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IB Maths AA HL – Topic 4: Statistics and Probability

Bayes’ theorem, continuous random variables, variance and expectation.

12 questions  ·  87 marks  ·  answers included  ·  IB AA HL
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This worksheet covers:

  • HL statistics and probability
  • Bayes’ theorem
  • Linear combinations of random variables
  • Continuous random variables (pdf, cdf, mode, median, expectation and variance)

Section A is non-calculator; a GDC is expected for Sections B and C. Allow about 90 minutes.

Questions

Section A – Skills

Non-calculator
1

A factory has two machines. Machine A makes $40\%$ of the output and $5\%$ of its items are defective; machine B makes the rest and $2\%$ of its items are defective. An item is chosen at random.(a) Find the probability that the item is defective.(b) Given that the item is defective, find the probability that it was made by machine A.

Show answer
(a) $0.4(0.05) + 0.6(0.02) = 0.032$(b) $\dfrac{0.02}{0.032} = \dfrac{5}{8} = 0.625$
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2

A discrete random variable $X$ has probability distribution $P(X = x) = kx$ for $x = 1, 2, 3, 4$.(a) Show that $k = \dfrac{1}{10}$.(b) Find $E(X)$ and $\text{Var}(X)$.

Show answer
(a) $k(1+2+3+4) = 1 \Rightarrow k = \dfrac{1}{10}$(b) $E(X) = \dfrac{1+4+9+16}{10} = 3$; $E(X^2) = \dfrac{1+8+27+64}{10} = 10$; $\text{Var}(X) = 10 – 9 = 1$
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3

The continuous random variable $X$ has probability density function $f(x) = kx(4 – x)$ for $0 \le x \le 4$, and $f(x) = 0$ otherwise.(a) Show that $k = \dfrac{3}{32}$.(b) Write down the median of $X$.(c) Find $\text{Var}(X)$.

Show answer
(a) $\displaystyle\int_0^4 x(4-x)\,dx = \left[2x^2 – \tfrac{x^3}{3}\right]_0^4 = \dfrac{32}{3}$, so $k = \dfrac{3}{32}$(b) By symmetry, median $= 2$ (and $E(X) = 2$)(c) $E(X^2) = \dfrac{3}{32}\displaystyle\int_0^4 x^3(4-x)\,dx = \dfrac{3}{32}\cdot\dfrac{256}{5} = \dfrac{24}{5}$; $\text{Var}(X) = \dfrac{24}{5} – 4 = \dfrac{4}{5} = 0.8$
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4

$X$ and $Y$ are independent random variables with $E(X) = 10$, $\text{Var}(X) = 4$, $E(Y) = 6$ and $\text{Var}(Y) = 9$. Find(a) $E(2X – 3Y)$(b) $\text{Var}(2X – 3Y)$(c) $\text{Var}(X_1 + X_2 + X_3)$, where $X_1, X_2, X_3$ are three independent observations of $X$.

Show answer
(a) $20 – 18 = 2$(b) $4(4) + 9(9) = 97$(c) $3 \times 4 = 12$
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5

The cumulative distribution function of a continuous random variable $X$ is $F(x) = \dfrac{x^2}{9}$ for $0 \le x \le 3$, with $F(x) = 0$ for $x < 0$ and $F(x) = 1$ for $x > 3$.(a) Find the probability density function $f(x)$.(b) Find $P(1 < X < 2)$.(c) Find $E(X)$.

Show answer
(a) $f(x) = \dfrac{2x}{9}$ for $0 \le x \le 3$, $0$ otherwise(b) $F(2) – F(1) = \dfrac{4}{9} – \dfrac{1}{9} = \dfrac{1}{3}$(c) $\displaystyle\int_0^3 \dfrac{2x^2}{9}\,dx = \left[\dfrac{2x^3}{27}\right]_0^3 = 2$
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Section B – Exam-style

Calculator allowed
6

A screening test for a disease has sensitivity $95\%$ (probability of a positive result given the disease is present) and specificity $90\%$ (probability of a negative result given the disease is absent). In the population tested, $1\%$ of people have the disease.(a) Find the probability that a randomly chosen person tests positive.(b) Find the probability that a person who tests positive actually has the disease.(c) Comment on what your answer to (b) suggests about the usefulness of a single positive result.

Show answer
(a) $0.01(0.95) + 0.99(0.10) = 0.1085$(b) $\dfrac{0.0095}{0.1085} = 0.0876$ (3 s.f.)(c) Only about $9\%$ of positive results are true positives because the disease is rare; a positive result should be followed by further testing.
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7

The continuous random variable $X$ has probability density function $f(x) = \dfrac{k}{x}$ for $1 \le x \le e^2$, and $f(x) = 0$ otherwise.(a) Show that $k = \dfrac{1}{2}$.(b) Find the exact value of the median of $X$.(c) Show that $\text{Var}(X) = \dfrac{e^2 – 1}{2}$.

