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IB Maths AI HL – Topic 4: Statistics and Probability

Poisson distribution, transition matrices, hypothesis tests, confidence intervals.

12 questions  ·  47 marks  ·  answers included  ·  IB Mathematics: Applications & Interpretation HL
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This worksheet covers:

  • The Poisson distribution
  • Transition matrices
  • Hypothesis testing
  • Confidence intervals from IB AI HL Topic 4

A GDC is required throughout; give answers to three significant figures unless stated otherwise.

Questions

Section A – Poisson Distribution

Calculator (GDC) required
1

The number of emails, $X$, received by an office in a one-hour period follows a Poisson distribution with mean $3.2$.(a) Find $P(X = 5)$.(b) Find $P(X \leq 2)$.(c) State $\text{Var}(X)$.

Show answer
(a) $P(X=5) = 0.114$.(b) $P(X \leq 2) = 0.380$.(c) $\text{Var}(X) = 3.2$ (mean = variance for a Poisson distribution).
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2

Calls to a helpline arrive independently at a mean rate of $2.1$ calls per $10$-minute period, and can be modelled by a Poisson distribution.Find the probability that at least one call is received in a randomly chosen $10$-minute period.

Show answer
$P(X \geq 1) = 1 – P(X=0) = 1 – e^{-2.1} = 0.878$.
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3

Let $X \sim \text{Po}(3.2)$ and $Y \sim \text{Po}(2.1)$ be independent random variables, representing the number of emails and calls received in an hour respectively. Let $T = X + Y$.(a) State the distribution of $T$, including its parameter.(b) Find $P(T = 4)$.

Show answer
(a) $T \sim \text{Po}(5.3)$.(b) $P(T=4) = 0.164$.
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4

The number of flaws, $W$, in a metre of fabric follows a Poisson distribution with mean $m$. Given that $P(W = 0) = 0.0498$, find the value of $m$.

Show answer
$e^{-m} = 0.0498 \Rightarrow m = -\ln(0.0498) = 3.00$.
[3]

Section B – Confidence Intervals and Hypothesis Testing

Calculator (GDC) required
5

A random sample of $40$ students sat a test. The sample mean score was $68$ and the sample standard deviation was $12$.Assuming the sample size is large enough to use a normal approximation, construct a $95\%$ confidence interval for the population mean score, giving the endpoints correct to one decimal place.

Show answer
$68 \pm 1.96 \times \dfrac{12}{\sqrt{40}} = 68 \pm 3.72$, giving the interval $(64.3,\ 71.7)$.
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6

A random sample of $15$ observations from a normal population has mean $52$ and standard deviation $6$. It is claimed the population mean is $48$. Test, at the $5\%$ significance level, whether the sample provides evidence that the population mean exceeds $48$.State your hypotheses, calculate the test statistic, and state your conclusion (critical value $t_{14} = 1.761$ at the $5\%$ level, one-tailed).

Show answer
$H_0: \mu = 48$, $H_1: \mu > 48$. Test statistic $t = \dfrac{52-48}{6/\sqrt{15}} = 2.58$. Since $2.58 > 1.761$, reject $H_0$: there is significant evidence at the $5\%$ level that the population mean exceeds $48$.
[5]
7

Explain what is meant by:(a) a Type I error,(b) a Type II error,in the context of a hypothesis test.

Show answer
(a) A Type I error is rejecting $H_0$ when $H_0$ is actually true.(b) A Type II error is failing to reject $H_0$ when $H_0$ is actually false.
[2]
8

A fair six-sided die is rolled $60$ times, so the expected frequency for each face is $10$. The observed frequencies for faces $1$ to $6$ are $4, 12, 9, 15, 11, 9$.Carry out a $\chi^2$ goodness-of-fit test at the $5\%$ significance level to determine whether the die is fair. The critical value is $\chi^2_5 = 11.07$.

Show answer
$H_0$: the die is fair; $H_1$: the die is not fair. $\chi^2_{\text{calc}} = \sum \dfrac{(O-E)^2}{E} = 6.80$. Since $6.80 < 11.07$, do not reject $H_0$: there is insufficient evidence that the die is biased.
[5]
9

A researcher wants to estimate a population mean, given that the population standard deviation is known to be $8$. Find the minimum sample size required so that a $95\%$ confidence interval for the mean has a margin of error of at most $2$.

Show answer
$n \geq \left(\dfrac{1.96 \times 8}{2}\right)^2 = 61.5$, so the minimum sample size is $n = 62$.
[4]

Section C – Transition Matrices

Calculator (GDC) required
10

In a simplified weather model, if a day is sunny the probability the next day is sunny is $0.7$ (otherwise rainy); if a day is rainy the probability the next day is sunny is $0.4$ (otherwise rainy). Today is sunny.(a) Write down the transition matrix $T$ for this Markov chain, using the state vector $\begin{pmatrix} P(\text{sunny}) \\ P(\text{rainy}) \end{pmatrix}$.(b) Find the probability that it is sunny two days from today.

Show answer
(a) $T = \begin{pmatrix} 0.7 & 0.4 \\ 0.3 & 0.6 \end{pmatrix}$ acting on the state column vector.(b) $P(\text{sunny in 2 days}) = 0.61$.
[5]
11

Using the weather model above, find the long-run (steady-state) probability that a day is sunny.

Show answer
Solving $\pi_1 = 0.7\pi_1 + 0.4\pi_2$ with $\pi_1 + \pi_2 = 1$ gives $\pi_1 = \dfrac{4}{7} = 0.571$ and $\pi_2 = \dfrac{3}{7} = 0.429$. Long-run $P(\text{sunny}) = \dfrac{4}{7} \approx 0.571$.
[4]
12

Explain, in the context of the weather model, what is meant by a steady-state (equilibrium) probability vector, and state one condition on the transition matrix needed for a unique steady-state to exist.

Show answer
A steady-state vector $\pi$ satisfies $\pi T = \pi$: once reached, the probabilities of being in each state no longer change from one step to the next, regardless of the initial state. A unique steady state exists provided the transition matrix is regular (some power of $T$ has all entries strictly positive).
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