Continuous probability density functions: VCE Mathematical Methods Units 3 & 4

Loav · VCE Mathematical Methods · Units 3 and 4

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A continuous probability model assigns probability to intervals through area. This connects probability with integration, but the meaning of the graph is different from a discrete probability table. This guide explains valid densities, interval and conditional probabilities, mean, variance, standard deviation and the effects of defining parameters. You should be able to integrate simple functions and interpret a definite integral as area.

A random variable is a numerical function of the outcome of an experiment. A continuous model is useful for quantities such as duration or length that can vary across an interval. Recorded measurements may be rounded to a finite number of decimals; the underlying continuous model and the rounded observation are different objects. A stated density is a mathematical model, not proof that every real measurement behaves exactly that way.

1. Density is not probability at a point

A probability density function f describes how probability is distributed along the number line. A tall density means that a short interval near that point carries relatively more probability. The probability itself is area, so it depends on both height and width.

f(x) ≥ 0 on its support.
The total area under f is 1.
P(a ≤ X ≤ b) = ∫_a^b f(x) dx, for a ≤ b.

The support is the set of values where the density can contribute probability. A piecewise rule should specify zero outside the stated support. When an interval extends beyond it, only the overlapping portion contributes. Never continue a formula outside its stated domain merely because the algebra still evaluates.

A density can exceed one. For example, height 4 over an interval of width 1/4 gives total area one. This does not violate a probability rule because the height is not a probability. If X has units of seconds, f has units of inverse seconds so that density multiplied by interval width gives a dimensionless probability.

For a continuous density, P(X = a) = 0 because a single point has zero width. Thus strict and inclusive endpoints give the same interval probability. This differs from a discrete count, where an endpoint may have positive probability. Probability zero in a continuous model does not mean that an observed value is logically impossible; each individual exact value has zero probability.

2. Construct a density from a nonnegative function

To turn a nonnegative function g into a density on a specified interval, multiply it by a constant k so that its total area is one. First check that the function really is nonnegative on the whole support. A total integral of one cannot repair a negative region.

If A = ∫_a^b g(x) dx is finite and positive, use f(x) = g(x)/A on [a,b], and zero elsewhere.

For a piecewise function, add the integrals of all pieces. Keep endpoints and partition points explicit. Overlap at a single endpoint does not change the probability, but assigning two conflicting values there makes the written rule ambiguous. A clean piecewise definition avoids that distraction.

3. Mean and spread use weighted integrals

The continuous mean plays the same role as the discrete expected value. Instead of adding x multiplied by individual probability masses, integrate x multiplied by density. The model’s balance point is μ = E(X). It need not be the location of the graph’s greatest height, and it need not split the total area equally.

E(X) = ∫ x f(x) dx
E(X²) = ∫ x² f(x) dx
Var(X) = E(X²) − [E(X)]²
sd(X) = √Var(X)

These integrals run over the entire support. Alternatively, variance can be calculated by integrating (x − μ)²f(x). The two methods agree. Probability, mean and second moment use different integrands: f(x), xf(x) and x²f(x). Write them separately to prevent using the normalizing integral as the expected value.

The mean and standard deviation have the same units as X; variance has squared units. Symmetry about a central value can check the mean when the relevant integrals exist, but does not by itself give the variance. Use exact fractions while integrating and round only the requested final numerical results.

Worked examples

Worked example 1: normalize, integrate and condition

An original model for a scaled waiting measure X has density f(x) = kx for 0 ≤ x ≤ 4 and zero elsewhere. Determine k, P(1 < X < 3), the mean and variance, and P(X > 2 | X > 1).

Normalization gives ∫₀⁴ kx dx = 8k = 1, so k = 1/8. The density is nonnegative on its support.
P(1 < X < 3) = ∫₁³ x/8 dx = [x²/16]₁³ = 1/2.
E(X) = ∫₀⁴ x²/8 dx = 8/3.
E(X²) = ∫₀⁴ x³/8 dx = 8, so Var(X) = 8 − (8/3)² = 8/9. Thus sd(X) = 2√2/3.
P(X > 2) = 3/4 and P(X > 1) = 15/16. The former event is contained in the latter, so the conditional probability is (3/4)/(15/16) = 4/5.

The conditional denominator is the remaining area beyond one, not the interval’s length. A uniform ratio of lengths would be wrong because this density rises with x.

4. Piecewise densities require a split integral

If an interval crosses a point where the density rule changes, integrate each relevant piece separately. Treating one formula as valid across the whole interval may produce a plausible-looking but incorrect result. Draw the support and partition points before writing the bounds.

Worked example 2: a piecewise probability model

Let f(x) = x/4 for 0 ≤ x ≤ 2, f(x) = (4 − x)/4 for 2 < x ≤ 4, and f(x) = 0 elsewhere. Find P(1 ≤ X ≤ 3), the mean and the variance.

The graph is a nonnegative triangle with base four and height one-half. Its area is one, so the density is valid.
P(1 ≤ X ≤ 3) = ∫₁² x/4 dx + ∫₂³ (4 − x)/4 dx = 3/8 + 3/8 = 3/4.
Symmetry about x = 2 gives E(X) = 2. Direct integration of xf(x) over the two pieces confirms this.
Var(X) = ∫₀² (x − 2)²x/4 dx + ∫₂⁴ (x − 2)²(4 − x)/4 dx = 2/3.

