AHL 8.3 Paper 3 modelling: a logistic investigation · AI HL · AHL 8: Statistics 2

AHL 8.3

Paper 3 modelling: a logistic investigation

AHL 8.3

Paper 3 modelling: a logistic investigation

By the end of this section

Key points

How a Paper 3 investigation works

It is one long question with about ten parts that build on each other. Early parts find the model, middle parts test it, later parts use it.
A "show that" gives you the result. If you cannot show it, use the result you are given and carry on.
Store k and C on the calculator and use the stored values. Rounding k or C early can change the third significant figure of later answers.
Give answers to 3 significant figures unless told otherwise, and write the units: thousand households, engineers per year.

The logistic model

\(N=\dfrac{L}{1+Ce^{-kt}}\), with C and k positive. N grows slowly, then quickly, then levels off.
L is the carrying capacity: the limit of N as t becomes large. In context it is the most the quantity can reach. N is always less than L.
To use a data point, put t and N into the model, then rearrange: \(1+Ce^{-kt}=\dfrac{L}{N}\), so \(Ce^{-kt}=\dfrac{L}{N}-1\), so \(C=\left(\dfrac{L}{N}-1\right)e^{kt}\).
Two data points give two expressions for C. Put them equal and solve for k with the calculator, then find C.

Judging a model

\(SS_{res}=\sum(y-\hat y)^2\): the sum of the squared differences between the data and the model's predictions.
The error function \(E=\sqrt{SS_{res}/n}\), where n is the number of predictions made, is a typical size of the error. A smaller E is a better fit.
In a show-that, write the value of E and the comparison with the limit: "E = 0.899 < 1.0, so the model will be used".

Using the model

Read how t is measured. If t is years since the start of 2018, then the end of 2033 is t = 16. A value such as t = 8.73 lies in the 9th year: 2018 + 8 = 2026.
A percentage and a ratio are two multiplications. Percentage of N, then divide by the number each unit can serve. Keep N in thousands, or convert.
Average rate of change = change in the quantity ÷ change in time, with units.
To find when one model first exceeds another, graph both and find the intersection with the calculator. Check which intersection is first for the t values you are told to use.

On the calculator

Solving for k

Type the equation with k as the unknown: both expressions for C set equal.
Use the numerical solver (Solve). Give it a starting guess if it asks.
Store the answer to k as a variable so you can use its unrounded value.

Screenshot and video to come

Storing values and defining the model

Store k and C. Define \(N(t)=\dfrac{L}{1+Ce^{-kt}}\) as a function using the stored values.
Evaluate N(t) for each t you need. Round only your final answer.

Screenshot and video to come

A table of residuals

In one list put the data, in a second the model values, and in a third the residuals.
Square the residuals, add them to get \(SS_{res}\), then take the square root of \(SS_{res}\) divided by n.

Screenshot and video to come

Intersection of two graphs

Graph the logistic function and the straight line on the same axes.
Set a window that includes t from 5 to the end of the range.
Find the intersection points and take the first one for t ≥ 5.

Screenshot and video to come

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.

Setting up the model

The number of households in a country that have a heat pump, N thousand, is recorded at the end of each year. The time t is in years since the start of 2018, so t = 1 is the end of 2018. The number is modelled by \(N=\dfrac{840}{1+Ce^{-kt}}\), where C and k are positive constants. The table gives the data. (a) In context, explain what the value 840 means in the model. (b) Use the value of N at t = 2 to show that \(C=34e^{2k}\).

Working

Carrying capacity
As t becomes large, \(e^{-kt}\to 0\), so \(N\to 840\)

840 thousand households is the maximum number of households that will have a heat pump (the carrying capacity). N gets closer to 840 as t increases, but never reaches it.
Substitute
At t = 2, N = 24: \(24=\dfrac{840}{1+Ce^{-2k}}\)

Use the data point exactly as given.
Rearrange
\(1+Ce^{-2k}=840\div 24=35\), so \(Ce^{-2k}=34\)

Swap the denominator and N, then take away 1.
Finish
\(C=34e^{2k}\)

Multiply both sides by \(e^{2k}\). This is the result you were asked to show, and you can use it in later parts even if you got stuck here.

A second expression, then k and C

For the model \(N=\dfrac{840}{1+Ce^{-kt}}\), the data point at t = 2 gives \(C=34e^{2k}\). The data point at t = 7 is N = 240. (a) Find a second expression for C. (b) Equate the two expressions and find k. (c) Find C.

Working

Second expression
\(240=\dfrac{840}{1+Ce^{-7k}}\), so \(1+Ce^{-7k}=3.5\) and \(C=2.5e^{7k}\)

The same steps as before with the second data point.
Equate
\(34e^{2k}=2.5e^{7k}\)

Both expressions equal C, so they equal each other.
Solve for k
From the calculator: Solve \(34e^{2k}=2.5e^{7k}\) for k: k = 0.522

Use the equation solver. Store the unrounded k.
Find C
\(C=34e^{2k}\) = 96.6

Substitute the unrounded k into either expression.

