Standardise measurements using the standard deviation
Show how subtracting the mean and dividing by standard deviation converts a normal measurement into a standard normal value.
Exam scope and assessment notes
Probability & Statistics 1 (Paper 5, AS & A Level) · Mathematics 9709, 2026–2027, section 5.5. N(μ,σ²) uses variance as its second parameter, with σ > 0; probabilities are areas, not density heights. Show the full standardisation Z = (X − μ)/σ. The MF19 table gives Φ(z) = P(Z ≤ z), with negative values found by symmetry. Model sketches and shaded probabilities may be required. Probability answers use quoted four-decimal table values where supplied; calculations with non-tabulated z use unrounded standardisation and numerical normal probabilities, so table interpolation can differ slightly in the last digits. Inverse examples use the quoted three-decimal MF19 critical values and are approximate. A binomial normal approximation requires both np > 5 and n(1 − p) > 5, and a continuity correction. It remains approximate, including for integer probability thresholds. A continuous normal variable has zero probability at a single exact value; a discrete binomial count does not. Sample means, the Central Limit Theorem and hypothesis testing belong to Statistics 2 and are not covered here. Model assumptions need evidence in real applications. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Read N(μ,σ²) correctly.
- Use cumulative normal probabilities.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For X ~ N(50,16), what is the denominator in standardisation?
What you will learn
- Show Z = (X − μ)/σ with the actual parameter values.
- Translate normal events without dividing by variance.
Centre, then scale
If X ~ N(50,16), then μ = 50 and σ = 4. Standardisation is Z = (X − 50)/4, giving Z ~ N(0,1). A measurement 58 is two standard deviations above the mean, so its z value is 2. Show the substituted fraction, not only a final probability.
Subtracting μ locates a measurement relative to the centre; dividing by the positive σ expresses that difference in standard-deviation units. Because σ is positive, inequality directions stay unchanged. Using 16 as the denominator would confuse variance with standard deviation and give the wrong shaded area.
Translate the whole event
For P(X < 46), write P(Z < (46 − 50)/4) = P(Z < −1) = 0.1587 using Φ(1) = 0.8413. For P(X > 56), the boundary becomes (56 − 50)/4 = 1.5, and the probability is 1 − Φ(1.5) = 0.0668 using Φ(1.5) = 0.9332.
For Y ~ N(120,144), σ = 12. Then P(Y < 138) = P(Z < (138 − 120)/12) = Φ(1.5) = 0.9332. Keep X or Y and Z distinct in your working: 138 is an original measurement, whereas 1.5 is its standardised value. A sketch can label both scales to make the correspondence clear.
Pause and explain
For X ~ N(50,16), what is the z value at X = 58?
Worked example
X ~ N(50,16). Find P(X < 58).
Show the worked solution
- Identify σ = √16 = 4.
- Standardise fully: P(X < 58) = P(Z < (58 − 50)/4) = P(Z < 2).
- Read Φ(2) = 0.9772 from the cumulative table.
Answer 0.9772 using Φ(2) = 0.9772.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
X ~ N(50,16). Find P(X < 46) using Φ(1) = 0.8413.
Give me a hint
Find σ first and show the fraction for the boundary.
Compare my reasoning
- σ = 4 and z = (46 − 50)/4 = −1.
- P(X < 46) = Φ(−1) = 1 − Φ(1).
- The answer is 1 − 0.8413 = 0.1587.
z = −1; probability 0.1587.
Look for these in your work
- Show full standardisation.
- Reflect the negative boundary correctly.
X ~ N(50,16). Find P(X > 56) using Φ(1.5) = 0.9332.
Give me a hint
The required probability is a right tail after standardisation.
Compare my reasoning
- z = (56 − 50)/4 = 1.5.
- P(X > 56) = 1 − Φ(1.5).
- Subtract to obtain 0.0668.
z = 1.5; probability 0.0668.
Look for these in your work
- Use standard deviation four.
- Take the correct complementary area.
Y ~ N(120,144). Find P(Y < 138) and explain why dividing by 144 is wrong.
Give me a hint
Convert the stated variance to the scale used by Z.
Compare my reasoning
- σ = √144 = 12.
- z = (138 − 120)/12 = 1.5, giving Φ(1.5) = 0.9332.
- 144 is variance in squared units; standardisation uses SD 12 in the same units as Y.
0.9332; divide by SD 12, not variance 144.
Look for these in your work
- Show the original-to-standard scale conversion.
- Explain the parameter and units distinction.
Common mistakes
- Divide by σ, not σ².
- Show the actual substituted standardisation.
- Keep original measurements and z values on their correct scales.
What can you explain now?
Why does standardising by a positive σ preserve an inequality’s direction?
Compare with the explanation
Division by a positive number preserves order.
A valid non-degenerate normal model has σ > 0.
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Make it stick
Identify σ, subtract μ, divide by σ, preserve the event and read the appropriate standard normal area.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.