Reverse normal inequalities under a negative scale
Translate reflected-variable events without keeping the wrong tail or reversing only one interval endpoint.
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
- Use normal affine parameters.
- Calculate cumulative areas and intervals.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What happens when an inequality is divided by −3?
What you will learn
- Reverse inequalities when dividing by a negative coefficient.
- Check probabilities on original and transformed normal scales.
Translate the event as well as the model
For X ~ N(10,4), U = 30 − 3X follows N(0,36). U > 6 means −3X > −24, hence X < 8 after division by minus three. The probability is Φ(−1) = 0.1587 using Φ(1) = 0.8413. Keeping X > 8 would select the wrong tail.
Using U directly gives z = (6 − 0)/6 = 1 and P(U > 6) = 1 − Φ(1), the same result. This independent event translation is a useful check on the reflection and the positive SD.
Reverse and reorder an interval
U < −12 translates to X > 14, giving right area 1 − Φ(2) = 0.0228 with Φ(2) = 0.9772. A negative scale changes ordering, not the sign of a probability.
For −6 < U < 6, substitution and division give 8 < X < 12 after correctly reordering the endpoints. Its probability is Φ(1) − Φ(−1) = 0.6826 using the quoted values. Continuous exact endpoints have zero mass, unlike the constant case of a zero scale.
Pause and explain
If U = 30 − 3X, which X event means U > 6?
Worked example
X ~ N(10,4), U = 30 − 3X. Find P(U > 6).
Show the worked solution
- U > 6 is −3X > −24.
- Dividing by minus three gives X < 8.
- z = (8 − 10)/2 = −1, giving probability 0.1587.
Answer 0.1587 using the quoted table value.
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.
For X ~ N(10,4), U = 30 − 3X, calculate P(U > 6), using Φ(1) = 0.8413.
Give me a hint
Translate the reflected event or use U’s positive SD.
Compare my reasoning
- U > 6 means X < 8.
- Original z is −1.
- The probability is 1 − 0.8413 = 0.1587.
0.1587.
Look for these in your work
- Reverse the inequality.
- Choose the correct tail.
For the same U, find P(U < −12), using Φ(2) = 0.9772.
Give me a hint
A lower reflected event becomes an upper original event.
Compare my reasoning
- U < −12 means X > 14.
- Original z is two.
- Right area is 0.0228.
0.0228.
Look for these in your work
- Keep the signed threshold.
- Take the upper complement.
Find P(−6 < U < 6), using the same model and Φ(1) = 0.8413.
Give me a hint
Transform and reorder both interval endpoints.
Compare my reasoning
- The event becomes 8 < X < 12.
- Original z limits are −1 and one.
- Subtract 0.8413 − 0.1587 = 0.6826.
0.6826.
Look for these in your work
- Reorder reflected endpoints.
- Subtract cumulative areas.
Common mistakes
- Negative division reverses order.
- Reorder both transformed interval endpoints.
- Use positive SD even for a reflection.
What can you explain now?
What event in X equals U > 6 here?
Compare with the explanation
X < 8.
The negative coefficient reverses the direction.
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Make it stick
Transform the complete event, reverse negative division and check the selected area on both scales.
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.