AS & A Level · Practice questions · Probability density functions, mean and variance

Integrate, then interpret.

Six original Cambridge-style Paper 6 questions on probability density functions, with full worked solutions.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Total area 1, then integrate what you need.

Unknown constant: set the integral over the whole domain equal to 1.
Probability: integrate \(f\) over the interval; end points do not matter.
Mean: \(\int x f(x)\,dx\), or use symmetry and say so.
Variance: \(\int x^{2} f(x)\,dx - \mu^{2}\).
Median or percentile: area to the left equals 0.5 (or the percentage); reject roots outside the domain.
Infinite domain: use \(x^{-n} \to 0\) as \(x \to \infty\).

Recall

[2 marks]

State two properties that a function \(f\) must have to be a probability density function on \(a \le x \le b\).

\(f(x) \ge 0\) for all \(x\) in \(a \le x \le b\). ✓
\(\displaystyle\int_a^b f(x)\,dx = 1\). ✓

Calculation

[3 marks]

\(f(x) = k(x + 1)\) for \(0 \le x \le 2\), and 0 otherwise. Find \(k\) and \(\text{P}(X > 1)\).

\(\displaystyle\int_0^2 (x + 1)\,dx = \left[\tfrac{x^{2}}{2} + x\right]_0^2 = 4\), so \(k = \tfrac{1}{4}\). ✓
\(\text{P}(X > 1) = \tfrac{1}{4}\left[\tfrac{x^{2}}{2} + x\right]_1^2 = \tfrac{1}{4}(4 - 1.5) = 0.625\). ✓

Calculation

[4 marks]

For the density in the previous question, \(f(x) = \tfrac{1}{4}(x + 1)\), \(0 \le x \le 2\), find \(\text{E}(X)\) and \(\text{Var}(X)\).

\(\text{E}(X) = \tfrac{1}{4}\displaystyle\int_0^2 (x^{2} + x)\,dx = \tfrac{1}{4}\left(\tfrac{8}{3} + 2\right) = \tfrac{7}{6}\). ✓
\(\text{E}(X^{2}) = \tfrac{1}{4}\displaystyle\int_0^2 (x^{3} + x^{2})\,dx = \tfrac{1}{4}\left(4 + \tfrac{8}{3}\right) = \tfrac{5}{3}\). ✓
\(\text{Var}(X) = \tfrac{5}{3} - \tfrac{49}{36} = \tfrac{11}{36} = 0.306\) (3 s.f.). ✓

Analysis

[3 marks]

For \(f(x) = \tfrac{1}{4}(x + 1)\), \(0 \le x \le 2\), find the median of \(X\), giving your answer in exact form.

\(\tfrac{1}{4}\left(\tfrac{m^{2}}{2} + m\right) = 0.5\), so \(m^{2} + 2m - 4 = 0\). ✓
\(m = \dfrac{-2 \pm \sqrt{20}}{2} = -1 \pm \sqrt{5}\). ✓
Only \(m = \sqrt{5} - 1 = 1.24\) lies in \(0 \le m \le 2\). ✓

Analysis

[5 marks]

\(f(x) = \dfrac{4}{x^{5}}\) for \(x \ge 1\), and 0 otherwise. Find \(\text{E}(X)\), \(\text{Var}(X)\) and the median of \(X\).

\(\text{E}(X) = \displaystyle\int_1^\infty 4x^{-4}\,dx = \left[-\tfrac{4}{3}x^{-3}\right]_1^\infty = \tfrac{4}{3}\). ✓
\(\text{E}(X^{2}) = \displaystyle\int_1^\infty 4x^{-3}\,dx = \left[-2x^{-2}\right]_1^\infty = 2\), so \(\text{Var}(X) = 2 - \tfrac{16}{9} = \tfrac{2}{9}\). ✓
Median: \(\left[-x^{-4}\right]_1^m = 1 - m^{-4} = 0.5\), so \(m^{4} = 2\) and \(m = \sqrt[4]{2} = 1.19\) (3 s.f.). ✓

Analysis

[8 marks]

The time, \(X\) hours, to repair a machine has density \(f(x) = k(9 - x^{2})\) for \(0 \le x \le 3\), and 0 otherwise.

(a) Show that \(k = \tfrac{1}{18}\).

(b) Find the probability that a repair takes more than 2 hours.

(d) Show that the median \(m\) satisfies \(m^{3} - 27m + 27 = 0\), and find \(m\) to 3 significant figures.

(a) \(\displaystyle\int_0^3 (9 - x^{2})\,dx = \left[9x - \tfrac{x^{3}}{3}\right]_0^3 = 27 - 9 = 18\), so \(18k = 1\), \(k = \tfrac{1}{18}\). ✓
(b) \(\tfrac{1}{18}\left[9x - \tfrac{x^{3}}{3}\right]_2^3 = \tfrac{1}{18}\left(18 - \tfrac{46}{3}\right) = \tfrac{1}{18} \times \tfrac{8}{3} = \tfrac{4}{27} = 0.148\). ✓
(c) \(\tfrac{1}{18}\displaystyle\int_0^3 (9x - x^{3})\,dx = \tfrac{1}{18}\left(\tfrac{81}{2} - \tfrac{81}{4}\right) = \tfrac{81}{72} = 1.125\) hours. ✓
(d) \(\tfrac{1}{18}\left(9m - \tfrac{m^{3}}{3}\right) = 0.5\), so \(9m - \tfrac{m^{3}}{3} = 9\); multiply by 3: \(27m - m^{3} = 27\), i.e. \(m^{3} - 27m + 27 = 0\). ✓
Solving (the root between 0 and 3): \(m = 1.04\) (3 s.f.). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.