IB Maths AI SL · Paper 2

The centre tests every patient who comes on one day, whoever they are. This is convenience sampling.

Complete the tree. Of all people, 5% have the disease, so 95% don't. Of those with the disease, 99% test positive, so 1% test negative. Of those without it, 2% test positive and 98% test negative.

(i) No disease and positive:

(ii) Negative (from either branch):

(iii) Disease given negative:

The patients on that day are not a random sample of the population, so the proportion who have the disease may differ from 5%. Even with a random sample the observed numbers would vary around the predicted ones.

Draw a rectangle with three overlapping circles M, N and D. In the centre (all three) write 9. Then M and D only: 18. M and N only: 11. D only: 2. Using the totals: N and D only is $30-9-18-2=1$, M only is $78-9-18-11=40$, and N only is $45-9-11-1=24$.

Add every region of the diagram.

Use the cosine rule with the side opposite C being AB.

In triangle DCE use the sine rule: $\frac{DE}{\sin C}=\frac{DC}{\sin E}$.

The third angle of triangle DCE is $180^{\circ }-100^{\circ }-32.0^{\circ }=48.0^{\circ }$, at D. Use $\frac{1}{2}ab\sin C$ with the two known sides DE and DC.

D lies on AC, so $AD=1005-210=795$. The angle $\angle ADF=180^{\circ }-48.0^{\circ }$ has the same sine as $48.0^{\circ }$. Set the area of ADF equal to the area from part (c).

A 1.2% increase multiplies by 1.012.

From 2019 to 2025 there are 7 years. Add the yearly averages with the geometric sum, then multiply by 50 concerts.

Sketch a bell-shaped curve centred on the mean 6.1 kg. Mark the standard deviation 0.5 kg on the axis, at 5.6 and 6.6.

With $X\sim \mathrm{N}(6.1,0.5^2)$:

The proportion below 5.3 kg times 80:

If 12 of 80 weigh more than $x$, the proportion is $\frac{12}{80}=0.15$, so $\mathrm{P}(X>x)=0.15$. Use the inverse normal.

The probability that a cat weighs over 6.25 kg is $\mathrm{P}(X>6.25)\approx 0.3821$. Among ten cats the number is $\mathrm{B}(10,0.3821)$.

A regular hexagon is made of 6 equilateral triangles with side $x$, each of area $\frac{1}{2}{x}^2\sin 60^{\circ }$.

The graph of $V$ is a cubic that starts at the origin, rises to a maximum near $x\approx 8.8$, then falls and crosses the $x$-axis near $x\approx 15.2$, within $0\le x\le 16$.

Differentiate term by term.

The maximum is where the derivative is zero.

Substitute into the volume formula.

Spheres cannot fill a box completely: there are gaps between them. The box's volume also uses its outside measurements, so the inside is a little smaller. So dividing the volumes does not give the exact number of chocolates.