Transform a normal distribution with a nonzero scale
State the complete transformed normal model and standardise using its new positive standard deviation.
Exam scope and assessment notes
Probability & Statistics 2 (Paper 6, A Level) · Mathematics 9709, 2026–2027, section 6.2. Paper 6 builds on Paper 5 and assumes Paper 3 calculus. Use E(aX + b) = aE(X) + b and Var(aX + b) = a²Var(X). Expectation is linear without independence, when the relevant moments exist. For independent X and Y, Var(aX + bY) = a²Var(X) + b²Var(Y); negative coefficients still give positive squared contributions. A constant changes the mean but not variance. SD scales by |a|, not a². Affine transforms of normal variables and linear combinations of independent normal variables are normal when the resulting variance is positive; zero variance is a constant, so handle it directly rather than standardising by zero. Independent Poisson counts have an exact Poisson sum with the parameters added; a scaled count or difference is not generally Poisson. Equal marginal distributions do not establish independence. Proofs are not required. Normal probabilities here use quoted cumulative values when supplied and otherwise unrounded standardisation with numerical areas. Exact distribution results are conditional on the stipulated models; real applications need evidence. Continuous density functions, sampling, estimation and hypothesis testing belong to later sections. Written practice is self-checked; cumulative assessment and human subject-teacher review remain pending.
Before you begin
- Read normal mean and variance.
- Use cumulative normal tables.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For X ~ N(10,4), what variance does 3X + 2 have?
What you will learn
- Find N(aμ + b,a²σ²) for nonzero a.
- Calculate transformed probabilities and handle a = 0 directly.
Transform both parameters
For X ~ N(10,4), U = 3X + 2 follows N(32,36). Its SD is six. Thus P(U < 38) = Φ((38 − 32)/6) = Φ(1) = 0.8413 using the quoted cumulative value. Equivalently U < 38 means X < 12; both calculations represent the same event.
For X ~ N(20,9), V = 0.5X − 4 follows N(6,2.25), with SD 1.5. P(V > 9) = 1 − Φ(2) = 0.0228 using Φ(2) = 0.9772. Use variance as the second parameter but SD as the standardisation denominator.
Normality has a constant-case limit
For a ≠ 0 and positive input variance, the affine transform remains a positive-variance normal variable. This is an exact distribution result under the model, not a count approximation; no continuity correction is needed for these continuous events.
For W = 0X + 7, variance is zero and the variable is constant seven. P(W ≤ 7) = 1, P(W < 7) = 0. Do not use a non-degenerate normal curve or divide by zero SD. A real quantity’s original normal model still needs evidence.
Pause and explain
For U ~ N(32,36), which denominator standardises U = 38?
Worked example
X ~ N(10,4). Find the distribution of 3X + 2 and P(3X + 2 < 38).
Show the worked solution
- Mean = 32 and variance = 36.
- Use SD six to standardise 38 to z = 1.
- Read Φ(1) = 0.8413.
Answer N(32,36); probability 0.8413 using the quoted table value.
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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(10,4). State U = 3X + 2’s model and find P(U < 38), using Φ(1) = 0.8413.
Give me a hint
Transform both parameters before using the new SD.
Compare my reasoning
- U ~ N(32,36).
- z = (38 − 32)/6 = 1.
- P(U < 38) = 0.8413.
N(32,36); probability 0.8413.
Look for these in your work
- State the full distribution.
- Show substituted standardisation.
X ~ N(20,9), V = 0.5X − 4. Find its model and P(V > 9), using Φ(2) = 0.9772.
Give me a hint
Square one half for variance and take a right-tail complement.
Compare my reasoning
- V ~ N(6,2.25), SD 1.5.
- z = (9 − 6)/1.5 = 2.
- P(V > 9) = 0.0228.
N(6,2.25); probability 0.0228.
Look for these in your work
- Scale variance correctly.
- Use the requested upper area.
For W = 0X + 7 with normal X, calculate P(W ≤ 7) and P(W < 7).
Give me a hint
Do not apply positive-variance standardisation to a constant.
Compare my reasoning
- W is always seven.
- The inclusive event holds in every outcome.
- The strict event holds in none.
P(W ≤ 7) = 1; P(W < 7) = 0.
Look for these in your work
- Recognise zero variance.
- Handle constant probabilities directly.
Common mistakes
- Transform variance, not only mean.
- Standardise with the new SD.
- A zero scale makes a constant.
What can you explain now?
Does this exact continuous transformation need continuity correction?
Compare with the explanation
No.
Continuity correction represents discrete counts by continuous areas; this variable is already continuous.
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Make it stick
Transform mean and variance, check positive variance, then standardise with the transformed SD.
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