OxfordAQA International AS and A-level Mathematics: Probability density and cumulative distribution functions (9660) – Revision Notes

These are condensed notes for the first part of section S2.2 (Continuous random variables) of the OxfordAQA International AS and A-level Mathematics specification (9660), Version 5.2. The content belongs to International A-level Unit S2 (Statistics) and is not examined at International AS. For full explanations and longer worked examples, read the study guide. Test yourself with the practice questions, tick topics off on the checklist, and see the course hub. For a quick check of where you stand, try the free diagnostics.

Discrete or continuous

	Discrete	Continuous
Values	Separate, countable	Any value in an interval
P(X = a)	Can be positive	Exactly 0
Described by	P(X = x)	pdf f(x)
Probability from	Adding	Area under f

Because P(X = a) = 0, the signs ≤ and < give the same probability for a continuous variable: P(X < a) = P(X ≤ a) and P(a < X < b) = P(a ≤ X ≤ b).

Key results

Result	Statement
Valid pdf	f(x) ≥ 0 everywhere, and total area = 1
Interval probability	P(a < X < b) = ∫ f(x) dx from a to b
Single value	P(X = a) = 0
cdf from pdf	F(x) = ∫ f(t) dt from −∞ to x = P(X ≤ x)
pdf from cdf	f(x) = dF/dx
Interval from cdf	P(a < X < b) = F(b) − F(a)
Upper tail	P(X > a) = 1 − F(a)
Ends of the cdf	F = 0 below the range, F = 1 above it

f(x) is a height, not a probability. It can exceed 1. Example: f(x) = 2 for 0 ≤ x ≤ 1/2 has area 1.

Method boxes

Finding an unknown constant in a pdf

1. Integrate f over its range, keeping the constant.
2. Set the result equal to 1.
3. Solve. Check that f stays non-negative with that constant.

Building a cdf from a piecewise pdf

1. Take the first piece: F(x) = ∫ f(t) dt from the lower end to x.
2. Find F at the end of that piece.
3. Next piece: F(x) = (F at the start of the piece) + ∫ f(t) dt from the start of the piece to x.
4. Check the last piece reaches 1 at the top of the range.
5. Write out F(x) = 0 below and F(x) = 1 above.

Finding a median or quartile

1. Median: F(m) = 1/2. Lower quartile: F = 1/4. Upper quartile: F = 3/4.
2. Compare the target with F at each break point to pick the piece.
3. Solve in that piece and reject any root outside its range.

Using a conditional probability

P(X > b | X > a) = P(X > b)/P(X > a) = (1 − F(b))/(1 − F(a)), for b > a.

Worked reminders

Constant. f(x) = k(x + 1) for 1 ≤ x ≤ 3. The area is k[x²/2 + x] from 1 to 3 = k(15/2 − 3/2) = 6k, so k = 1/6.

cdf to pdf. F(x) = x²/25 for 0 ≤ x ≤ 5. Then f(x) = 2x/25 on that range, and P(X > 3) = 1 − 9/25 = 16/25.

Median from a pdf. f(x) = 2(1 − x) for 0 ≤ x ≤ 1. F(x) = 2x − x². Solve 2m − m² = 1/2, i.e. m² − 2m + 1/2 = 0, so m = 1 ± 1/√2. Only m = 1 − 1/√2 ≈ 0.293 lies in the range.

Upper quartile. f(x) = 3x²/8 for 0 ≤ x ≤ 2. F(x) = x³/8. Solve x³/8 = 3/4, so x³ = 6 and Q₃ = ∛6 ≈ 1.817.

A full two-piece reminder

f(x) = 1/6        for 0 ≤ x < 3
f(x) = (5 − x)/4  for 3 ≤ x ≤ 5
f(x) = 0          otherwise

Areas: 3 × 1/6 = 1/2, and (1/4)[5x − x²/2] from 3 to 5 = (1/4)(25/2 − 21/2) = 1/2. The total is 1, so f is a valid pdf.

0 ≤ x < 3:   F(x) = x/6, so F(3) = 1/2
3 ≤ x ≤ 5:   F(x) = 1/2 + (1/4)(5x − x²/2 − 21/2) = −x²/8 + 5x/4 − 17/8
F(5) = −25/8 + 25/4 − 17/8 = 1  (check)

Median: F(3) = 1/2 exactly, so the median is 3, the break point itself. Upper quartile: F(x) = 3/4 lies in the second piece, giving x² − 10x + 23 = 0, so x = 5 ± √2. Only 5 − √2 ≈ 3.59 is in 3 ≤ x ≤ 5. Interval: P(2 < X < 4) = F(4) − F(2) = 7/8 − 1/3 = 13/24.

