SCSA WACE 2024-style practice: Calculator-free. A continuous random variable has f(x)=38x^2 for 0 x 2. (a) Verify that f is a valid pdf. (b) Find P(X 1).

A verify-and-compute question. (a) f(x) 0 on [0,2], and _0^238x^2\,dx=38[x^33]_0^2=38·83=1. Both conditions hold, so f is a valid pdf. (b) P(X 1)=_1^238x^2\,dx=38[x^33]_1^2=18(8-1)=78. Markers reward checking both validity conditions and integrating over [1,2].

§-Syllabus dot point

WAMath MethodsSyllabus dot point

How does a probability density function describe a continuous random variable, and how do we find probabilities from it?

Use probability density functions to find probabilities as areas, determine an unknown constant, and compute the mean and variance by integration

WACE Year 12 Mathematics Methods Unit 4 probability density functions: the validity conditions, probability as area, finding an unknown constant, the median, and mean and variance by integration, with worked SCSA-style examples.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Validity conditions
Finding an unknown constant
Mean and variance by integration
Variance by integration
The median

What this dot point is asking

SCSA Unit 4 introduces continuous random variables, where outcomes fill an interval rather than a list. This dot point asks you to verify a probability density function (pdf), find probabilities and an unknown constant, and compute the mean and variance by integration. It is examined in both sections, with the calculator-assumed section favouring integrals that need numerical evaluation and the calculator-free section favouring polynomial densities with exact answers.

Validity conditions

Because area over a single point is zero, $P(X=a)=0$ for any value $a$ , so $P(a\le X\le b)=P(a<X<b)$ . The inequality endpoints do not change a continuous probability. The density $f(x)$ is not itself a probability and can exceed $1$ ; only areas under it represent probabilities. The support of the variable is the set of $x$ where $f(x)>0$ , and outside the support $f$ is zero, contributing nothing to any integral. Always integrate only across the support.

Finding an unknown constant

If a pdf contains an unknown coefficient, the total-area condition determines it.

Finding the constant and a probability

$X$ has $f(x)=k x^{2}$ for $0\le x\le 3$ and $0$ otherwise. Find $k$ , then $P(X\le 2)$ .

Step 1: Use the total-area condition to find $k$

Every pdf must integrate to $1$ over its support. Setting up this equation is the standard method for finding any unknown constant in a pdf.

\int_0^3 k x^{2}\,dx = k\left[\frac{x^{3}}{3}\right]_0^3 = k\cdot 9 = 1,

so $k=\dfrac{1}{9}$ .

Step 2: Integrate the pdf over the required interval for the probability

Now that $k$ is known, substitute and integrate from $0$ to $2$ . The probability equals the area under the curve over that interval.

P(X\le 2) = \int_0^2 \frac{1}{9}x^{2}\,dx = \frac{1}{9}\left[\frac{x^{3}}{3}\right]_0^2 = \frac{1}{9}\cdot\frac{8}{3} = \frac{8}{27}

Final answer: $k=\dfrac{1}{9}$ and $P(X\le 2)=\dfrac{8}{27}\approx 0.296$ .

Mean and variance by integration

The discrete sums become integrals for a continuous variable.

Mean by integration

With $f(x)=\dfrac{1}{9}x^{2}$ on $[0,3]$ , find $E(X)$ .

Step 1: Set up the mean integral

The mean of a continuous variable is $E(X)=\int x\,f(x)\,dx$ . Compared to finding a probability, we multiply the pdf by an extra factor of $x$ before integrating.

E(X) = \int_0^3 x\cdot\frac{1}{9}x^{2}\,dx = \frac{1}{9}\int_0^3 x^{3}\,dx

Step 2: Evaluate the integral

Integrating $x^3$ and applying the limits gives the mean directly.

\frac{1}{9}\left[\frac{x^{4}}{4}\right]_0^3 = \frac{1}{9}\cdot\frac{81}{4} = \frac{9}{4}

Final answer: $E(X)=\dfrac{9}{4}=2.25$ . The mean lies toward the right of the interval because the density grows with $x$ .

Variance by integration

The variance uses the same idea with an extra factor of $x^{2}$ , then subtracts the square of the mean, exactly mirroring the discrete shortcut.

