The shape of the lesson
By the end of the lesson, learners can
- What a density is: For a continuous variable, \(f(x)\) is not a probability: it is a height, and probabilities are areas under it.
- Mean and variance: The formulas copy the discrete ones from Paper 5 with the sum replaced by an integral: \(\text{E}(X) = \int x f(x)\,dx\), \(\text{E}(X^{2}) = \int x^{2} f(x)\,dx\), \(\text{Var}(X) = \text{E}(X^{2}) - [\text{E}(X)]^{2}\).
- Median and percentiles: The median \(m\) splits the area in half: \(\int_{\text{lower limit}}^{m} f(x)\,dx = 0.5\).
- 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 \(x \to \infty\).
One picture every idea returns to
A probability density function satisfies \(f(x) \ge 0\) and \(\displaystyle\int f(x)\,dx = 1\) over its domain. \(\text{P}(a < X < b) = \displaystyle\int_a^b f(x)\,dx\), \(\text{E}(X) = \displaystyle\int x f(x)\,dx\), \(\text{Var}(X) = \displaystyle\int x^{2} f(x)\,dx - \mu^{2}\), and the median \(m\) has half the area to its left.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Probability as area.. Paper 6. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: What a density is | For a continuous variable, \(f(x)\) is not a probability: it is a height, and probabilities are areas under it. Model the worked example Finding the constant and a probability on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Mean and variance | The formulas copy the discrete ones from Paper 5 with the sum replaced by an integral: \(\text{E}(X) = \int x f(x)\,dx\), \(\text{E}(X^{2}) = \int x^{2} f(x)\,dx\), \(\text{Var}(X) = \text{E}(X^{2}) - [\text{E}(X)]^{2}\). Model the worked example A symmetrical density on the board, one line at a time. Your turn: \(f(x) = \tfrac{3}{8}x^{2}\) for \(0 \le x \le 2\). Find \(\text{E}(X)\). | Pairs |
| 21 to 30 min | Teach and check: Median and percentiles | The median \(m\) splits the area in half: \(\int_{\text{lower limit}}^{m} f(x)\,dx = 0.5\). Model the worked example Median and upper quartile on the board, one line at a time. | Anticipation / Reaction Guide |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 25 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Median and percentiles, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Anticipation / Reaction Guide
Before the lesson, students mark five statements true or false. After the teaching they mark them again and justify any change of mind in one sentence each. The statements come from the examiner traps below.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| \(f(2) = \tfrac{3}{8}\) is not \(\text{P}(X = 2)\). | For a continuous variable that probability is 0; a value of \(f\) only becomes a probability after you integrate it over an interval. |
| \(\text{Var}(X) = \int x^{2} f(x)\,dx\) is NOT the variance until you subtract \(\mu^{2}\). | Leaving the answer as 4.8 above is the most frequent lost mark. |
| The median is not half-way along the domain. | For \(f(x) = \tfrac{x}{8}\) on \([0, 4]\) it is \(2.83\), not \(2\): more of the area sits on the right where the density is higher. |
Probability density functions, mean and variance: worksheet
6 questions, 25 marks. Written in the style of the papers; no past paper question is reproduced.
State two properties that a function \(f\) must have to be a probability density function on \(a \le x \le b\).
\(f(x) = k(x + 1)\) for \(0 \le x \le 2\), and 0 otherwise. Find \(k\) and \(\text{P}(X > 1)\).
For the density in the previous question, \(f(x) = \tfrac{1}{4}(x + 1)\), \(0 \le x \le 2\), find \(\text{E}(X)\) and \(\text{Var}(X)\).
For \(f(x) = \tfrac{1}{4}(x + 1)\), \(0 \le x \le 2\), find the median of \(X\), giving your answer in exact form.
\(f(x) = \dfrac{4}{x^{5}}\) for \(x \ge 1\), and 0 otherwise. Find \(\text{E}(X)\), \(\text{Var}(X)\) and the median of \(X\).
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.
(c) Find the mean repair time.
(d) Show that the median \(m\) satisfies \(m^{3} - 27m + 27 = 0\), and find \(m\) to 3 significant figures.
Answer key and marking notes
- \(f(x) \ge 0\) for all \(x\) in \(a \le x \le b\).
- \(\displaystyle\int_a^b f(x)\,dx = 1\).
- \(\displaystyle\int_0^2 (x + 1)\,dx = \left[\tfrac{x^{2}}{2} + x\right]_0^2 = 4\), so \(k = \tfrac{1}{4}\).
