.
Why this example first: the direct read is the move every other move is built from, and 0.56 is the value textbooks reach for because its row (0.5) and column (0.06) are easy to confuse with each other — a deliberate early test of the split. If you can say, without hesitation, "row is the first decimal, column is the second", the remaining nine examples are variations, not new skills.
Example 2 — area to the right (1 minus the read)
Question: Find P(Z > 1.50).
Move: read the left area, subtract from 1. Φ(1.50) = 0.9332.
Answer: 1 − 0.9332 = 0.0668. Only 6.68% of values lie more than 1.5 standard deviations above the mean.
In words a report would use: about 1 value in 15 lies beyond z = 1.50 on the high side. "Top X%" questions are always this move — a right tail — no matter how the story is dressed, and the phrase to listen for is any variant of "above", "greater", "exceeds" or "more extreme than". When the story instead asks about a cut-off for the top 6.68%, run it backwards and you have found a reverse lookup (Example 8).
Example 3 — between two positives
Question: Find P(0.44 < Z < 1.09).
Move: two reads and a subtraction. Φ(1.09) = 0.8621, Φ(0.44) = 0.6700.
Answer: 0.8621 − 0.6700 = 0.1921. Sanity check: the band is 0.65 wide, in the curve's dense middle, so a share near 0.19 is plausible.
That plausibility habit deserves a name, because it catches real errors: width × density. The curve's height around z = 0.4–1.1 is roughly 0.29 per unit, so a band 0.65 wide should hold about 0.65 × 0.29 ≈ 0.19. If your subtraction had produced 0.45 or 0.02, the estimate would flag it before any marker does. You will not always estimate — but on the two or three problems per paper where the subtraction feels slippery, thirty seconds of width × density is the cheapest insurance in the exam.
Example 4 — spanning the mean
Question: Find P(−0.78 < Z < 0.78).
Move: same subtraction; signs do not change it. Φ(0.78) = 0.7823, Φ(−0.78) = 0.2177.
Answer: 0.7823 − 0.2177 = 0.5646. 56.46% of values fall within 0.78 standard deviations of the mean, either side.
Symmetric bands like this one power the empirical rule: at ±1 the same computation gives 0.6826, at ±2 it gives 0.9544, at ±3 it gives 0.9974. A faster route for symmetric bands exists — double the mean-to-z entry (2 × 0.2823 = 0.5646 here) — and noticing that both routes agree is a good mid-exam confidence check. The empirical rule page lays out every standard band in this family.
Example 5 — between two negatives
Question: Find P(−2.00 < Z < −1.00).
Move: read both from the negative table and subtract: Φ(−1.00) = 0.1587, Φ(−2.00) = 0.0228.
Answer: 0.1587 − 0.0228 = 0.1359. The smaller-read-first instinct is the trap: always larger cumulative minus smaller cumulative.
Quantify the trap so you recognise its smell: subtracting in the wrong order gives 0.0228 − 0.1587 = −0.1359, a negative probability — impossible on its face, yet scripts submit it every year because the digits look familiar. A negative area, or any area outside 0 to 1, is not an answer to reconsider; it is an instruction to redo the subtraction. The same reflex protects the two-tailed move in Example 6, where doubling the wrong number sails past 1 without slowing down.
Example 6 — outside two values (both tails)
Question: Find P(|Z| > 1.96) — the proportion more extreme than ±1.96.
Move: one tail, doubled. Upper tail: 1 − 0.9750 = 0.0250. By symmetry the lower tail matches.
Answer: 2 × 0.0250 = 0.0500. This is why 1.96 is the 95% fence: exactly 5% of the distribution lives outside it.
This is the computation the entire confidence-interval industry stands on: the interval from −1.96 to +1.96 is trusted with 95% confidence precisely because this example leaves exactly 5% outside it. When you meet "outside" phrasing in a question — "more extreme than", "either below −a or above +a" — the shape is two tails, and symmetry means you only ever compute one of them. If the two cutoffs differ (below −1.50 or above 2.10, say), drop the doubling and add the two tails separately: 0.0668 + 0.0179 = 0.0847.
Example 7 — a raw score end-to-end (exam)
Question: In a class with mean 70 and SD 10, what proportion of students scored below 85 — and what proportion scored above 85?
Move: standardise first: z = (85 − 70) / 10 = 1.50. Then Examples 1–2's moves.
Answer: below = 0.9332 (93.32%); above = 0.0668 (6.68%). The pair sums to 1 — always check that it does.
