PDFs and probability as area | CAIE 9709 Probability & Statistics 2 Notes

Chapters

Probability & Statistics 2

Mathematics9709

Read probability from area, not height

Definition

1 mark

What is meant by a continuous random variable?

Model answer: A random variable that can take any real value within an interval.

Definition

2 marks

What is meant by a probability density function?

Model answer: A non-negative function f whose total area over its support is 1, with interval probabilities given by areas under f.

Probability is area, not height

whole area = 1

orange = P(0.8 < X < 1.6)

A PDF gives probability through area, not through the height at one point.

P(a<X<b)=\int_a^b f(x)\,dx

Symbols

$f(x)$ = probability density at x (inverse units of X)
$a,b$ = interval boundaries (same as X)

The support is the interval of allowed values of X. Outside that interval, $f(x)=0$ .
A valid PDF has $f(x)\ge 0$ everywhere and total area $\int f(x)\,dx=1$ over its support.
For a continuous variable, $P(X=c)=0$ . Therefore changing $<$ to $\le$ at an endpoint does not change the probability.
A density height may be greater than 1. It is the area, not a single height, that must be between 0 and 1.

Make the whole area 1 before taking a part

Worked example

Start with a simple increasing density

The PDF is $f(x)=kx$ for $0\le x\le2$ . Find k and then find $P(0.5<X<1.5)$ .

\int_0^2 kx\,dx=1

2k=1,\qquad k=\frac12

\int_{0.5}^{1.5}\frac{x}{2}\,dx=0.5

Answer

k=\frac12

P(0.5<X<1.5)=0.5

Key idea

The equation “whole area = 1” finds an unknown constant. Only after that should you integrate over the interval named in the question.

Respect the support and the event boundaries

Worked example

Current-paper curved density

X has PDF $f(x)=1+\cos(\pi x)$ for $0\le x\le1$ . Show that $P(X<\tfrac12)=\tfrac12+\tfrac1\pi$ .

Use the area from 0 to one half:

\int_0^{1/2}(1+\cos(\pi x))\,dx

=\left[x+\frac{\sin(\pi x)}{\pi}\right]_0^{1/2}

Answer

P(X<\tfrac12)=\frac12+\frac1\pi

(9709/61/M/J/25 Q7(a))

Your turnPDF constant and upper tail [5 marks]

The PDF is $f(x)=k(4-x)$ for $0\le x\le4$ . Find k and $P(X>3)$ .

Show worked answer

\int_0^4 k(4-x)\,dx=8k=1

P(X>3)=\int_3^4\frac{4-x}{8}\,dx

Answer

k=\frac18

P(X>3)=\frac1{16}

Common mistake

Outside the stated support, the density is 0. Do not integrate from negative infinity unless the support actually starts there.

Read probability from area, not height

Definition

1 mark

What is meant by a continuous random variable?

Model answer: A random variable that can take any real value within an interval.

Definition

2 marks

What is meant by a probability density function?

Model answer: A non-negative function f whose total area over its support is 1, with interval probabilities given by areas under f.

Probability is area, not height

whole area = 1

orange = P(0.8 < X < 1.6)

A PDF gives probability through area, not through the height at one point.

P(a<X<b)=\int_a^b f(x)\,dx

Symbols

$f(x)$ = probability density at x (inverse units of X)
$a,b$ = interval boundaries (same as X)

The support is the interval of allowed values of X. Outside that interval, $f(x)=0$ .
A valid PDF has $f(x)\ge 0$ everywhere and total area $\int f(x)\,dx=1$ over its support.
For a continuous variable, $P(X=c)=0$ . Therefore changing $<$ to $\le$ at an endpoint does not change the probability.
A density height may be greater than 1. It is the area, not a single height, that must be between 0 and 1.

Make the whole area 1 before taking a part

Worked example

Start with a simple increasing density

The PDF is $f(x)=kx$ for $0\le x\le2$ . Find k and then find $P(0.5<X<1.5)$ .

\int_0^2 kx\,dx=1

2k=1,\qquad k=\frac12

\int_{0.5}^{1.5}\frac{x}{2}\,dx=0.5

Answer

k=\frac12

P(0.5<X<1.5)=0.5

Key idea

The equation “whole area = 1” finds an unknown constant. Only after that should you integrate over the interval named in the question.

Respect the support and the event boundaries

Worked example

Current-paper curved density

X has PDF $f(x)=1+\cos(\pi x)$ for $0\le x\le1$ . Show that $P(X<\tfrac12)=\tfrac12+\tfrac1\pi$ .

Use the area from 0 to one half:

\int_0^{1/2}(1+\cos(\pi x))\,dx

=\left[x+\frac{\sin(\pi x)}{\pi}\right]_0^{1/2}

Answer

P(X<\tfrac12)=\frac12+\frac1\pi

(9709/61/M/J/25 Q7(a))

Your turnPDF constant and upper tail [5 marks]

The PDF is $f(x)=k(4-x)$ for $0\le x\le4$ . Find k and $P(X>3)$ .

Show worked answer

\int_0^4 k(4-x)\,dx=8k=1

P(X>3)=\int_3^4\frac{4-x}{8}\,dx

Answer

k=\frac18

P(X>3)=\frac1{16}

Common mistake

Outside the stated support, the density is 0. Do not integrate from negative infinity unless the support actually starts there.