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Contents · 0% complete
  1. Learning objectives
  2. Core idea
  3. Key formulae
  4. Worked examples
  5. Explore the graph
  6. Common mistakes
  7. Exam tip
  8. Guided practice
  9. Practice questions
  10. Challenge
  11. Summary
  12. Quick quiz
  13. Same topic, different board
  14. Past paper questions
Continuous random variables · Subtopic 1 of 3

Probability Density Functions

For a continuous variable, probability lives in area. A probability density function is a curve whose total area is 11; the probability of landing between aa and bb is the area under the curve between them, found by integration.

Cambridge 9709 Paper 6Cambridge 9231 Paper 4Edexcel YMA01 WST02Edexcel 9FM0 OptionAQA 7367 Paper 3≈ 35 minRevision 6 · updated 2026-10-27
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Section 1 of 14

Learning objectives

  • Know the properties of a pdf: f(x)≥0f(x)\ge0 and ∫f(x) dx=1\int f(x)\,dx=1.
  • Find an unknown constant kk in a pdf.
  • Find probabilities as P(a<X<b)=∫abf(x) dxP(a<X<b)=\int_a^b f(x)\,dx.
  • Sketch a pdf and find its mode.
Pause and think

A bus arrives at a random moment in a 1010-minute window. What is the probability it arrives at exactly 33 minutes 0.000…0.000\ldots seconds? What about between 33 and 44 minutes?

Section 2 of 14

Core idea

Start hereThe probability of exactly 1.5 seconds

A runner's reaction time could be 0.1510.151 s, 0.15120.1512 s, 0.151170.15117 s... infinitely many values. The chance of hitting any one exactly is zero. So for continuous quantities, probability lives in intervals, measured as area under a curve.

The big idea

A probability density function f(x)≥0f(x)\ge0 has total area 11. P(a<X<b)=∫abf(x) dxP(a<X<b)=\int_a^bf(x)\,dx. P(X=a)=0P(X=a)=0, so << and ≤\le are interchangeable.

1. Area is probability

f(x) = 3x²/8 for 0 ≤ x ≤ 2: area up to b
0.511.522.50.20.40.60.811.21.41.60xf(x)f(x) = 3x²/8P(X < b)
P(X<b)=∫0b3x28 dx=b38P(X<b)=\displaystyle\int_0^b\frac{3x^2}{8}\,dx=\frac{b^3}{8}. At b=2b=2 the whole area is 11.
P(a<X<b)=∫abf(x) dx,∫−∞∞f(x) dx=1P(a<X<b)=\int_a^bf(x)\,dx,\qquad\int_{-\infty}^{\infty}f(x)\,dx=1
where
f(x)f(x)
density: probability per unit of xx (can exceed 1).
Worked examplef(x)=kx2f(x)=kx^2, 0≤x≤30\le x\le3

Find kk and P(X>2)P(X>2).

Goal
kk; probability.
Strategy
Total area 11.
  1. k[x33]03=9k=1⇒k=19k\left[\tfrac{x^3}3\right]_0^3=9k=1\Rightarrow k=\tfrac19
    Normalise.
  2. P(X>2)=19[x33]23=27−827=1927P(X>2)=\tfrac19\left[\tfrac{x^3}3\right]_2^3=\tfrac{27-8}{27}=\tfrac{19}{27}
    Integrate.
Answerk=19k=\frac19; 1927\frac{19}{27}.
✓ CheckMost area is near 33 (density grows), so >13>\frac13 of the range having 70%70\% is plausible ✓.
Edge caseDensity above 1 f(x)=2f(x)=2 on [0,0.5][0,0.5] is valid: area 2×0.5=12\times0.5=1.
Common misconceptionReading f as a probability f(1)f(1) is not P(X=1)P(X=1) (that's 00).
Pause and think

f(x)=38x2f(x)=\frac38x^2 on [0,2][0,2]. P(X<1)P(X<1)?

Summary

  • Probability = area.
  • Total area 1 fixes constants.
Key words
pdf
Probability density function.
A probability density function: f(x) = ¾x(2 − x) for 0 ≤ x ≤ 2
−0.50.511.522.50.20.40.60.810xf(x)y = f(x)total area = 1
A pdf is never negative and the total area under it is 11. Probabilities are areas: P(a<X<b)=∫abf(x) dxP(a<X<b)=\int_a^b f(x)\,dx.
HistoryLaplace and the density. Pierre-Simon Laplace's Théorie analytique des probabilités (1812) treated probability with calculus, using smooth density curves and integrals instead of counting. Laplace called probability theory "common sense reduced to calculation".
Fun factProbability zero is not impossible. Your exact height to infinitely many decimal places has probability 00 of being any particular number, yet you have a height. For continuous variables "probability 00" and "impossible" are different things, a subtlety that troubled mathematicians until Andrey Kolmogorov put probability on firm foundations in 1933.
Section 3 of 14

