Unit 19 Review Exercises
Continuous Random Variables and the Normal Distribution
These questions mix the ideas from the whole unit. Paper 1 means no calculator; Paper 2 means a GDC is allowed. AHL marks questions for HL students only.
1 Paper 1 AHL \(X\) has probability density function (p.d.f.) \(f(x) = k(4 - x)\) for \(0 \leq x \leq 4\), and \(f(x) = 0\) otherwise.
(a) Find \(k\).
(b) Find \(\mathrm{P}(X > 2)\).
(c) Find \(\mathrm{E}(X)\).
(a) The graph is a triangle with base \(4\) and height \(4k\): \(\;\frac{1}{2} \times 4 \times 4k = 8k = 1\), so \(k = \dfrac{1}{8}\).
(b) A triangle with base \(2\) and height \(f(2) = \frac{1}{4}\): \(\;\mathrm{P}(X > 2) = \frac{1}{2} \times 2 \times \frac{1}{4} = \dfrac{1}{4}\).
(c) \(\mathrm{E}(X) = \dfrac{1}{8}\displaystyle\int_0^4 \left(4x - x^2\right)\mathrm{d}x = \frac{1}{8}\left(32 - \frac{64}{3}\right) = \frac{4}{3}\). See 19.1.
2 Paper 1 AHL \(X\) has p.d.f. \(f(x) = 6x(1 - x)\) for \(0 \leq x \leq 1\), and \(f(x) = 0\) otherwise.
(a) Write down \(\mathrm{E}(X)\).
(b) Find \(\mathrm{Var}(X)\).
(c) Find \(\mathrm{E}(5X - 2)\) and \(\mathrm{Var}(5X - 2)\).
(a) Symmetrical about \(x = \frac{1}{2}\), so \(\mathrm{E}(X) = \dfrac{1}{2}\).
(b) \(\mathrm{E}(X^2) = 6\displaystyle\int_0^1 \left(x^3 - x^4\right)\mathrm{d}x = 6\left(\frac{1}{4} - \frac{1}{5}\right) = \frac{3}{10}\), so \(\mathrm{Var}(X) = \dfrac{3}{10} - \dfrac{1}{4} = \dfrac{1}{20}\).
(c) \(\mathrm{E}(5X - 2) = \dfrac{5}{2} - 2 = \dfrac{1}{2}\); \(\;\mathrm{Var}(5X - 2) = 25 \times \dfrac{1}{20} = \dfrac{5}{4}\). See 19.1 and 19.2.
3 Paper 1 AHL \(X\) has p.d.f. \(f(x) = \dfrac{3}{8}(2 - x)^2\) for \(0 \leq x \leq 2\), and \(f(x) = 0\) otherwise. Find the exact value of the median of \(X\).
\(\displaystyle\int_0^m \frac{3}{8}(2 - x)^2\,\mathrm{d}x = \frac{1}{8}\Big[-(2 - x)^3\Big]_0^m = \frac{1}{8}\left(8 - (2 - m)^3\right) = \frac{1}{2}\).
So \((2 - m)^3 = 4\), \(\;2 - m = \sqrt[3]{4}\), and \(m = 2 - \sqrt[3]{4}\) \(\;(\approx 0.413)\). See 19.2.
4 Paper 1 AHL \(X\) has p.d.f. \(f(x) = \dfrac{1}{4}x(4 - x^2)\) for \(0 \leq x \leq 2\), and \(f(x) = 0\) otherwise.
(a) Find \(\mathrm{P}(X < 1)\).
(b) Show that the median \(m\) satisfies \(m^4 - 8m^2 + 8 = 0\), and hence find the exact value of \(m\).
\(\displaystyle\int_0^t \left(x - \frac{x^3}{4}\right)\mathrm{d}x = \frac{t^2}{2} - \frac{t^4}{16}\).
(a) \(\mathrm{P}(X < 1) = \dfrac{1}{2} - \dfrac{1}{16} = \dfrac{7}{16}\).
(b) \(\dfrac{m^2}{2} - \dfrac{m^4}{16} = \dfrac{1}{2}\); multiply by \(16\): \(8m^2 - m^4 = 8\), so \(m^4 - 8m^2 + 8 = 0\).
This is a quadratic in \(m^2\): \(m^2 = \dfrac{8 \pm \sqrt{64 - 32}}{2} = 4 \pm 2\sqrt{2}\). Since \(0 \leq m \leq 2\), we need \(m^2 \leq 4\), so \(m^2 = 4 - 2\sqrt{2}\) and \(m = \sqrt{4 - 2\sqrt{2}}\) \(\;(\approx 1.08)\). See 19.2.