Examiners build questions like this one in pairs deliberately: the two halves of the question reuse one standardisation and one table read, so a candidate who sets up the first half properly collects both marks for one piece of thinking. The reverse is also true — an error in the z step taxes you twice. That asymmetry is why the standardisation line deserves its own visible row in your working, never buried inside the table step: it is the load-bearing line of the whole problem, and it is also where a marker can award a method mark when the final digit slips.
Example 8 — reverse lookup (area → z)
Question: Which z score has 90% of the distribution below it?
Move: read the table backwards. 0.9000 in the body sits between 1.28 (0.8997) and 1.29 (0.9015); the exact z is 1.282.
Answer: z ≈ 1.28. On a test with mean 1050 and SD 200, that is a score of 1050 + 1.282 × 200 ≈ 1306 for the 90th-percentile threshold.
Reverse lookups feel harder than forward ones for a simple reason: the answer lives in the table's margins rather than its body, so your eye works in the opposite direction. The technique that keeps it clean is to write the target area first ("need Φ(z) = 0.9000"), then hunt for the closest body entry, then read the row and column labels as the answer. Interpolation earns its keep here too: the body holds 0.8997 at 1.28 and 0.9015 at 1.29, and the true 0.9000 sits about a fifth of the way across — 1.282 — which is why the calculator reports 1.282 while a table-only answer of 1.28 is accepted.
Example 9 — a p-value from the table
Question: A two-tailed z-test gives a test statistic of z = −1.84. Find the p-value.
Move: single tail beyond |z|, doubled. Φ(−1.84) = 0.0329 (the tail itself, read from the negative table).
Answer: p = 2 × 0.0329 = 0.0658. Not significant at 0.05 — and note how close it is: at z = −1.96 the same calculation gives exactly 0.0500. Borderline results deserve reporting, not rounding into significance.
There is an ethics footnote hiding in this arithmetic, and working statisticians take it seriously: a p-value of 0.0658 reported as "approaching significance" and a p-value of 0.0496 reported as "significant" can describe nearly identical evidence, because the 0.05 line is a convention, not a cliff in nature. The table cannot teach judgement, but it can refuse to hide the raw material: report the p-value itself, not only the verdict it produced, and a reader can see exactly how close the call was. Example 9's 0.0658 and a hypothetical 0.0500 are different answers to the same question — but they are neighbours, and honest reporting says so.
Example 10 — temperatures (a between question in disguise)
Question: Daily July temperatures average 22°C with SD 3°C, roughly normally. On what share of days does the temperature fall between 19°C and 28.5°C?
Move: standardise both ends: (19 − 22) / 3 = −1.00 and (28.5 − 22) / 3 = 2.17. Then the between move: Φ(2.17) = 0.9850, Φ(−1.00) = 0.1587.
Answer: 0.9850 − 0.1587 = 0.8263 — about 82.6% of days, roughly 26 days in a 31-day month. Real weather is only approximately normal, so treat this as the model's estimate, not a promise.
Notice the full shape of what just happened, because it is the template for nearly every applied question you will ever meet: a story produced two raw values, the formula converted both to z scores, one table move (between) produced the probability, and one multiplication by 31 turned it into a count of days. Raw → z → table → real-world units. When a problem resists you, it is almost always because one of those four stations was skipped — most often the last one, where a perfectly correct 0.8263 is left hanging instead of becoming "about 26 days", the form the question actually asked for.
The pattern underneath all ten
Every example above is one of four moves: read (left), subtract from 1 (right), subtract two reads (between), or read backwards (area → z). Exam questions dress these moves in stories — exams, heights, temperatures, quality lines — but the dressing never changes the move. The fastest diagnostic is the sketch: draw the curve, mark the mean, place z, shade what is being asked. Ten seconds of drawing prevents the two errors that actually cost marks: answering the complement of the question, and subtracting the reads in the wrong order. Then the arithmetic is a lookup, a subtraction, or both.
How markers award these problems
Understanding the marking scheme changes how you write. In most schemes, the standardisation line (raw → z) earns a method mark, the correct table move earns a second, and the final value — stated with its interpretation ("so 87.65% of values fall between…") — earns the accuracy mark. A wrong final digit with visible, correct method typically loses one mark; a correct digit with no visible method can lose all of them when the marker cannot see how it arose, and it forfeits the partial credit that rescues borderline scripts. Write the move's name, the two table reads, and the subtraction as separate lines: it takes ten extra seconds and it is the cheapest marks insurance available anywhere in the paper.
Now do it unaided: 20 practice problems with worked solutions. The guide behind the moves is how to read a z score table.