Distinguish density height from continuous probability
Read probability as area, allow density heights above one and give exact points zero probability.
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
- Calculate areas and definite integrals.
- Distinguish continuous measurements from counts.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which quantity represents continuous probability?
What you will learn
- Interpret a probability density and its units.
- Distinguish an interval from an exact point.
Height must be integrated
Let X have density 2 on 0 ≤ x ≤ 0.5 and zero elsewhere. The rectangle has width 0.5 and height 2, so total probability is one. The height being greater than one is valid: it is area, not height, that is a probability.
For 0.1 < X < 0.3, area is 2 × 0.2 = 0.4. The density units are reciprocal to the measurement units, making height multiplied by width dimensionless.
Exact values have no area
P(X = 0.2) = 0, even though the density there is 2. An interval of zero width has zero area. Including or excluding endpoints therefore does not change an interval probability for this continuous model.
A rounded reading is different from an exact value. If a measurement is reported to one decimal place as 0.2, its rounding interval is [0.15,0.25), with probability 0.2 under this stipulated density.
Pause and explain
What is P(X = 0.2) for density 2 on [0,0.5]?
Worked example
Density is 2 on [0,0.5]. Find P(0.1 < X < 0.3).
Show the worked solution
- The interval lies in the support.
- Its width is 0.2.
- Probability is 2 × 0.2 = 0.4.
Answer 0.4.
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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 P(0.1 < X < 0.3) for density 2 on [0,0.5].
Give me a hint
Multiply height by the supported width.
Compare my reasoning
- Both bounds lie within [0,0.5].
- Width = 0.3 − 0.1 = 0.2.
- Area = 2 × 0.2 = 0.4.
0.4.
Look for these in your work
- Use interval area.
- Keep probability at most one.
Find P(X ≤ 0.25) and P(X = 0.25) for this density.
Give me a hint
One event has width; the other does not.
Compare my reasoning
- For X ≤ 0.25, intersect with support to obtain [0,0.25].
- Area from zero to 0.25 is 2 × 0.25 = 0.5.
- The exact point has zero width and therefore zero area.
0.5 and 0 respectively.
Look for these in your work
- Distinguish a tail from a point.
- Include the support limit.
X is rounded to one decimal place. Find the probability it is reported as 0.2.
Give me a hint
Use the whole rounding interval, not X = 0.2.
Compare my reasoning
- The interval is [0.15,0.25).
- Its width is 0.1.
- Area = 2 × 0.1 = 0.2.
0.2, using [0.15,0.25).
Look for these in your work
- Translate the reporting rule.
- Do not assign a mass to an exact point.
Common mistakes
- Density height may exceed one.
- Exact points have zero probability.
- A rounded reading represents an interval.
What can you explain now?
Is density 2 invalid?
Compare with the explanation
No; it can have total area one.
Check width and area, not height alone.
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Make it stick
Density is probability per unit width; area is probability.
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