- \(\text{P}(X > 1) = \tfrac{1}{4}\left[\tfrac{x^{2}}{2} + x\right]_1^2 = \tfrac{1}{4}(4 - 1.5) = 0.625\).
- \(\text{E}(X) = \tfrac{1}{4}\displaystyle\int_0^2 (x^{2} + x)\,dx = \tfrac{1}{4}\left(\tfrac{8}{3} + 2\right) = \tfrac{7}{6}\).
- \(\text{E}(X^{2}) = \tfrac{1}{4}\displaystyle\int_0^2 (x^{3} + x^{2})\,dx = \tfrac{1}{4}\left(4 + \tfrac{8}{3}\right) = \tfrac{5}{3}\).
- \(\text{Var}(X) = \tfrac{5}{3} - \tfrac{49}{36} = \tfrac{11}{36} = 0.306\) (3 s.f.).
- \(\tfrac{1}{4}\left(\tfrac{m^{2}}{2} + m\right) = 0.5\), 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 \(0 \le m \le 2\).
- \(\text{E}(X) = \displaystyle\int_1^\infty 4x^{-4}\,dx = \left[-\tfrac{4}{3}x^{-3}\right]_1^\infty = \tfrac{4}{3}\).
- \(\text{E}(X^{2}) = \displaystyle\int_1^\infty 4x^{-3}\,dx = \left[-2x^{-2}\right]_1^\infty = 2\), 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.).
- (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.).
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
The random variable \(X\) has probability density function \(f(x) = kx^{3}\) for \(0 \le x \le 2\), and \(f(x) = 0\) otherwise. Find the value of \(k\).
- A\(\dfrac{1}{4}\)
- B\(\dfrac{1}{8}\)
- C\(\dfrac{1}{16}\)
- D\(4\)
\(X\) has probability density function \(f(x) = \dfrac{x}{18}\) for \(0 \le x \le 6\), and \(f(x) = 0\) otherwise. Find \(\text{P}(2 < X < 4)\).
- A\(\dfrac{1}{3}\)
- B\(\dfrac{1}{9}\)
- C\(\dfrac{4}{9}\)
- D\(\dfrac{1}{2}\)
\(X\) has probability density function \(f(x) = \dfrac{x}{18}\) for \(0 \le x \le 6\), and \(f(x) = 0\) otherwise. Find \(\text{E}(X)\).
- A\(4\)
- B\(3\)
- C\(18\)
- D\(2\)
\(X\) has probability density function \(f(x) = \dfrac{3}{64}x^{2}\) for \(0 \le x \le 4\), and \(f(x) = 0\) otherwise. Given that \(\text{E}(X) = 3\), find \(\text{Var}(X)\).
- A\(0.6\)
- B\(9.6\)
- C\(6.6\)
- D\(0.775\)
\(X\) has probability density function \(f(x) = \dfrac{3}{64}x^{2}\) for \(0 \le x \le 4\), and \(f(x) = 0\) otherwise. Find the median of \(X\), correct to 3 significant figures.
- A\(3.17\)
- B\(2\)
- C\(3\)
- D\(2.52\)
\(X\) has probability density function \(f(x) = \dfrac{k}{x^{4}}\) for \(x \ge 2\), and \(f(x) = 0\) otherwise. Find \(\text{E}(X)\).
- A\(3\)
- B\(0.375\)
- C\(12\)
- D\(2.52\)
\(X\) has probability density function \(f(x) = a + bx\) for \(0 \le x \le 2\), and \(f(x) = 0\) otherwise, where \(a\) and \(b\) are constants. Given that \(\text{E}(X) = \dfrac{7}{6}\), find \(a\) and \(b\).
- A\(a = \dfrac{1}{4}\), \(b = \dfrac{1}{4}\)
- B\(a = \dfrac{1}{2}\), \(b = 0\)
- C\(a = \dfrac{1}{8}\), \(b = \dfrac{3}{8}\)
- D\(a = 0\), \(b = \dfrac{1}{2}\)
\(X\) has probability density function \(f(x) = \dfrac{3}{64}x^{2}\) for \(0 \le x \le 4\), and \(f(x) = 0\) otherwise. Two independent observations of \(X\) are taken. Find the probability that both are greater than 3.
