IB Maths AA HL 4 Statistics and Probability Questions

Practise IB Maths AA HL statistics and probability through shared-core and HL distributions, inference, regression and probability models with clear interpretation.

Exam points

analyse sampling methods, bias, missing data, outliers and representativeness for a target population
construct and interpret frequency tables, histograms, cumulative-frequency graphs and box plots using quartiles and IQR
calculate and interpret mean, variance, standard deviation and effects of transforming all data values
use scatter diagrams, correlation and the correct regression direction to make cautious predictions
calculate probabilities with sample spaces, Venn/tree diagrams, conditional independence, binomial or normal models
apply Bayes’ theorem and continuous probability-density methods, including expectation, variance and transformations

Covered syllabus points

4.1 Statistics and probability - SL content 4.2 Statistics and probability - AHL content

[Maximum number: 6]

A supermarket analyses the shopping habits of its customers.
The number of times, X, each customer visits the supermarket in a week is given by the following probability distribution.

(a)

Find the value of a.

[ 2 ]

(b)

Write down the mode of X.

[ 1 ]

(c)

Find the mean of X.

The manager wants to know why customers come to their supermarket. They survey the first 50 customers to arrive at the supermarket on a particular day.

[ 2 ]

(d)

Identify which one of the following best describes the manager's sampling method. Circle your answer.

Simple random / Systematic / Convenience / Quota / Stratified

[ 1 ]

[Maximum number: 6]

A new test has been developed to identify whether a particular gene, G, is present in a population of parrots. The test returns a correct positive result 95 % of the time for parrots with the gene, and a false positive result 2 % of the time for parrots without the gene.
The proportion of the population with the gene is p.

(a)

Complete the tree diagram below.

Figure for Question (a) — IB Maths AA HL

[ 2 ]

award A1 for branches correctly labelled 0.95 and 0.05
award A0 for $G^{\prime}$ branch labelled $p^{\prime}$
award A0 for $G^{\prime}$ branch labelled q unless explicitly defined as 1-p

Figure for Answer to Question (a) — IB Maths AA HL

Note: award A1 for branch correctly labelled 1-p

(b)

A random sample of the population was tested. It was found that 150 tests returned a positive result. Out of the 150 parrots with a positive test result, 18 did not actually have the gene. Find an estimate for p.

[ 4 ]

METHOD 1 recognizing conditional probability
$\mathrm{P}\left(G^{\prime} \mid p o s\right)$ OR $\mathrm{P}(G \mid p o s)$ $\frac{0.02(1-p)}{0.95 p+0.02(1-p)}\left(=\frac{18}{150}\right)$ OR $\frac{0.95 p}{0.95 p+0.02(1-p)}\left(=\frac{132}{150}\right)$

Note: Award A1 for a correct numerator and A1 for a correct denominator.
p=0.133738
p=0.134

METHOD 2
attempt to set up a system of equations ( S= sample size)
p(0.95 S)=132 and (1-p)(0.02 S)=18
attempt to solve for p or S
$\frac{0.95 p}{0.02(1-p)}=\frac{132}{18}$
OR $S=p S+(1-p) S=\frac{132}{0.95}+\frac{18}{0.02}=138.947 \ldots+900=1038.94 \ldots$ $p=0.133738 \ldots$
p=0.134

METHOD 3
attempt to find the number of parrots with the gene and the number without number of parrots with the gene $\approx \frac{132}{0.95}=138.947 .$ . AND
number of parrots without the gene $\approx \frac{18}{0.02}=900$
number of parrots in the sample $\approx 138.947 \ldots+900=1038.94 \ldots$
attempt to find proportion of sample with the gene
$p \approx \frac{138.947 \ldots}{1038.94 \ldots}=0.133738 \ldots$
p=0.134

[Maximum number: 15]

The continuous random variable X has a probability density function given by

f(x)=\left\{\begin{array}{cc} k \sin \left(\frac{\pi x}{6}\right), & 0 \leq x \leq 6 \\ 0, & \text { otherwise } \end{array}\right.

(a)

Find the value of k.

[ 4 ]

attempt to equate integral to 1 (may appear later)
$k \int_{0}^{6} \sin \left(\frac{\pi x}{6}\right) \mathrm{d} x=1$
correct integral
$k\left[-\frac{6}{\pi} \cos \left(\frac{\pi x}{6}\right)\right]_{0}^{6}=1$
substituting limits
$-\frac{6}{\pi}(-1-1)=\frac{1}{k}$ $k=\frac{\pi}{12}$

(b)

By considering the graph of f write down

[ 3 ]

(i)

the mean of X;

[ 1 ]

(ii)

the median of X;

[ 1 ]

(iii)

the mode of X.

[ 1 ]

(c)

Show that $P(0 \leq X \leq 2)=\frac{1}{4}$ .

[ 4 ]

$\frac{\pi}{12} \int_{0}^{2} \sin \left(\frac{\pi x}{6}\right) \mathrm{d} x$

=\frac{\pi}{12}\left[-\frac{6}{\pi} \cos \left(\frac{\pi x}{6}\right)\right]_{0}^{2}

Note: Accept without the $\frac{\pi}{12}$ at this stage if it is added later.

\begin{aligned} & \frac{\pi}{12}\left[-\frac{6}{\pi}\left(\cos \frac{\pi}{3}-1\right)\right] \\ & =\frac{1}{4} \end{aligned}

(d)

Hence state the interquartile range of X.

[ 2 ]

from (c)(i) $Q_{1}=2$
as the graph is symmetrical about the middle value $x=3 \Rightarrow Q_{3}=4$
so interquartile range is
4-2
=2

(e)

Calculate $P(X \leq 4 \mid X \geq 3)$ .

[ 2 ]

$\quad P(X \leq 4 \mid X \geq 3)=\frac{P(3 \leq X \leq 4)}{P(X \geq 3)}$ $=\frac{\frac{1}{4}}{\frac{1}{2}}$ $=\frac{1}{2}$

All question bank results loaded

Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)

Question (d)

Question 2

Question (a)

Question (b)

Question 3

Question (a)

Question (b)

Question (i)

Question (ii)

Question (iii)

Question (c)

Question (d)

Question (e)