A value from the fitted model

For the model \(N=\dfrac{840}{1+Ce^{-kt}}\) with C = 96.6 and k = 0.522, find the value the model gives at t = 5, correct to 1 decimal place.

Working

Substitute
\(N=\dfrac{840}{1+Ce^{-5k}}\)

t = 5. Use the stored values of C and k, not the rounded ones.
Evaluate
N(5) = 103.7

Give 1 decimal place, as the question asks. This is the missing value in the next example.

SS_res and the error function

The model's predictions for t = 3 to 6 are compared with the data in the table, with a = 103.7. The government will use the model if E < 1.0, where \(E=\sqrt{SS_{res}/n}\) and n is the number of predictions. (a) Find \(SS_{res}\). (b) By finding E, show that the model will be used.

Working

Residuals
Data − model at t = 3, 4, 5, 6: 0.3, −0.8, 1.3, 0.9

Use the 1 decimal place values shown.
SS_res
\(SS_{res}=\) \(0.09 + 0.64 + 1.69 + 0.81 = 3.23\)

Square each residual, then add.
E
\(E=\sqrt{3.23\div 4}\) = 0.899

n = 4 predictions.
Compare
0.899 < 1.0, so the model will be used

Write the value of E and the comparison. That finishes a show-that.

Predicting further ahead

For the model \(N=\dfrac{840}{1+Ce^{-kt}}\) with C = 96.6 and k = 0.522, where t is the number of years since the start of 2018, predict the number of households with a heat pump at the end of 2033. Give your answer to 3 significant figures.

Working

Find t
End of 2033 is 2033 − 2018 + 1 = 16 years after the start of 2018: t = 16

t = 1 is the end of 2018, so the end of 2033 is t = 16.
Evaluate
N(16) = 821

Use the stored C and k.
Check
840 is the carrying capacity

The prediction is just below 840, as expected for a logistic model late in its growth. It is beyond the data (t ≤ 8), so it depends on the model keeping the same shape.

Percentages and a ratio, then a rate

The model is \(N=\dfrac{840}{1+Ce^{-kt}}\) with C = 96.6 and k = 0.522. It is estimated that 12% of households with a heat pump need a service visit each year, and that one engineer can complete 150 service visits in a year. (a) Use the model to find an expression for the number of engineers, R, needed to service the heat pumps at time t. (b) At the end of 2022 there were 110 qualified engineers, and at the end of 2024 there were 230. Find the average number of engineers added per year between the end of 2022 and the end of 2024.

Working

Visits per year
12% of the households need a visit each year: 0.12 × 1000N = 120N visits

N is in thousands of households, so multiply by 1000.
Engineers
One engineer does 150 visits: R = 120N ÷ 150 = 0.8N

Divide the number of visits by the visits one engineer can do.
The expression
\(R=0.8\times\dfrac{840}{1+96.6e^{-0.522t}}=\dfrac{672}{1+96.6e^{-0.522t}}\)

Replace N by the model.
Average rate
\(\dfrac{230-110}{2}\) = 60 engineers per year

Change in the number of engineers divided by the change in time. Give the units.

When supply first falls short

Demand is \(R=\dfrac{672}{1+96.6e^{-0.522t}}\) engineers. Supply is 110 engineers at t = 5 and increases by 60 engineers per year. Find the value of t, where t ≥ 5, at which the number of engineers first becomes insufficient, and the year in which this happens.

Working

Supply
\(S=110+60(t-5)=60t-190\)

A straight line through (5, 110) with gradient 60.
Graph both
From the calculator: graph R and S. First intersection with t ≥ 5: t = 8.73

Choose a window that shows t from 0 to 16 and R up to 700.
Read the graph
Before t = 8.73, supply is above demand: R < S

At t = 5, R = 82.9 is below S = 110. Demand first passes supply at P.
The year
t = 8.73: 8 complete years after the start of 2018, so in 2026

t = 8.73 is part way through the ninth year, which is 2026.

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Statistics 2

Paper 3 modelling: a logistic investigation

By the end of this section

Key points

How a Paper 3 investigation works

The logistic model

Judging a model

Using the model

On the calculator

Solving for k

Storing values and defining the model

A table of residuals

Intersection of two graphs

Worked examples

Setting up the model

Working

A second expression, then k and C

Working

A value from the fitted model

Working

SS_res and the error function

Working

Predicting further ahead

Working

Percentages and a ratio, then a rate

Working

When supply first falls short

Working

Choose a skill

Quiz