Key formulae

Propertiesf(x)≥0,∫−∞∞f(x) dx=1f(x)\ge0,\qquad\int_{-\infty}^{\infty}f(x)\,dx=1
ProbabilitiesP(a<X<b)=∫abf(x) dxP(a<X<b)=\int_a^bf(x)\,dx
Modef′(x)=0 (or an end point if f is monotonic)f'(x)=0\ \text{(or an end point if } f \text{ is monotonic)}
TrickUse shapes before integrals. If the pdf is a straight line or constant, the area is a triangle, rectangle or trapezium: no integration needed. For f(x)=x8f(x)=\tfrac x8 on [0,4][0,4], P(X<2)P(X<2) is a triangle of base 22 and height 14\tfrac14: area 14\tfrac14. And if the pdf is symmetric, half the area lies on each side of the centre.
Section 4 of 14

Worked examples

Example 1 · Finding k0 / 2 steps
f(x)=k(x+1)f(x)=k(x+1) for 0≤x≤20\le x\le2, and 00 otherwise. Find kk.
    Example 2 · A probability0 / 2 steps
    f(x)=38x2f(x)=\tfrac38x^2 for 0≤x≤20\le x\le2. Find P(X>1)P(X>1).
      Example 3 · The mode0 / 2 steps
      f(x)=332x(4−x)f(x)=\tfrac3{32}x(4-x) for 0≤x≤40\le x\le4. Find the mode.
        Section 5 of 14

        Explore the graph

        Area is probability
        Loading the interactive graph…
        Try this
        1. The curve is f(x)=kx2f(x)=kx^2 on [0,3][0,3]; the shaded area is P(X<a)=ka33P(X<a)=\tfrac{k a^3}{3}.
        2. Only one value of kk makes the total area 11. Set a=3a=3: the area is 9k9k, so k=19≈0.11k=\tfrac19\approx0.11.
        3. With k≈19k\approx\tfrac19 and a=2a=2, the shaded area is 827≈0.30\tfrac{8}{27}\approx0.30. Does that match the picture?

        A pdf must enclose exactly one unit of area.

        In the real worldWaiting for the bus. Transport planners model waiting times with pdfs. For buses every 1010 minutes arriving at random times, the waiting time is uniform on [0,10][0,10]: f(x)=0.1f(x)=0.1. If buses bunch up, the pdf changes shape and the average wait gets longer, even with the same number of buses per hour.
        Section 6 of 14

        Common mistakes

        IncorrectP(X=2)=f(2)P(X=2)=f(2)
        CorrectP(X=2)=0P(X=2)=0
        ff is a density; probabilities need an interval.
        Incorrect"f(x)>1f(x)>1, so it can't be a pdf"
        CorrectDensities can exceed 11
        Only the total area must be 11.
        IncorrectIntegrating outside the stated interval
        CorrectUse only where f(x)≠0f(x)\ne0
        f(x)=0f(x)=0 elsewhere.
        IncorrectMode == value where f(x)=0f(x)=0
        CorrectWhere f(x)f(x) is greatest
        The mode is the peak of the curve.

        f(2)=1.5f(2)=1.5. Me: "a probability of 1.51.5?!"

        The pdf: "I'm a density, not a probability. Relax."

        Section 7 of 14

        Exam tip

        Exam tipLabel sketches. A pdf sketch needs the shape, the end points of the interval labelled on the xx-axis, and the key ff values (for example the maximum). Draw f(x)=0f(x)=0 outside the interval too, or say so.
        Section 8 of 14

        Guided practice

        SA-9709-P6-CR-G012 marksL1 FoundationCalculatorPDFs
        XX has probability density function f(x)=kx2f(x)=kx^2 for 0≤x≤30\le x\le3, and 00 otherwise. Find kk.

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        Salam Academy Original Practice · Proposed mark scheme
        Section 9 of 14

        Practice questions

        SA-9709-P6-CR1-0012 marksL1 FoundationCalculatorPDFs
        f(x)=kxf(x)=kx for 0≤x≤40\le x\le4, and 00 otherwise, is a probability density function. Find kk.

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        Salam Academy Original Practice · Proposed mark scheme
        SA-9709-P6-CR-0012 marksL1 FoundationCalculatorPDFs
        For f(x)=x29f(x)=\tfrac{x^2}9 on 0≤x≤30\le x\le3, find P(X<2)P(X<2).

