DSE Maths | Normal Distribution Lesson Prep | From z-scores to percentiles and sampling error

2026-11-04

Tutor lesson-prep guides · Secondary Maths

Students can write z equals X minus mu over sigma, then freeze over whether to subtract from one, and stall completely on the ninetieth percentile. This lesson fixes three moves: draw the curve, standardise, and state which area is wanted. Percentiles and sample means then run on the same rails.

Objectives — Students can: (1) standardise any normal variable and use the sketch to identify a left tail, right tail or middle area; (2) reverse a probability into a z-score and restore the raw score, and explain what the ninetieth percentile means; (3) explain why the standard error is sigma over root n and apply the normal approximation to a large binomial count.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Mean and standard deviation	What does the standard deviation measure?	Spread around the mean
Symmetry	What share lies within one standard deviation?	About 68 per cent
Probability	What is the total area under the curve?	One, or 100 per cent

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-8 min	Three-question check	Spread, the 68 rule, total area	Answer orally
8-22 min	Standardise and sketch	The z formula and three regions	Sketch before calculating
22-36 min	Table and percentiles	Probability to z, then back to X	One forward, one reverse
36-50 min	Sample mean and approximation	Standard error, binomial approximation	Change n and watch the error
50-57 min	Timed practice	One forward, one reverse, one sample mean	Work alone
57-60 min	Exit check	Three checkpoints, one question each	Self-mark and circle slips

03 Teaching script (read it out)

Scene 1: Standardising: sketch first, then write z

Let X be normal with mean 60 and standard deviation 10. For P(X > 75), z is (75 – 60) divided by 10, which is 1.5. The table gives Phi(1.5) about 0.9332, so 6.68 per cent lie above 75, roughly one student in fifteen. For the range 50 to 70 the bounds are one standard deviation either side, and Phi(1) minus Phi(-1) is 68.27 per cent. The formula never changes; only the last step does, which is why the sketch comes first.

Fixed routine: sketch on the left, shade the wanted area, write z, then reach for the calculator.

Scene 2: Percentiles: reverse the table, then restore

The ninetieth percentile is the z-score with 0.90 to its left, about 1.2816. Restoring gives 60 plus 1.2816 times 10, which is 72.82. It means ninety per cent of candidates score below 72.82, not that 72.82 is ninety per cent. The usual failures are reading the percentile as a mark and forgetting to multiply by sigma. Always sanity check: a positive z must land above the mean, so an answer below 60 signals an error.

Have them say the area, the z and the raw score aloud; all three must appear.

Scene 3: Sample mean: standard error and the binomial shortcut

For samples of 25 the standard error is 10 divided by root 25, which is 2. So P(sample mean above 63) uses z equal to (63 – 60) divided by 2, again 1.5, and the probability is 6.68 per cent. Larger samples shrink the standard error and narrow the distribution, which is why big samples estimate better. For a binomial with 100 trials and success probability 0.5, the mean is 50 and the standard deviation 5; P(X at least 60) needs the continuity correction to 59.5, giving z equal to 1.9 and about 2.87 per cent. Skipping the correction overstates the tail every time.

Put the two questions side by side: the same 1.5 standard deviations, one on sigma and one on sigma over root n.

04 Question types and worked steps

Type	Example	Steps	Answer
Right tail	X normal with mean 60, sd 10, find P(X > 75)	Standardise, then one minus Phi	6.68 per cent
Middle area	Find the probability between 50 and 70	Two z-scores, subtract	68.27 per cent
Reverse percentile	Find the ninetieth percentile	Reverse lookup, then multiply by sigma and add mu	72.82
Sample mean	n = 25, find P(mean > 63)	Standard error is 10 over 5, so 2	6.68 per cent

Sanity check (teach this): Three consistency checks: the middle area 68.27 per cent must sit near 68, because that is one standard deviation either side; the right tail 6.68 per cent must be under a half, because 75 is above the mean; and the ninetieth percentile 72.82 must exceed 60, because z is positive. If any fails, the lookup direction or the addition is wrong and nothing needs recomputing from scratch.

05 Independent practice (3 questions)

Question	Answer	What to watch
X normal with mean 50, sd 8, find P(X > 62)	z = 1.5, about 6.68 per cent	Subtract from one
Same distribution, find the ninety-fifth percentile (z about 1.6449)	About 63.16	Multiply by 8 when restoring
n = 64, sigma 8, find the standard error	1	Root 64 is 8, not 64

06 Common mistakes and what to say

Mistake	Student writes	Say this
Forgetting the one minus	Reports P(X > 75) as 93.3 per cent	The table gives the left area, so the right tail needs one minus it.
Reading a percentile as a mark	Says the ninetieth percentile is 90 marks	A percentile is a rank, so reverse the lookup and restore the score.
Using sigma as the standard error	Uses 10 when n is 25	The standard error is sigma over root n, and it shrinks with sample size.
Skipping the continuity correction	Uses 60 instead of 59.5	Moving from counts to a continuous curve needs a half-step at the boundary.

07 Homework (paper, section and question numbers)

Maths chapter index — All chapters
Maths practice papers — Statistics: normal distribution and sampling questions

08 Exit check (last 5 minutes)

Question	Answer	Decision
Standardise	X normal with mean 60, sd 10, find z for 75	Cannot: reteach the formula
Percentile	What does the ninetieth percentile mean	Says 90 marks: add the reverse lookup
Standard error	What is the standard error when n is 25	Says 10: add root n

09 Tutor FAQ

May students use the built-in normal function?

Yes, but the standardising line must still be written, or the method marks are lost.

Why insist on a sketch?

The same z-score answers four different questions, and the sketch prevents reading the wrong tail.

Why is the sample mean still normal?

If the population is normal the sample mean is normal; otherwise the central limit theorem gives an approximation once n is large.

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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.