IB Maths AI HL 3.6 Voronoi diagrams

[Maximum number: 12]

This question considers how the assessment of the Air Quality Index (AQI) for a school depends on the method chosen by the person doing the assessing.
Air quality for a district is measured at three monitoring stations. The positions of these stations on a coordinate system with units in kilometres are A(0,5), B(8,9) and C(8,1).
A Voronoi diagram is constructed with the three stations as sites.

Figure for Question SL 3.6—Voronoi diagrams question 1 — IB Maths AI HL

(a)

Given that the equation of the perpendicular bisector of $[\mathrm{AB}]$ is y=15-2 x, find the coordinates of vertex V.

A school, S, is situated in the district at the point with coordinates (5,6).

[ 2 ]

equating the correct lines
5=15-2 x OR sketch graph

x=5 \quad y=5 \quad \text { OR }(5,5)

(b)

State which station is closest to the school.

The principal of the school is concerned about the air quality in the area. Air quality is measured by the Air Quality Index (AQI). In this district, values less than 50 are taken to indicate good air quality.

The principal contacts the local environmental agency requesting an AQI value for her school. They tell her the mean AQI reading from the closest station to the school.

[ 1 ]

(c)

Write down the type of interpolation being used by the environmental agency.

The principal obtains the mean AQI value from each of the three stations.

[ 1 ]

(d)

Explain why the principal might not accept that the air quality around the school can be classed as "good".

The principal decides to obtain an expected value for the AQI at the school that uses all the available information. To do this, she uses an alternative method: the natural neighbour algorithm. This algorithm has two stages.

The first stage is to create a new Voronoi diagram with the school as an extra site. This is shown in the following diagram with the edges of the previous diagram shown by dashed lines.

Figure for Question (d) — IB Maths AI HL

The second stage is to estimate the AQI value at the school, W, by using the formula

W=\frac{w_{A} a_{A}+w_{B} a_{B}+w_{C} a_{C}}{T}

In the formula, $a_{A}$ is the area within the new cell that has been taken from the cell surrounding site A , shown as region P on the diagram, and $w_{A}$ is the mean AQI value from site A . This is given as 132 in the table of mean AQI values above. Similarly for sites B and C . T is the total area of the new cell around S .

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(e)

Hence find $a_{B}$ (the area of region Q on the diagram).

The areas of regions P and R are $a_{A}=13.7 \mathrm{~km}^{2}$ and $a_{C}=6.9 \mathrm{~km}^{2}$ respectively.

[ 5 ]

attempt to find where this line meets the other two lines
(9,5)
14-x=15-2 x
(1,13)

Note: These two marks are un-implied due to the "hence" in the question. At least one needs to be seen for the (M1) to be awarded.

\begin{aligned} \text { Area } & =\frac{1}{2} \times 4 \times 8 \\ & =16\left(\mathrm{~km}^{2}\right) \end{aligned}

Note: Award (M1) for any valid method which is completed (for example cosine rule, followed by area of triangle formula or using one of the other sides as base and correctly finding the perpendicular height) which leads to an answer, even if the answer is not correct due to a calculation error. (the side lengths of the triangle are 4, 11.3, 8.94 and the angles $45^{\circ}, 18.4^{\circ}, 116.6^{\circ}$ )

Note: Award (M0)A0 for an unsupported answer of 16.

(f)

Use the natural neighbour algorithm to show that an estimate for the expected AQI value at the school, W, is 94.4.

The principal is still concerned that this method is underestimating the AQI value at the school, as the school is situated close to a busy traffic intersection. She decides to take her own readings ( x ) over a period of 60 days. Her results are summarized as

\bar{x}=97.8, s_{n-1}=17.2, n=60 .

The principal assumes that the daily AQI values at the school can be modelled by a normal distribution and that each value is independent of any other value.

[ 2 ]

attempt to substitute the values into given formula M1

\begin{aligned} & W=\frac{132 \times 13.7+49 \times 16+125 \times 6.9}{16+13.7+6.9} \\ & W=94.4(94.3961 \ldots) \end{aligned}

IB Maths AI HL 3.6 Voronoi diagrams Question Bank

Exam points

SL 3.6—Voronoi diagrams question 1

Question (a)

Question (b)

Question (c)

Question (d)

Question (e)

Question (f)