5 Paper 2 AHL \(X\) has p.d.f. \(f(x) = kx\,\mathrm{e}^{-x}\) for \(0 \leq x \leq 3\), and \(f(x) = 0\) otherwise.
(a) Find the value of \(k\).
(b) Find the mode of \(X\).
(c) Find the median of \(X\).
(a) GDC: \(\displaystyle\int_0^3 x\,\mathrm{e}^{-x}\,\mathrm{d}x = 0.80085\ldots\) (exactly \(1 - 4\mathrm{e}^{-3}\)), so \(k = \dfrac{1}{0.80085\ldots} = 1.25\) (3 s.f.).
(b) \(f'(x) = k\,\mathrm{e}^{-x}(1 - x) = 0\) gives \(x = 1\); \(f'\) changes from positive to negative there, so the mode is \(1\).
(c) Solve \(\displaystyle\int_0^m kx\,\mathrm{e}^{-x}\,\mathrm{d}x = 0.5\) with nSolve: \(\;m = 1.38\) (3 s.f.). See 19.1 and 19.2.
6 Paper 1 The heights of plants are normally distributed with mean \(170\) mm and standard deviation \(5\) mm. Use the 68–95–99.7% rule to estimate
(a) \(\mathrm{P}(160 < H < 175)\)
(b) \(\mathrm{P}(H < 165)\).
(a) \(160 = \mu - 2\sigma\), \(175 = \mu + \sigma\): \(\;0.475 + 0.34 = 0.815\).
(b) \(165 = \mu - \sigma\): \(\;\dfrac{1 - 0.68}{2} = 0.16\). See 19.3.
7 Paper 2 The masses of newborn babies in a hospital are normally distributed with mean \(3.2\) kg and standard deviation \(0.4\) kg.
(a) Find the probability that a baby has a mass less than \(2.5\) kg.
(b) Find the probability that a baby has a mass between \(2.8\) kg and \(4.0\) kg.
(c) The heaviest \(5\%\) of babies have a mass of more than \(w\) kg. Find \(w\).
(a) normCdf(-1E99,2.5,3.2,0.4) \(= 0.0401\) (3 s.f.).
(b) normCdf(2.8,4,3.2,0.4) \(= 0.819\) (3 s.f.).
(c) \(\mathrm{P}(M < w) = 0.95\): \(\;w = \mathtt{invNorm}(0.95, 3.2, 0.4) = 3.86\) kg (3 s.f.). See 19.3.
8 Paper 2 \(X \sim \mathrm{N}(40, 36)\). Find
(a) the \(z\)-value of \(x = 49\),
(b) \(k\) such that \(\mathrm{P}(40 - k < X < 40 + k) = 0.8\).
9 Paper 2 The lengths of fish in a lake are normally distributed with mean \(\mu\) cm and standard deviation \(\sigma\) cm. \(20\%\) of the fish are shorter than \(6.2\) cm, and \(25\%\) are longer than \(8.0\) cm. Find \(\mu\) and \(\sigma\).
\(\mathtt{invNorm}(0.2) = -0.84162\ldots\) and \(\mathtt{invNorm}(0.75) = 0.67449\ldots\)
\(\mu - 0.84162\sigma = 6.2\) and \(\mu + 0.67449\sigma = 8.0\). Subtracting: \(1.51611\sigma = 1.8\).
\(\sigma = 1.19\) cm and \(\mu = 7.20\) cm (3 s.f.). See 19.4.
10 Paper 2 The masses of apples are normally distributed with mean \(\mu\) g and standard deviation \(8\) g. \(15\%\) of the apples have a mass greater than \(120\) g.
(a) Find \(\mu\).
(b) \(8\) apples are chosen at random. Find the probability that at most one of them has a mass greater than \(120\) g.
(a) \(\mathrm{P}(A < 120) = 0.85\): \(\;\dfrac{120 - \mu}{8} = \mathtt{invNorm}(0.85) = 1.03643\ldots\), so \(\mu = 120 - 8.2915 = 112\) g (3 s.f.).
(b) \(Y \sim \mathrm{B}(8, 0.15)\): \(\;\mathrm{P}(Y \leq 1) = \mathtt{binomCdf}(8, 0.15, 0, 1) = 0.657\) (3 s.f.). See 19.4.