Therefore sd(X) = √(2/3). The density is continuous at the joining point, but its two algebraic pieces still require separate treatment. Equal left and right area is a useful check on the symmetric graph.

5. Parameter changes must preserve total probability

In the family f(x) = 2x/b² for 0 ≤ x ≤ b, with b > 0, the parameter changes both the support and the height. The total area remains one because ∫₀^b 2x/b² dx = 1. Increasing b widens the support while lowering the maximum height, 2/b. Simply extending the old graph without changing its coefficient would increase the area and invalidate the density.

The mean of this family is 2b/3 and its variance is b²/18, found by the usual integrals. These expressions let you predict how location and spread change with the defining parameter. Always check b’s permitted values before solving a parameter equation.

Worked example 3: use symmetry without losing the calculation

A centred measurement-error model has f(x) = k(1 − x²) on −1 ≤ x ≤ 1 and zero elsewhere. Find k, the probability of an error between −0.5 and 0.5, and its mean and variance.

The factor 1 − x² is nonnegative on this support. Its integral is 4/3, so k = 3/4.
P(−0.5 ≤ X ≤ 0.5) = ∫_−1/2^1/2 (3/4)(1 − x²) dx = 11/16 = 0.6875.
The positive and negative contributions to ∫ xf(x) dx cancel by symmetry, giving E(X) = 0.
E(X²) = (3/4)∫₋₁¹(x² − x⁴) dx = 1/5. Thus Var(X) = 1/5 and sd(X) = 1/√5.

Zero mean error does not mean every error is zero. Positive and negative values balance, while the positive variance records their spread.

6. Check the intersection in a conditional interval

Two interval events may overlap only partly. If A describes values between one and three, while B describes values between two and four, their intersection covers only values between two and three. The conditional probability P(A | B) divides the integral over that overlap by the integral over all of B. Integrating over all of A in the numerator would include outcomes excluded by the condition.

Before evaluating either integral, intersect its bounds with the support of the density. If the conditioning interval has no probability mass, the elementary conditional ratio is undefined. A positive interval length alone does not ensure a positive denominator when the density is zero there.

Practice and reasoned answers

1. A density has constant height c on [2,2.5] and is zero elsewhere. Find c and P(2.1 < X < 2.3).

Total area is 0.5c = 1, so c = 2. The requested interval has width 0.2 and area 2(0.2) = 0.4. A height greater than one is valid because the total area, rather than the height, must equal one.

2. f(x) = 4x³ on [0,1], zero elsewhere. Find E(X) and Var(X).

The density integrates to one and is nonnegative. E(X) = ∫₀¹4x⁴ dx = 4/5. E(X²) = ∫₀¹4x⁵ dx = 2/3. Therefore Var(X) = 2/3 − 16/25 = 2/75.

3. f(x) = k(2 − x) on [0,2], zero elsewhere. Find k and P(X > 1).

Normalization gives k[2x − x²/2]₀² = 2k = 1, so k = 1/2. The area from one to two is ∫₁²(2 − x)/2 dx = 1/4. The decreasing density puts more probability on the lower half.

4. A proposed density is f(x) = x on [−1,1]. Why is it invalid?

It is negative for −1 ≤ x < 0, which already violates the density requirement. Its integral is also zero, not one. Multiplying by a normalizing constant cannot make both halves nonnegative; the proposed shape itself must change.

5. A uniform density is 1/4 on [1,5]. Find the mean, variance and P(X = 3).

E(X) = (1/4)∫₁⁵x dx = 3. E(X²) = 31/3, so Var(X) = 31/3 − 9 = 4/3. P(X = 3) = 0 because one point has zero area, even though the density there is 1/4.

6. For f(x) = x/8 on [0,4], find c such that P(X > c) = 0.25.

The desired area gives ∫_c⁴x/8 dx = 1 − c²/16 = 0.25. Therefore c² = 12. Only c = √12 lies in the support, so c ≈ 3.4641. Reject the negative algebraic root.

7. For f(x) = 2x/b² on [0,b], b > 0, the mean is 6. Find b and the standard deviation.

The family mean is 2b/3, so 2b/3 = 6 gives b = 9. The variance is b²/18 = 81/18 = 4.5 and the standard deviation is √4.5 = 3/√2. Use the positive root because standard deviation is nonnegative.

Learning check

Explain the difference between density height and probability. Check both validity conditions, split an integral at every rule change, and calculate mean and second moment using different integrands. For a conditional question, identify the overlap and the conditioning area. Next, apply the same area reasoning to the normal density, using numerical probability tools when elementary integration is unavailable.

Curriculum and study route

Loav Mathematical Methods Units 3 and 4. These module divisions are a teaching sequence. Use the current VCAA Mathematics study design, Mathematical Methods Units 3 and 4, Area of Study 4, and the VCAA Mathematical Methods FAQ for course requirements. Scope checked 15 October 2026. All worked scenarios, questions and explanations in this guide are original.