Continuous random variables

• Probability is area, never height: the area is 1, the median is where the area reaches \(\tfrac{1}{2}\), and the variance always ends with \(-\{\mathrm{E}(X)\}^2\).
• Area 1, never height; median: area from the lower end equals ½; Var(X) = ∫x2 f(x) dx − {E(X)}2.

• 6.3.1 I can understand — understand the concept of a continuous random variable, and recall and use properties of a probability density function
• 6.3.2 I can use — use a probability density function to solve problems involving probabilities, and to calculate the mean and variance of a distribution

Accuracy for this chapter. Keep constants and moments as exact fractions through the working (\(\tfrac{3}{32}\), \(\tfrac{24}{5}\), \(\tfrac{11}{36}\)); square \(\mathrm{E}(X)\) as a fraction, not as a rounded decimal, before subtracting it. Give probabilities and moments to 3 significant figures only at the end, or leave them exact. Carry a median or percentile root unrounded (1.5874, 1.2599) until the final line. Write every integral with its limits substituted before its value: an integral read off a calculator, with no working, is an unsupported answer.

• “The median is where \(f(m) = \tfrac{1}{2}\).” Correct The median is where the area reaches \(\tfrac{1}{2}\): \(\displaystyle\int_{\text{lower end}}^{m} f(x)\,dx = \tfrac{1}{2}\). The height \(f(m)\) is a density, not a probability, and setting it to \(\tfrac{1}{2}\) answers a question nobody asked.
• “\(\mathrm{Var}(X) = \displaystyle\int x^2 f(x)\,dx\).” Correct That integral is \(\mathrm{E}(X^2)\). The variance subtracts the square of the mean: \(\mathrm{Var}(X) = \displaystyle\int x^2 f(x)\,dx - \{\mathrm{E}(X)\}^2\), exactly as \(\Sigma x^2p - \{\mathrm{E}(X)\}^2\) did for a discrete variable.
• “\(f(2) = 1.5\), so this cannot be a probability density function.” Correct \(f(x)\) can exceed 1. Only areas under \(f\) are probabilities, so the conditions are \(f(x) \ge 0\) and total area 1. \(f(x) = \tfrac{3}{8}x^2\) on \(0 \le x \le 2\) reaches 1.5 and is a valid density.
• “\(\mathrm{P}(X \le 3)\) is bigger than \(\mathrm{P}(X < 3)\).” Correct For a continuous variable \(\mathrm{P}(X = 3) = 0\): the area over a single point is zero. So \(\le\) and \(<\) give the same probability, and \(\mathrm{P}(X = 3)\) is never \(f(3)\).
• “\(\displaystyle\int_1^\infty 3x^{-4}\,dx = \Big[-x^{-3}\Big]_1^\infty = -\infty^{-3} + 1 = 1\).” Correct \(\infty\) is not a number to substitute. Integrate to \(t\), then state the limit: \(\Big[-x^{-3}\Big]_1^t = 1 - t^{-3} \to 1\) as \(t \to \infty\), because \(t^{-3} \to 0\).
• “\(\mathrm{P}(X < 2) = \displaystyle\int_0^2 \dfrac{3}{x^4}\,dx\)” for a density defined on \(x \ge 1\). Correct Below \(x = 1\) the density is 0, so the limits are cut to the interval of \(f\): \(\displaystyle\int_1^2 3x^{-4}\,dx = \tfrac{7}{8}\). The integral from 0 does not even exist, because \(3x^{-4}\) is unbounded near 0.
• Wrong: “\(f(2) = 1.5\), which is impossible for a probability.” Repair \(f(x)\) is a density, not a probability. Only areas under \(f\) are probabilities, and a density may exceed 1 provided its total area is 1.
• Wrong: “\(\mathrm{P}(X = 2) = f(2)\).” Repair For a continuous variable \(\mathrm{P}(X = 2) = \displaystyle\int_2^2 f(x)\,dx = 0\).
• Wrong: “\(\mathrm{P}(X \le 3)\) is greater than \(\mathrm{P}(X < 3)\).” Repair They are equal, because \(\mathrm{P}(X = 3) = 0\).
• Wrong: “\(\mathrm{E}(X) = \displaystyle\int f(x)\,dx\).” Repair That integral, over the whole interval, is 1. The mean weights each value by \(x\): \(\mathrm{E}(X) = \displaystyle\int x f(x)\,dx\).
• Wrong: “\(\mathrm{Var}(X) = \displaystyle\int x^2 f(x)\,dx\).” Repair That is \(\mathrm{E}(X^2)\). Subtract the square of the mean: \(\mathrm{Var}(X) = \displaystyle\int x^2 f(x)\,dx - \{\mathrm{E}(X)\}^2\).
• Wrong: “\(\mathrm{Var}(X) = \displaystyle\int \left(x f(x)\right)^2 dx - \{\mathrm{E}(X)\}^2\).” Repair The integrand is \(x^2\) times \(f(x)\), not the square of \(x f(x)\): \(f\) appears once, as the weight.
• Wrong: “The median is where \(f(m) = \tfrac{1}{2}\).” Repair The median is where the area from the lower end of the interval reaches \(\tfrac{1}{2}\): \(\displaystyle\int_{\text{lower end}}^{m} f(x)\,dx = \tfrac{1}{2}\).
• Wrong: “The median of \(\tfrac{3}{8}x^2\) on \([0, 2]\) is 1, halfway along the interval.” Repair Only a symmetric density has its median at the centre of its interval. This one rises, so more area lies to the right: \(\tfrac{m^3}{8} = \tfrac{1}{2}\) gives \(m = 1.59\), not 1.
• Wrong: “\(m^2 + 2m - 4 = 0\), so \(m = 1.24\) or \(m = -3.24\).” Repair A median must lie in the interval \(0 \le x \le 2\). Reject \(-3.24\), say why, and give one answer, 1.24.
• Wrong: “\(\displaystyle\int_1^\infty 3x^{-4}\,dx = \Big[-x^{-3}\Big]_1^\infty = -\infty^{-3} + 1\).” Repair Write the limit: \(\displaystyle\int_1^t 3x^{-4}\,dx = 1 - t^{-3}\), and as \(t \to \infty\), \(t^{-3} \to 0\), so the integral is 1.
• Wrong: integrating from 0 for a density defined on \(x \ge 1\). Repair \(f\) is zero below 1, so every integral starts at 1, the lower end of the interval of \(f\). (From 0, the integral of \(3x^{-4}\) does not even exist.)
• Wrong: “the mode of \(\tfrac{3}{8}x^2\) on \([0, 2]\) is where \(f'(x) = 0\), so it is 0.” Repair \(f'(x) = \tfrac{3}{4}x\) is zero only at \(x = 0\), where \(f\) is least. \(f\) increases throughout the interval, so the mode is the right-hand end, \(x = 2\).
• Wrong: “\(k = \tfrac{3}{32}\) and \(\displaystyle\int_0^4 x \cdot x(4 - x)\,dx = \tfrac{64}{3}\), so \(\mathrm{E}(X) = \tfrac{64}{3}\).” Repair The constant belongs to the density and must multiply every integral taken from it: \(\mathrm{E}(X) = \tfrac{3}{32} \times \tfrac{64}{3} = 2\). A mean of \(\tfrac{64}{3}\) lies outside \([0, 4]\), which is the check that catches it.

• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.
• Interleave with the chapters that use this one. Chapter 32 (hypothesis tests) and chapter 31 (sampling and estimation) read probabilities as areas under a density again, this time the normal. When you reach them, re-answer worked example 3 here: it is the same habit of integrating, or reading a table, over the right region and never at a single point. Recalling a method inside a new problem is worth more than another pass over this chapter on its own.

• Chapter 30 · Probability & Statistics 2 · How it is assessed
• Cambridge International AS & A Level Mathematics 9709 has six components, and a candidate takes two of them for the AS Level and four for the A Level. This chapter is Probability & Statistics 2 content, examined in Paper 6. Paper 6 (Probability & Statistics 2) is offered only as part of the A Level, where it is 20%. It assumes the whole of the Paper 5 content and the calculus of Paper 3. Every paper is a written examination of compulsory structured questions, answered on the question paper, with MF19 (the list of formulae and statistical tables) supplied. Examinations are available in the June and November series, and in March in India.
• Across the whole qualification the assessment objectives are weighted AO1 55% (knowledge and understanding: concepts, terminology, notation and accurate manipulative technique) and AO2 45% (application and communication: choosing the procedure, combining techniques to solve problems, and presenting the work clearly and logically) at AS Level, and AO1 52%, AO2 48% at A Level. AS candidates are graded a–e; A Level candidates A*–E.
• A question on this section can give one density, perhaps with an unknown constant, and ask a sequence of parts about it: show that the constant has a stated value, then a probability, the mean and variance, a median or percentile, perhaps a mode or a sketch. Each part is one of the five integrals of this chapter, so the reasoning that is marked is the integral with its limits, written before the number.
• MF19 prints \(\mathrm{E}(X) = \displaystyle\int x f(x)\,dx\) and \(\mathrm{Var}(X) = \displaystyle\int x^2 f(x)\,dx - \{\mathrm{E}(X)\}^2\), with no limits: you supply the interval of \(f\). It also prints \(\displaystyle\int x^n\,dx\). The properties of a density, the median and percentile condition and the meaning of the mode are not printed and must be known (MF19 card).
• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.