Probability density functions and total area | Year 12

01 / LEARN

Understand the idea.

Probability is area

For a variable described by a probability density, probability is area under the curve over an interval. A height $f (x)$ describes how densely probability is distributed near $x$ ; it is not the probability of that exact value.

A density is nonnegative everywhere and has total area $1$ . Its support identifies where it can be nonzero. An individual point has zero area, so including or excluding a single endpoint does not change an interval probability.

\begin{array}{c} f (x) \geq 0 \\ \int_{- \infty}^{\infty} f (x) d x = 1 \end{array}

The density f(x)=2x is drawn on its supplied support from zero to one, with horizontal labels 0, 0.5 and 1. Density heights remain unnumbered and no probability area is shaded.

Normalise, then check

If a proposed density contains an unknown constant, set its integral over the support equal to $1$ and solve. Then check that the resulting function is nonnegative throughout that support. Both requirements are necessary for a valid density.

There is no upper bound of $1$ on a density's height. A tall curve can have area $1$ if its support is narrow enough. Probabilities are constrained to $[0, 1]$ ; density heights are constrained only to be nonnegative.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Find the scale factor

Find k if $f (x) = k x$ on [0,2] is a density.

Worked solution

1
Require total area one on the stated support.
$\int_{0}^{2} k x d x = 1$
The density is zero outside this interval, so all probability is included.
2
Integrate and solve for the constant.
$\begin{aligned} {[\frac{k x^{2}}{2}]}_{0}^{2} & = 1 \\ 2 k & = 1 \\ k & = \frac{1}{2} \end{aligned}$
The constant scales the entire area.
3
Check nonnegativity.
$f (x) = \frac{x}{2} \geq 0 (0 \leq x \leq 2)$
The normalised function satisfies the second density condition as well.

Answer

$k = 1 / 2$ .

EXAMPLE 02

A tall density

Is $f (x) = 2$ on [0,0.5], zero elsewhere, a density?

Worked solution

1
Check that the height is nonnegative.
$f (x) = 2 \geq 0 (0 \leq x \leq 0.5)$
Outside the support the density is zero, which is also allowed.
2
Calculate the rectangle area.
$\int_{0}^{0.5} 2 d x = 2 (0.5 - 0) = 1$
The integral equals height times width for a constant density.
3
Conclude that the model is valid.
$P (0 \leq X \leq 0.5) = 1$
Both density conditions hold; a height above one does not invalidate an area-based probability model.

Answer

Yes.

03 / YOUR TURN

Try it yourself.

Check your understanding

Find c if $f (x) = c$ on [1,5] is a density.

Show answer & reasoning

$c = 1 / 4$ .

The support has width $5 - 1 = 4$ , so its area is $4 c = 1$ . Thus $c = \frac{1}{4}$ , which is nonnegative.

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Is $f (x) = 0.8$ on $[0, 2]$ a density? Explain.
2 marks
Find $k$ if $f (x) = k x$ on $[0, 3]$ is a density.
2 marks
Check both conditions for $f (x) = 3 x^{2}$ on $[0, 1]$ .
2 marks
For any continuous density, find $P (X = 0.7)$ .
2 marks
A nonnegative triangular density has width 1 and height 2. Is its area correct?
2 marks
A function has integral one but is negative on part of its support. Is it a density?
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Is $f (x) = 0.8$ on $[0, 2]$ a density? Explain.
2 marks
Total area $0.8 (2) = 1.6$ , not one.
No.
Find $k$ if $f (x) = k x$ on $[0, 3]$ is a density.
2 marks
$\int_{0}^{3} k x d x = \frac{9 k}{2} = 1$ .
$k = \frac{2}{9}$ .
Check both conditions for $f (x) = 3 x^{2}$ on $[0, 1]$ .
2 marks
$3 x^{2} \geq 0$ and $\int_{0}^{1} 3 x^{2} d x = 1$ .
It is valid.
For any continuous density, find $P (X = 0.7)$ .
2 marks
One point has zero width and zero area.
0.
A nonnegative triangular density has width 1 and height 2. Is its area correct?
2 marks
$\frac{1}{2} (1) (2) = 1$ .
Yes.
A function has integral one but is negative on part of its support. Is it a density?
2 marks
Densities must be nonnegative on their support.
No.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Mathematical Methods: continuous random variables, normal distributions and sampling distributions of sample means. Original lessons and practice.

Official curriculum source ↗

In this lesson

Understand the idea.

Probability is area

Normalise, then check

See the method in action.

Find the scale factor

Require total area one on the stated support.

Integrate and solve for the constant.

Check nonnegativity.

A tall density

Check that the height is nonnegative.

Calculate the rectangle area.

Conclude that the model is valid.

Try it yourself.

Check your understanding

Keep practising