Use density symmetry without equating every mean and median
Use a symmetric density for centre checks and contrast it with a skewed model.
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 polynomial moments.
- Identify reflection symmetry.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Can a continuous variable take negative values?
What you will learn
- Use symmetry to check mean and median.
- Explain why skewed models may have different centres.
Symmetry must belong to the density
Density (3/4)(1 − x²) on [−1,1] is symmetric about zero, non-negative and has area one. Equal areas on either side give median zero. The mean integral is odd, so it is also zero. The second moment is 1/5, giving variance 1/5.
The central interval |X| < 0.5 has area 2∫₀⁰·⁵(3/4)(1 − x²) dx = 11/16. Negative supported values are allowed; negative density values are not.
Support symmetry alone is insufficient
Density x/2 on [0,2] rises across its interval. Its mean 4/3 and median √2 differ even though the interval has a midpoint. Equal distances to endpoints do not prove equal probability areas.
For the symmetric variable and T = 2X + 7, mean and median are both seven and variance is 4/5. Reflection and scaling preserve the relevant symmetry but scale spread separately.
Pause and explain
Does the midpoint of [0,2] determine the median of density x/2?
Worked example
Find centre and variance for (3/4)(1 − x²) on [−1,1].
Show the worked solution
- Symmetry gives mean and median zero.
- E(X²) = (3/4)(2/3 − 2/5) = 1/5.
Answer Mean 0; median 0; variance 1/5.
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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.
Find mean, median and variance for (3/4)(1 − x²) on [−1,1].
Give me a hint
Use symmetry and integrate the second moment.
Compare my reasoning
- Mean and median are zero.
- E(X²) = 1/5.
- Variance is 1/5.
Mean 0; median 0; variance 1/5.
Look for these in your work
- Check density symmetry.
- Keep the full supported interval.
Find P(|X| < 0.5) for this density.
Give me a hint
Double the area on the positive half.
Compare my reasoning
- Translate |X| < 0.5 to the symmetric interval (−0.5,0.5).
- Integral from zero to 0.5 is (3/4)(1/2 − 1/24) = 11/32.
- Double by symmetry to obtain 11/16.
11/16.
Look for these in your work
- Use symmetric bounds.
- Interpret the absolute-value interval.
For T = 2X + 7, find mean, median and variance.
Give me a hint
The symmetric centre moves while variance scales by four.
Compare my reasoning
- The original density is symmetric about zero, with variance 1/5.
- The affine transform moves that centre to seven, giving both mean and median seven.
- Variance = 2² × 1/5 = 4/5.
Mean 7; median 7; variance 4/5.
Look for these in your work
- Move the centre.
- Square the spread coefficient.
Common mistakes
- Support symmetry is not density symmetry.
- Negative outcomes can have positive density.
- Scale variance independently of centre.
What can you explain now?
Must every continuous mean equal its median?
Compare with the explanation
No.
The rising triangular model provides a counterexample.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Use symmetry of the density, not just symmetry of its support.
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.