IB Maths AA HL 4.2 Statistics and Probability Ahl Content Questions

Practise IB Mathematics AA HL 4.2 by solving advanced probability, distributions, inference, hypothesis and statistical-modelling problems.

Exam points

HL syllabus concepts and methods
HL extended reasoning

Covered syllabus points

AHL 4.13 (HL)—Bayes' theorem AHL 4.14 (HL)—Random variables and probability density

[Maximum number: 4]

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 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.

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: 11]

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}

[Maximum number: 3]

5.
Two species of plant, A and B , are identical in appearance though it is known that the mean length of leaves from a plant of species A is 5.2 cm , whereas the mean length of leaves from a plant of species B is 4.6 cm . Both lengths can be modelled by normal distributions with standard deviation 1.2 cm .
In order to test whether a particular plant is from species A or species B, 16 leaves are collected at random from the plant. The length, x, of each leaf is measured and the mean length evaluated. A one-tailed test of the sample mean, $\bar{X}$ , is then performed at the 5 % level, with the hypotheses: $H_{0}: \mu=5.2$ and $H_{1}: \mu<5.2$ .

It is now known that in the area in which the plant was found $90 \%$ of all the plants are of species A and $10 \%$ are of species B.

If, having done the test, the sample mean is found to lie within the critical region, find the probability that the leaves came from a plant of species A .

Attempt to use the conditional probability formula:
$\frac{0.9\times0.05}{0.108875997\ldots}$ $=0.41334\ldots=0.413.$

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Covered syllabus points

Question 1

Question 2

Question (a)

Question (b)

Question (i)

Question (ii)

Question (iii)

Question (c)

Question 3