Cambridge International AS & A Level Mathematics 9709 (2026-2027) · Paper 6: Probability & Statistics 2 · official syllabus
CONTINUOUS RANDOM VARIABLES LAB
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Four short missions on density normalisation, probability areas, expectation and variance. Integrate over the support and keep a density height distinct from a probability.
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Mission 1 of 4
Normalise the density
A candidate density is f(x)=kx for 0<x<1, and zero elsewhere. Set k so the total area is 1. A valid density must also be non-negative over its support.
Can a valid density exceed 1 at some points?
The line follows your candidate k. Its total shaded area is k/2; only area 1 gives the required normalisation. Non-negativity must also hold.
Your total area is 0.50. Set k/2=1.
Study notes
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A density gives probabilities through areas 6.3
A continuous random variable has probabilities represented by areas under a probability density function f. A valid density is non-negative on its support and has total integral 1. State the support and that the density is zero outside it.
A density height can exceed 1. A probability cannot. For a continuous density, P(X=a)=0 because a single point has no area, so strict and non-strict endpoints give the same probability. Do not read f(a) as P(X=a).
The syllabus considers densities on a single interval, which may be infinite. Identify that interval before integrating. An algebraic formula that becomes negative elsewhere is not a problem if those other points are outside the stated support and f is zero there.
Find a normalising constant 6.3
If f contains an unknown constant k, integrate f over its full support and set the result to 1. Then check non-negativity throughout that support. The integral condition alone does not rescue a function that is negative on part of its support.
Worked example: f(x)=kx on 0<x<1 and zero elsewhere. Its integral is k/2, so k=2. The density is non-negative, and its height approaches 2 near the upper end, which is allowed.
A useful check is geometric when the density is linear: f(x)=2x on (0,1) forms a triangle of base 1 and height 2, with area 1. The endpoints themselves have zero probability, so assigning a different finite density value at one endpoint does not change any probability.
Integrate over the required event 6.3
P(a<X<b) is the integral of f(x) from a to b within the support. Sketch the event and clip integration limits to the support. Complementary areas can be easier than a direct integral, but retain the original inequality in the working.
For f(x)=2x on (0,1), P(X>1/2)=integral from 1/2 to 1 of 2x dx=1-1/4=3/4. P(X=1/2)=0, not f(1/2)=1. For an interval inside the support, P(a<X<b)=b squared minus a squared.
A result outside [0,1] signals a mistake in the model, limits, antiderivative or subtraction. Integrating f over its entire support must return 1. Integrating over an interval outside the support returns 0.
Locate a median or another percentile by area 6.3
A median m divides the probability into two equal areas: the integral from the lower support limit to m equals 1/2. For another percentile with lower-tail probability q, set that area equal to q. Solve and choose a root inside the support.
For f(x)=2x on (0,1), the area from 0 to m is m squared. The median is therefore m=1/sqrt(2), approximately 0.7071 to 4 d.p. The negative square root is outside the support and must be rejected.
For this density, the lower quartile is sqrt(1/4)=1/2 and the upper quartile is sqrt(3/4)=sqrt(3)/2. These are value cutoffs, not probability heights. A median need not equal the expectation.
Find expectation and variance by weighted integrals 6.3
E(X) is the integral of x f(x) over the support, and E(X squared) is the integral of x squared f(x). Then Var(X)=E(X squared)-[E(X)] squared. Standard deviation is the square root of that variance.
For f(x)=2x on (0,1), E(X)=integral of 2x squared=2/3, while E(X squared)=integral of 2x cubed=1/2. Thus Var(X)=1/2-4/9=1/18. Here the mean 2/3 differs from the median 1/sqrt(2).
Check the expectation lies within the bounded support and the variance is non-negative. Do not replace a negative variance by its absolute value; recheck the squared integrand, exact mean and subtraction. Keep exact fractions until the final rounding step.
Handle an infinite interval with a limit 6.3
On an infinite support, an integral to infinity is evaluated by a limit. The density integral must converge to 1. Existence of total probability does not by itself guarantee that every moment integral converges.
Worked example: f(x)=3/x^4 for x>=1 and zero elsewhere has total integral 1. Its mean is 3 times the integral of x^-3 from 1 to infinity, giving 3/2; its second moment is 3 times the integral of x^-2, giving 3. Thus its variance is 3-(3/2)^2=3/4.
For that example P(X>b)=b^-3 for b>=1. The median satisfies 1-m^-3=1/2, so m=2^(1/3), approximately 1.2599 to 4 d.p. Show the support restriction and limiting calculation rather than substituting infinity as an ordinary number.
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