Find medians and percentiles from density areas
Set the left area to the requested probability and retain only roots within the support.
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 density over a variable upper bound.
- Solve power equations.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What left area locates the 80th percentile?
What you will learn
- Locate median and other percentiles directly by area.
- Check the root against the support.
Use the area target
For density 3x² on [0,1], the area from zero to q is q³. A percentile with left-area probability p therefore satisfies q³ = p, giving q = p^(1/3). The median uses p = 1/2 and is 2^(−1/3), approximately 0.7937.
This is a direct density-area calculation. Explicit cumulative distribution function knowledge is not required in section 6.3. Show the integral and area target rather than treating the median as a mean or a peak.
Roots and quartiles
The 80th percentile is 0.8^(1/3), approximately 0.9283. The interquartile range is 0.75^(1/3) − 0.25^(1/3), approximately 0.2786. It is a difference of locations, not a difference of probabilities.
For density x/2 on [0,2], the median solves m²/4 = 1/2. Only m = √2 lies in support; the negative root is rejected. A probability target outside [0,1] cannot define a percentile.
Pause and explain
Which median root is valid for density x/2 on [0,2]?
Worked example
Find the median for density 3x² on [0,1].
Show the worked solution
- Set ∫₀ᵐ3x² dx = 1/2.
- Then m³ = 1/2.
- Take the supported cube root.
Answer 2^(−1/3) ≈ 0.7937.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
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.
Find the median for 3x² on [0,1].
Give me a hint
Set the integral to one half.
Compare my reasoning
- Integrate 3x² from zero to m to get left area m³.
- Solve m³ = 1/2.
- The supported root is 2^(−1/3), approximately 0.7937.
2^(−1/3) ≈ 0.7937.
Look for these in your work
- Use half-area.
- Check support.
Find the 80th percentile and IQR for this density.
Give me a hint
Solve the same area equation at 0.8, 0.75 and 0.25.
Compare my reasoning
- Set q³ to each target probability 0.8, 0.75 and 0.25.
- q₀.₈ = 0.8^(1/3), approximately 0.9283.
- IQR = 0.75^(1/3) − 0.25^(1/3), approximately 0.2786.
80th percentile ≈ 0.9283; IQR ≈ 0.2786.
Look for these in your work
- Keep unrounded roots until subtraction.
- Interpret location units.
Find the median for density x/2 on [0,2], retaining the valid root only.
Give me a hint
The left area is m²/4.
Compare my reasoning
- m²/4 = 1/2.
- m = ±√2 algebraically.
- Only the positive root belongs to [0,2].
√2; reject −√2 outside support.
Look for these in your work
- Solve the area equation.
- Check every candidate root.
Common mistakes
- Use the requested probability, not a percentage as a whole number.
- Reject roots outside support.
- Median and mean need not agree.
What can you explain now?
Does the median split the density height?
Compare with the explanation
No; it splits total area.
The target is half the probability, not half the peak.
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Make it stick
A percentile is a supported location whose left density area reaches the target.
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.