Reading the question

Command words point to a method. Check each part of the question against this table before you start, and write the probability statement (for example F(m) = 1/2) before substituting.

The question says	You do
“Show that f is a probability density function”	Show f ≥ 0 and that the area equals 1
“Find the cumulative distribution function”	Integrate piece by piece, then state 0 below and 1 above
“Find P(a < X < b)”	Integrate f from a to b, or use F(b) − F(a)
“Given the cdf, find the pdf”	Differentiate each piece
“Find the median” or “quartile”	Solve F = 1/2, or 1/4, or 3/4
“Find the value c such that P(X > c) = 0.2”	Solve F(c) = 0.8
“Given that P(X < c) = 0.35, find the constant”	Write F(c) = 0.35 and solve for the unknown
“Find the probability that X > a, given that X > b”	Use 1 − F on top and bottom, as in the method box above

Sketching the shapes

A pdf that rises then falls has its highest point where the probability density is greatest. A cdf always starts at 0, never decreases, and ends at 1. Where the pdf is large, the cdf is steep; where the pdf is zero, the cdf is flat. Use these facts to check a sketch.

Must-know distinctions

• Density versus probability: f(x) is a height; probability is area.
• pdf versus cdf: the pdf is the derivative of the cdf, and the cdf is the running area under the pdf.
• P(X ≤ a) versus P(X = a): the second is 0 for a continuous variable.
• Piece boundaries: F must be continuous, so the two expressions agree at each break point.
• A quantile equation F(x) = p has p on the right-hand side as a probability, not as a percentage.

Quick self-test

1. f(x) = kx for 0 ≤ x ≤ 4, and 0 otherwise. Find k.
2. F(x) = x²/25 for 0 ≤ x ≤ 5 (0 below, 1 above). Find P(X > 3).
3. f(x) = 2(1 − x) for 0 ≤ x ≤ 1. Find the median.
4. State why f(x) = 2 on 0 ≤ x ≤ 1/2 can be a pdf even though 2 > 1.
5. X is continuous. Write down P(X = 0.3), and say whether P(X < 0.3) and P(X ≤ 0.3) differ.
6. f(x) = 3x²/8 for 0 ≤ x ≤ 2. Find the upper quartile to 3 significant figures.
7. Explain why f(x) = x − 1 for 0 ≤ x ≤ 2 is not a pdf.
8. f(x) = k(x + 1) for 1 ≤ x ≤ 3. Find k.
9. F(x) = x²/9 for 0 ≤ x ≤ 3. Find the value a with P(X < a) = 0.9, to 3 significant figures.
10. f(x) = x/18 for 0 ≤ x ≤ 6. Find P(X > 4 | X > 2).

Answers

1. ∫ kx dx = 8k = 1, so k = 1/8.
2. 1 − F(3) = 1 − 9/25 = 16/25.
3. F(x) = 2x − x²; 2m − m² = 1/2 gives m = 1 − 1/√2 ≈ 0.293.
4. The area is 2 × 1/2 = 1 and f ≥ 0. f is a height, not a probability, so it can exceed 1.
5. P(X = 0.3) = 0, and the two probabilities are equal.
6. x³/8 = 3/4 gives x³ = 6, so Q₃ ≈ 1.82.
7. f(x) < 0 for 0 ≤ x < 1, and a density cannot be negative. (Its integral over 0 ≤ x ≤ 2 is also 0, not 1.)
8. k = 1/6.
9. a²/9 = 0.9 gives a² = 8.1, so a ≈ 2.85.
10. F(x) = x²/36. P(X > 4 | X > 2) = (1 − 16/36)/(1 − 4/36) = (20/36)/(32/36) = 5/8.

Where marks are usually lost

• Writing f(x) = 1 or “total probability = 1” instead of showing that the integral equals 1.
• Integrating a piecewise pdf over the whole range with a single expression.
• Omitting the probability already accumulated when building a later piece of F.
• Not stating F(x) = 0 and F(x) = 1 outside the range in a full cdf.
• Accepting a root of the quantile equation that lies outside the piece being solved.
• Using a percentage (50) instead of a probability (0.5) in F(m) = …
• Differentiating F and forgetting to state the range, or that f = 0 elsewhere.
• Rounding an exact answer early and then giving an inaccurate final value.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), specification Version 5.2, published by OxfordAQA: section 3.4 International A-level Unit S2 (Statistics), S2.2 Continuous random variables.