Variance of a continuous variable

For the pdf $f(x)=\dfrac{1}{9}x^{2}$ on $[0,3]$ (mean $E(X)=2.25$ from the previous example), find $\mathrm{Var}(X)$ and the standard deviation.

Step 1: Compute $E(X^{2})$

The shortcut formula $\mathrm{Var}(X)=E(X^{2})-\mu^{2}$ requires $E(X^{2})$ first. This integral multiplies the pdf by $x^{2}$ (one more power of $x$ than for the mean).

E(X^{2})=\int_{0}^{3}x^{2}\cdot\frac{1}{9}x^{2}\,dx=\frac{1}{9}\int_{0}^{3}x^{4}\,dx=\frac{1}{9}\left[\frac{x^{5}}{5}\right]_{0}^{3}=\frac{1}{9}\cdot\frac{243}{5}=\frac{27}{5}=5.4

Step 2: Apply the shortcut formula for variance

Subtracting the square of the mean converts $E(X^{2})$ into the variance. Then the standard deviation is the square root.

\mathrm{Var}(X)=E(X^{2})-\mu^{2}=5.4-2.25^{2}=5.4-5.0625=0.3375

Final answer: $\mathrm{Var}(X)=0.3375$ and standard deviation $\sigma=\sqrt{0.3375}\approx 0.581$ .

The median

The median $m$ splits the area in half: $\int_{-\infty}^{m} f(x)\,dx = 0.5$ . For a symmetric pdf the median equals the mean, but for a skewed density such as the one above they differ, and each must be found from its own integral equation. To find the median of $f(x)=\dfrac{1}{9}x^{2}$ on $[0,3]$ , solve $\dfrac{1}{9}\cdot\dfrac{m^{3}}{3}=0.5$ , giving $m^{3}=13.5$ and $m\approx 2.38$ , which lies above the median position you might guess because the density is heaviest near $x=3$ .

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20227 marksCalculator-assumed. A continuous random variable

X

has probability density function

f(x)=k(4-x)

for

0\le x\le 4

and zero otherwise. (a) Find

k

. (b) Find

P(X\le 1)

. (c) Find

E(X)

Show worked answer →

A full pdf question.

(a) Total area $1$ : $\displaystyle\int_{0}^{4}k(4-x)\,dx=k\left[4x-\dfrac{x^{2}}{2}\right]_{0}^{4}=k(16-8)=8k=1$ , so $k=\dfrac{1}{8}$ .

(b) $P(X\le 1)=\displaystyle\int_{0}^{1}\dfrac{1}{8}(4-x)\,dx=\dfrac{1}{8}\left[4x-\dfrac{x^{2}}{2}\right]_{0}^{1}=\dfrac{1}{8}\left(4-\dfrac{1}{2}\right)=\dfrac{7}{16}$ .

(c) $E(X)=\displaystyle\int_{0}^{4}x\cdot\dfrac{1}{8}(4-x)\,dx=\dfrac{1}{8}\int_{0}^{4}(4x-x^{2})\,dx=\dfrac{1}{8}\left[2x^{2}-\dfrac{x^{3}}{3}\right]_{0}^{4}=\dfrac{1}{8}\left(32-\dfrac{64}{3}\right)=\dfrac{4}{3}$ .

Markers reward the area-equals-one equation, integrating over the correct interval, and the extra factor of $x$ for the mean.

WACE 20245 marksCalculator-free. A continuous random variable has

f(x)=\dfrac{3}{8}x^{2}

for

0\le x\le 2

. (a) Verify that

f

is a valid pdf. (b) Find

P(X\ge 1)

Show worked answer →

A verify-and-compute question.

(a) $f(x)\ge 0$ on $[0,2]$ , and $\displaystyle\int_{0}^{2}\dfrac{3}{8}x^{2}\,dx=\dfrac{3}{8}\left[\dfrac{x^{3}}{3}\right]_{0}^{2}=\dfrac{3}{8}\cdot\dfrac{8}{3}=1$ . Both conditions hold, so $f$ is a valid pdf.

(b) $P(X\ge 1)=\displaystyle\int_{1}^{2}\dfrac{3}{8}x^{2}\,dx=\dfrac{3}{8}\left[\dfrac{x^{3}}{3}\right]_{1}^{2}=\dfrac{1}{8}(8-1)=\dfrac{7}{8}$ .

Markers reward checking both validity conditions and integrating over $[1,2]$ .