Show answer
(a) $\displaystyle\int_1^{e^2}\dfrac{k}{x}\,dx = k[\ln x]_1^{e^2} = 2k = 1$(b) $\dfrac{1}{2}\ln m = \dfrac{1}{2} \Rightarrow m = e$(c) $E(X) = \dfrac{1}{2}\displaystyle\int_1^{e^2} 1\,dx = \dfrac{e^2-1}{2}$; $E(X^2) = \dfrac{1}{2}\displaystyle\int_1^{e^2} x\,dx = \dfrac{e^4-1}{4}$; $\text{Var}(X) = \dfrac{e^4-1}{4} – \dfrac{(e^2-1)^2}{4} = \dfrac{2e^2-2}{4} = \dfrac{e^2-1}{2}$ ($\approx 3.19$)
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8

The random variable $X \sim B(n, p)$ has $E(X) = 12$ and $\text{Var}(X) = 9$.(a) Find the values of $n$ and $p$.(b) Find $P(X \ge 15)$.(c) Find $P(X \ge 15 \mid X \ge 10)$.

Show answer
(a) $np = 12$, $np(1-p) = 9 \Rightarrow 1 – p = 0.75$, $p = 0.25$, $n = 48$(b) $P(X \ge 15) = 1 – P(X \le 14) = 0.200$ (3 s.f.)(c) $\dfrac{P(X \ge 15)}{P(X \ge 10)} = \dfrac{0.1999}{0.7952} = 0.251$ (3 s.f.)
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9

Bags of sugar are filled by a machine so that the mass, $M$ grams, is normally distributed with mean $1005$ and standard deviation $4$. A bag is rejected if its mass is less than $1000\,\text{g}$.(a) Find the probability that a randomly chosen bag is rejected.(b) Ten bags are chosen at random. Find the probability that at least one is rejected.(c) Given that a bag is not rejected, find the probability that its mass is greater than $1010\,\text{g}$.

Show answer
(a) $P(Z < -1.25) = 0.106$ (3 s.f.)(b) $1 – (0.8944)^{10} = 0.673$ (3 s.f.)(c) $\dfrac{P(M > 1010)}{P(M \ge 1000)} = \dfrac{0.1056}{0.8944} = 0.118$ (3 s.f.)
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10

The continuous random variable $X$ has probability density function$f(x) = kx$ for $0 \le x \le 2$, $f(x) = k(4 – x)$ for $2 < x \le 4$, and $f(x) = 0$ otherwise.(a) Find the value of $k$.(b) Write down $E(X)$.(c) Find $P(X > 3)$.(d) Find $\text{Var}(X)$.

Show answer
(a) Total area of triangle $= \dfrac{1}{2}(4)(2k) = 4k = 1 \Rightarrow k = \dfrac{1}{4}$(b) $E(X) = 2$ (symmetry)(c) $\dfrac{1}{4}\displaystyle\int_3^4 (4 – x)\,dx = \dfrac{1}{8}$(d) $E(X^2) = \dfrac{1}{4}\left[\displaystyle\int_0^2 x^3\,dx + \int_2^4 x^2(4-x)\,dx\right] = \dfrac{1}{4}\left[4 + \dfrac{44}{3}\right] = \dfrac{14}{3}$; $\text{Var}(X) = \dfrac{14}{3} – 4 = \dfrac{2}{3}$
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Section C – Challenge

Calculator allowed
11

The continuous random variable $X$ has probability density function $f(x) = 12x^2(1 – x)$ for $0 \le x \le 1$, and $f(x) = 0$ otherwise.(a) Verify that $f$ is a valid probability density function.(b) Find the mode of $X$.(c) Find $E(X)$ and show that $\text{Var}(X) = \dfrac{1}{25}$.(d) Find $P(X < 0.5)$.(e) Explain, without further calculation, whether the median is greater than or less than the mean.

Show answer
(a) $f(x) \ge 0$ on $[0,1]$ and $12\left(\dfrac{1}{3} – \dfrac{1}{4}\right) = 1$(b) $f'(x) = 24x – 36x^2 = 0 \Rightarrow x = \dfrac{2}{3}$(c) $E(X) = 12\left(\dfrac{1}{4} – \dfrac{1}{5}\right) = \dfrac{3}{5}$; $E(X^2) = 12\left(\dfrac{1}{5} – \dfrac{1}{6}\right) = \dfrac{2}{5}$; $\text{Var}(X) = \dfrac{2}{5} – \dfrac{9}{25} = \dfrac{1}{25}$(d) $12\left(\dfrac{1}{24} – \dfrac{1}{64}\right) = \dfrac{5}{16}$(e) The distribution is negatively skewed (mode $\dfrac{2}{3} >$ mean $\dfrac{3}{5}$), so the median lies between them: median $>$ mean.
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12

In a company’s email system, $30\%$ of incoming messages are spam. The word ‘free’ appears in $60\%$ of spam messages and in $5\%$ of genuine messages.(a) Find the probability that a randomly chosen message contains the word ‘free’.(b) Find the probability that a message containing the word ‘free’ is spam.(c) Find the probability that a message which does not contain the word ‘free’ is spam.(d) The filter marks a message as spam if it contains the word ‘free’. Find the probability that a randomly chosen message is classified incorrectly.

Show answer
(a) $0.3(0.6) + 0.7(0.05) = 0.215$(b) $\dfrac{0.18}{0.215} = 0.837$ (3 s.f.)(c) $\dfrac{0.3(0.4)}{0.785} = \dfrac{0.12}{0.785} = 0.153$ (3 s.f.)(d) $P(\text{spam without ‘free’}) + P(\text{genuine with ‘free’}) = 0.12 + 0.035 = 0.155$
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