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A Level Maths 9709 · Paper 6 Probability & Statistics 2 · 6.3 · ~3 min read

Continuous random variables

Probability density functions, the area condition ∫f(x) dx = 1, probabilities as areas, the mean and variance by integration, and the median and other percentiles of a continuous random variable.

Probability density functionTotal area = 1Probabilities as areasSketching pdfsE(X)Var(X)MedianPercentiles
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Paper tag = where it is examinedFORMULA LIST (MF19) given in the examM1A1B1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • A pdf f(x) is never negative and the total area under it is 1. Probabilities are areas: P(a < X < b) = ∫ₐᵇ f(x) dx.
  • For a continuous variable, P(X = a) = 0, so < and ≤ give the same probability.
  • E(X) = ∫x f(x) dx and Var(X) = ∫x² f(x) dx − [E(X)]², over the whole range.
  • Median m: solve ∫ from the lower limit to m of f(x) dx = 0.5.

By the end you can

  • Use a pdf, including finding an unknown constant from the total area.
  • Find probabilities by integration, and sketch simple pdfs.
  • Calculate E(X) and Var(X).
  • Find the median and other percentiles.

1Probability density functions

The area rule

xf(x)01230.51area 0.5median 2.38f(x) = x²/9, 0 ≤ x ≤ 3total area = 1E(X) = 2.25
f(x) = x²/9 on [0, 3]: area under the curve is 1; the shaded area is a probability.
∫ f(x) dx = 1 (over the whole range) ; P(a < X < b) = ∫ₐᵇ f(x) dx
f(x) = 0 outside the stated range
Worked example

f(x) = k(4 − x²) for 0 ≤ x ≤ 2, and 0 otherwise. Find k and E(X).

  1. k[4x − x³/3]₀² = k(8 − 8/3) = 16k/3 = 1 ⇒ k = 3/16 M1 A1
  2. E(X) = (3/16)∫₀² (4x − x³) dx = (3/16)(8 − 4) = 0.75 M1 A1
Common mistakeIgnoring the domain. Integrate only over the range where f(x) is non-zero.

2Mean, variance and median

By integration

E(X) = ∫ x f(x) dx ; Var(X) = ∫ x² f(x) dx − [E(X)]² FORMULA LIST (MF19)
integrate over the whole range
Worked example

f(x) = x²/9 for 0 ≤ x ≤ 3. Find (a) E(X), (b) Var(X), (c) the median, (d) P(X < 1).

  1. (a) ∫₀³ x³/9 dx = 81/36 = 2.25 M1 A1
  2. (b) ∫₀³ x⁴/9 dx = 243/45 = 5.4; Var = 5.4 − 2.25² = 0.3375 M1 A1
  3. (c) m³/27 = 0.5 ⇒ m = 3 × 0.51/3 = 2.38 M1 A1
  4. (d) 1/27 = 0.0370 B1
  • The median here (2.38) exceeds the mean (2.25): the pdf is increasing, with a long tail to the left (negative skew).
  • Percentiles: for the upper quartile solve the integral = 0.75.
  • Mode: the x where f(x) is greatest (look at the sketch, or set f′(x) = 0).
Common mistakeForgetting to subtract the mean squared, or using f(x) instead of xf(x) for the mean.
Linking questionThe integration is from Integration and Further integration; the normal distribution in The normal distribution is the most important continuous distribution.

Test yourself — cover the answers first

Quick check

  1. What must the area under a pdf equal?
  2. Find P(X = 1.5) for a continuous X.
  3. f(x) = ½ for 0 ≤ x ≤ 2. Find E(X).
  4. Write the equation for the median m of f on [0, 4].
  5. Can a pdf take the value 1.4 at some point?

Exam-style

  1. f(x) = k/x² for 1 ≤ x ≤ 4. Find k and the median.
  2. f(x) = 2x for 0 ≤ x ≤ 1. Find Var(X).
  3. For f(x) = x²/9 on [0, 3], find the upper quartile.
Show answers
  1. 1
  2. 0
  3. 1
  4. ∫₀ᵐ f(x) dx = 0.5
  5. yes, only areas must be at most 1
  6. k(1 − ¼) = 1 ⇒ k = 4/3; (4/3)(1 − 1/m) = 0.5 ⇒ m = 1.6
  7. E(X) = 2/3, E(X²) = 1/2, Var = 1/2 − 4/9 = 1/18
  8. q³/27 = 0.75 ⇒ q = 3 × 0.751/3 = 2.73
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Independent notes written by Riyaz Sir. Not produced, endorsed or approved by Cambridge University Press & Assessment. Syllabus codes 9709 and 9702 refer to the Cambridge International AS & A Level syllabuses.