- A\(0.334\)
- B\(0.578\)
- C\(0.178\)
- D\(0.25\)
Quiz key
\(\displaystyle\int_0^2 kx^{3}\,dx = k\left[\dfrac{x^{4}}{4}\right]_0^2 = 4k = 1\), so \(k = \dfrac{1}{4}\). The tempting \(\dfrac{1}{8}\) sets \(f(2) = 8k = 1\), treating the height as a probability instead of using area. \(\dfrac{1}{16}\) forgets to divide \(x^{4}\) by 4 when integrating, and 4 turns the answer upside down.
\(\displaystyle\int_2^4 \dfrac{x}{18}\,dx = \left[\dfrac{x^{2}}{36}\right]_2^4 = \dfrac{16 - 4}{36} = \dfrac{1}{3}\). The tempting \(\dfrac{1}{9}\) is \(f(4) - f(2) = \dfrac{4}{18} - \dfrac{2}{18}\), a difference of heights, not an area. \(\dfrac{4}{9}\) is \(\text{P}(X < 4)\), and \(\dfrac{1}{2}\) assumes the probability is spread evenly, but the density increases with \(x\).
\(\text{E}(X) = \displaystyle\int_0^6 \dfrac{x^{2}}{18}\,dx = \left[\dfrac{x^{3}}{54}\right]_0^6 = \dfrac{216}{54} = 4\). The tempting 3 is the midpoint of the domain; that is only the mean when the graph is symmetrical, and here more of the area is on the right. 18 is \(\text{E}(X^{2})\), and 2 is \(\text{Var}(X) = 18 - 4^{2}\).
\(\text{E}(X^{2}) = \displaystyle\int_0^4 \dfrac{3x^{4}}{64}\,dx = \left[\dfrac{3x^{5}}{320}\right]_0^4 = \dfrac{3072}{320} = 9.6\). \(\text{Var}(X) = 9.6 - 3^{2} = 0.6\). The tempting 9.6 stops at \(\text{E}(X^{2})\) and never subtracts \(\mu^{2}\). 6.6 subtracts \(\mu\) instead of \(\mu^{2}\), and 0.775 is the standard deviation \(\sqrt{0.6}\).
\(\displaystyle\int_0^m \dfrac{3x^{2}}{64}\,dx = \dfrac{m^{3}}{64} = 0.5\), so \(m^{3} = 32\) and \(m = \sqrt[3]{32} = 3.17\). The tempting 2 is half-way along the domain, but most of the area lies on the right where the density is higher. 3 is the mean, not the median, and 2.52 solves \(\dfrac{m^{3}}{64} = 0.25\), which is the lower quartile.
\(\displaystyle\int_2^{\infty} kx^{-4}\,dx = \left[-\dfrac{k}{3x^{3}}\right]_2^{\infty} = \dfrac{k}{24} = 1\), so \(k = 24\). \(\text{E}(X) = \displaystyle\int_2^{\infty} \dfrac{24}{x^{3}}\,dx = \left[-\dfrac{12}{x^{2}}\right]_2^{\infty} = \dfrac{12}{4} = 3\). The tempting \(0.375\) uses \(k = 3\), the value that would be right for the domain \(x \ge 1\). 12 is \(\text{E}(X^{2})\), and 2.52 is the median \(\sqrt[3]{16}\), which is less than the mean because of the long right tail.
Area: \(2a + 2b = 1\). Mean: \(\displaystyle\int_0^2 (ax + bx^{2})\,dx = 2a + \dfrac{8b}{3} = \dfrac{7}{6}\). Subtracting, \(\dfrac{2b}{3} = \dfrac{1}{6}\), so \(b = \dfrac{1}{4}\) and \(a = \dfrac{1}{4}\). Every other option also has total area 1, which is why they tempt, but they fail the mean condition: \(a = \dfrac{1}{2}, b = 0\) gives \(\text{E}(X) = 1\); \(a = \dfrac{1}{8}, b = \dfrac{3}{8}\) gives \(\dfrac{5}{4}\); \(a = 0, b = \dfrac{1}{2}\) gives \(\dfrac{4}{3}\).
\(\text{P}(X > 3) = \displaystyle\int_3^4 \dfrac{3x^{2}}{64}\,dx = \dfrac{64 - 27}{64} = \dfrac{37}{64}\). Both: \(\left(\dfrac{37}{64}\right)^{2} = 0.334\). The tempting \(0.578\) is for one observation only. \(0.178\) is \(\left(\dfrac{27}{64}\right)^{2}\), both LESS than 3, and \(0.25 = 0.5^{2}\) assumes 3 is the median; it is the mean, and the median is 3.17.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.