Use continuous complements and central density areas
Choose direct areas or complements for a curved density and keep exact interval bounds.
Exam scope and assessment notes
Probability & Statistics 2 (Paper 6, A Level) · Mathematics 9709, 2026–2027, section 6.3. Paper 6 builds on Paper 5 and assumes Paper 3 calculus. A probability density is non-negative and has total area one over a single interval; its support may be infinite. Density height is not probability, and individual points have probability zero. Calculate interval probabilities, median and other percentiles directly from density areas. Explicit cumulative distribution function knowledge is not required by this section. Calculate E(X) and E(X²) by integration, then Var(X) = E(X²) − E(X)² when the needed moments are finite. Infinite-support densities require convergent improper integrals; a valid density need not have finite moments. Only single-interval densities are used here, with zero density outside the stated support. Model conclusions are conditional on the stipulated distribution; examples do not establish a real population model. 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
- Integrate quadratic expressions.
- Subtract disjoint probabilities from one.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the complement of 0.5 < X < 1.5?
What you will learn
- Calculate tails and central intervals.
- Check symmetry without confusing height with area.
Normalise the curved model
The density f(x) = (3/4)x(2 − x) on [0,2] is non-negative and integrates to one. Its primitive is (3/4)(x² − x³/3). The symmetry about one gives equal lower and upper areas at reflected bounds.
The lower area to 0.5 is 5/32. Therefore the upper area above 1.5 is also 5/32. This statement uses matching reflected bounds, not just the fact that the graph has a maximum.
Remove both tails
P(0.5 < X < 1.5) = 1 − 5/32 − 5/32 = 11/16. Alternatively evaluate the primitive directly at 1.5 and 0.5. Both routes must describe the same event.
P(X = 1) is zero, while density at one is 3/4. A narrow interval around the peak has positive but small area; the peak height is never the probability of that exact value.
Pause and explain
If each excluded tail has probability 5/32, what is the central probability?
Worked example
Find P(0.5 < X < 1.5) for this parabolic density.
Show the worked solution
- Each excluded tail has area 5/32.
- Subtract both from total area one.
Answer 11/16.
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 f(x) = (3/4)x(2 − x) on [0,2], find P(X < 0.5).
Give me a hint
Evaluate the quadratic primitive at 0.5.
Compare my reasoning
- Primitive is (3/4)(x² − x³/3).
- The lower area is (3/4)(1/4 − 1/24).
- Subtract the primitive at zero and simplify to 5/32.
5/32.
Look for these in your work
- Evaluate the whole expression.
- Use exact fractions.
Find P(X > 1.5) and P(0.5 < X < 1.5).
Give me a hint
Use reflected tails, then a complement.
Compare my reasoning
- The upper tail is 5/32 by reflection of the lower tail around one.
- Central area is 1 − 10/32.
- Simplify 22/32 to 11/16; central and both tails sum to one.
5/32 and 11/16.
Look for these in your work
- Check reflection around one.
- Remove both tails.
Find P(X < 1) and P(X = 1), and explain why they differ.
Give me a hint
Symmetry divides area, while a point has no width.
Compare my reasoning
- Density is symmetric about one.
- Half the total area lies below one, so the lower probability is 1/2.
- The exact point has zero area, so its probability is zero.
1/2 and 0 respectively.
Look for these in your work
- Use area symmetry.
- Do not use the peak height as probability.
Common mistakes
- Use reflected bounds when invoking symmetry.
- Subtract both tails for a central interval.
- A density maximum is not a point probability.
What can you explain now?
Are the two excluded tails disjoint?
Compare with the explanation
Yes.
Their probabilities may be added in the complement.
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Make it stick
Central area plus both outside tails must equal one.
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.