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        Salam Academy Original Practice · Proposed mark scheme
        SA-9709-P6-CR1-0022 marksL1 FoundationCalculatorPDFs
        f(x)=38x2f(x)=\tfrac38x^2 for 0≤x≤20\le x\le2, and 00 otherwise. Find P(X>1)P(X>1).

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        Salam Academy Original Practice · Proposed mark scheme
        SA-9709-P6-CR-0044 marksL2 StandardCalculatorPDFs
        f(x)=k(4−x)f(x)=k(4-x) for 0≤x≤40\le x\le4, and 00 otherwise. Find kk and P(X>1)P(X>1).

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        Salam Academy Original Practice · Proposed mark scheme
        SA-9709-P6-CR1-0053 marksL2 StandardCalculatorPDFs
        f(x)=k(x+1)f(x)=k(x+1) for 0≤x≤20\le x\le2, and 00 otherwise. Find kk.

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        Salam Academy Original Practice · Proposed mark scheme
        SA-9709-P6-CR1-0031 markL1 FoundationCalculatorSketching pdfs
        f(x)=0.25f(x)=0.25 for 1≤x≤51\le x\le5, and 00 otherwise. Find P(2<X<3.5)P(2<X<3.5).

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        Salam Academy Original Practice · Proposed mark scheme
        SA-9709-P6-CR-0073 marksL2 StandardCalculatorPDFs
        f(x)=kx(4−x)f(x)=kx(4-x) for 0≤x≤40\le x\le4, and 00 otherwise. Find kk, and state E(X)E(X).

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        Salam Academy Original Practice · Proposed mark scheme
        SA-9709-P6-CR1-0062 marksL2 StandardCalculatorSketching pdfs
        f(x)=332x(4−x)f(x)=\tfrac3{32}x(4-x) for 0≤x≤40\le x\le4, and 00 otherwise. Find the mode of XX.

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        Salam Academy Original Practice · Proposed mark scheme

        More questions on every skill in this subtopic are in the question bank under Practise.

        Section 10 of 14

        Challenge

        SA-9709-P6-CR-0083 marksL3 ExamCalculatorPDFs
        For f(x)=332x(4−x)f(x)=\tfrac3{32}x(4-x) on 0≤x≤40\le x\le4, find P(1<X<3)P(1<X<3).

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        Salam Academy Original Practice · Proposed mark scheme
        Section 11 of 14

        Summary

        • f(x)≥0f(x)\ge0 and total area 11.
        • P(a<X<b)=∫abf(x) dxP(a<X<b)=\int_a^bf(x)\,dx; P(X=a)=0P(X=a)=0.
        • Find kk by setting the total integral equal to 11.
        • Mode: where ff is largest.
        Section 12 of 14

        Quick quiz

        Q1 · SA-9709-P6-CR-Q011 markL1 FoundationCalculatorPDFs
        Which is a property of every probability density function?
        Choose one option

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        Salam Academy Original Practice · Proposed mark scheme
        Q2 · SA-9709-P6-CR-0111 markL1 FoundationCalculatorSketching pdfs
        For a continuous random variable XX, P(X=2)P(X=2) is
        Choose one option

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        Salam Academy Original Practice · Proposed mark scheme
        Q3 · SA-9709-P6-CR1-0041 markL1 FoundationCalculatorSketching pdfs
        Which of these cannot be a probability density function on the given interval?
        Choose one option

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        Salam Academy Original Practice · Proposed mark scheme
        Section 13 of 14

        Same topic, different board

        Specification
        IAL Mathematics
        Where
        WST02 (Statistics 2) · Continuous random variables
        Expected
        Probability density functions, including piecewise ones, and the cumulative distribution function.
        Calculator
        Permitted. Statistical tables are provided in the formula booklet.
        Typical style
        Edexcel also uses the cumulative distribution function F(x) and piecewise pdfs.
        Open in
        YMA01 · WST029FM0 · Option
        Section 14 of 14

        Past paper questions

        5 questions (47 marks) from real exams test this subtopic. Try each one under timed conditions, then mark it with the official mark scheme.

        1. 9709/62 · Oct/Nov 2025Question 59 marks

          Find a constant in a probability density function, then use the median for a probability.

        2. 9709/62 · May/June 2025Question 79 marks

          Probability density function: find a constant, a parameter from the mean, and the median.

        3. 9709/62 · Oct/Nov 2024Question 611 marks

          Continuous distribution: constants and the median.

        4. 9709/62 · May/June 2024Question 710 marks

          Probability density function: a constant, the median and E(X).

        5. 9709/62 · May/June 2023Question 78 marks

          Linear density function: constant, expectation and a limit from the median.

        All 9709 past paper questions by topic

        Salam Academy original content · Content status: reviewed and published. Syllabus mapping: Pearson Edexcel YMA01, 2018 specification (official specification).