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MST-12-S2-05 ~35 min ⚡ +90 XP available

Capture-Recapture and the Cost of Electricity

Two very different questions, one idea. A proportion you can measure stands in for a whole you cannot, whether that whole is a lake full of fish or a quarterly power bill.

Today's hook, Nobody can count every fish in a lake. A researcher tags $40$, waits a week, then catches $50$ and finds $8$ of them tagged. How many fish are in the lake?

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Orient and prepare

Capture your first estimate and preview the lesson language.

Worksheets

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Three printable worksheets that build from foundations to mastery, or build your own from any question in this focus area.

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Think first, your gut answer

+5 XP warm-up

A researcher tags $40$ fish and releases them. A week later she catches $50$ fish, and $8$ of them are tagged.

Without calculating write down your estimate of how many fish are in the lake, and say what the $8$ out of $50$ is telling you.

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What you will be able to do

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Estimate the size of a population using the capture-recapture method
State the assumptions the method depends on, and say what happens when one fails
Convert between watts and kilowatts
Calculate energy used in kilowatt hours from a power rating and a time
Work out what an appliance costs to run, and compare two appliances

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Preview the vocabulary

Review the essential terms before using them in the worked methods.

Key terms

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Capture-recapture: A way of estimating a population you cannot count, by marking some, releasing them, then seeing what fraction of a later sample carries a mark. Like this: if $8$ of $50$ caught are marked, then marked animals are about one sixth of the population.
Power, in watts: How fast an appliance uses energy, at every instant it is switched on. Like this: a $2400\text{ W}$ heater draws energy twice as fast as a $1200\text{ W}$ one, whether it runs for a minute or a month.
Kilowatt: One thousand watts, the unit electricity bills actually use. Like this: a $2400\text{ W}$ heater is a $2.4\text{ kW}$ heater, found by dividing by $1000$.
Kilowatt hour: A unit of energy, not power: the energy a $1\text{ kW}$ appliance uses in one hour. Like this: a $2\text{ kW}$ heater running for $3$ hours uses $2 \times 3 = 6\text{ kWh}$, and that is what you are billed for.

Estimate with capture-recapture

Set up the matching proportion and interpret the population estimate.

Capture-recapture, a proportion in disguise

+10 XP

You cannot count every fish in a lake, but you can count the ones in your net. The method rests on a single idea: the fraction marked in your sample should match the fraction marked in the whole population.

Write that as two equal fractions. If $n_1$ were marked altogether, $n_2$ were caught the second time and $m_2$ of those carried a mark, then $\frac{m_2}{n_2} = \frac{n_1}{N}$. Rearranging for the unknown population $N$ gives the formula.

$N \approx \dfrac{n_1 \times n_2}{m_2}$

Worked example 1, the hook

$n_1 = 40$, $\;n_2 = 50$, $\;m_2 = 8$

Marked first, caught second, and how many of the second were marked.

$\dfrac{8}{50} = \dfrac{40}{N}$

The sample is $16\%$ marked, so the lake should be about $16\%$ marked too.

$N \approx \dfrac{40 \times 50}{8} = \dfrac{2000}{8}$

The recaptured number $m_2$ is always the denominator.

$N \approx 250$ fish

Check: $40$ out of $250$ is $0.16$, and $8$ out of $50$ is also $0.16$.

The assumptions matter, and they are examinable. The method only works if the marked animals mix back in evenly, if being marked does not change how easily an animal is caught, and if the population does not change between the two samples through births, deaths or migration.

Watch what happens when an assumption fails. If marked animals become easier to catch, then $m_2$ comes out too big, and because $m_2$ is the denominator the estimate $N$ comes out too small. Being able to say which way the error runs is worth a mark.

Worked example 2, possums

$120$ tagged, later $90$ caught, $15$ tagged

Same three numbers, different animal.

$N \approx \dfrac{120 \times 90}{15} = \dfrac{10\,800}{15}$

Substitute, keeping $m_2 = 15$ underneath.

$N \approx 720$ possums

Check: $15 \div 90 = 0.1\dot{6}$ and $120 \div 720 = 0.1\dot{6}$.

In your book. Write the two equal fractions above the formula, so you can rebuild it if you forget it. Underneath, list the three assumptions in your own words.

Quick check. $30$ birds are banded. Later $45$ are caught and $9$ carry a band. Estimate the population.

Connect power and energy

Convert watts to kilowatts and calculate kilowatt hours.

Watts, kilowatts and kilowatt hours

+10 XP

Appliances are labelled in watts, but electricity is billed in kilowatt hours. Getting between the two is two steps, and confusing power with energy is what costs marks.

Power is how fast energy is used, and it does not depend on time. Energy is power multiplied by how long it ran. A $2\text{ kW}$ heater is always $2\text{ kW}$, whether it runs for one hour or ten.

So convert the label to kilowatts first, then multiply by the hours, then multiply by the price. Three multiplications, in that order, and the units tell you when each one is done.

$\text{kW} = \dfrac{\text{W}}{1000} \qquad \text{kWh} = \text{kW} \times \text{hours}$

Worked example 3, running a heater

A $2400\text{ W}$ heater runs for $5$ hours. Electricity costs $\$0.32$ per kWh.

Three quantities: a power, a time and a price.

$2400 \div 1000 = 2.4\text{ kW}$

Convert to kilowatts first. Bills never use watts.

$2.4 \times 5 = 12\text{ kWh}$

Power times hours gives energy. This is the number you are billed on.

$12 \times 0.32 = \$3.84$

Check: $3.84 \div 0.32 = 12$, the energy used.

Read the time carefully. Questions give hours per day and a number of days. A $2.4\text{ kW}$ heater run $5$ hours a day for $30$ days uses $2.4 \times 5 \times 30 = 360\text{ kWh}$, not $12\text{ kWh}$.

Worked example 4, lots of small things

Ten $60\text{ W}$ bulbs run for $8$ hours at $\$0.30$ per kWh

Add the power first, then convert once.

$10 \times 60 = 600\text{ W} = 0.6\text{ kW}$

Small appliances give a power under one kilowatt, which is fine.

$0.6 \times 8 = 4.8\text{ kWh}$

Energy, in the unit the bill uses.

$4.8 \times 0.30 = \$1.44$

Check: $1.44 \div 0.30 = 4.8$.

In your book. Write "power does not include time, energy does" and beside it the chain: watts, then kilowatts, then kilowatt hours, then dollars.

Did you get this? True or false: one kilowatt hour is the energy used by a $1\text{ kW}$ appliance running for one hour.

Compare electricity costs

Read appliance information and calculate practical running costs.

Comparing appliances and reading a bill

+10 XP

The interesting questions ask you to compare. A more powerful appliance is not always the more expensive one to run, because the time it is switched on matters just as much as the rating on its label.

Take a $2.4\text{ kW}$ heater used $5$ hours a day against a $1.8\text{ kW}$ heater used $7$ hours a day. The first uses $12\text{ kWh}$ a day and the second uses $12.6\text{ kWh}$, so the less powerful heater costs more to run.

Convert before multiplying

Multiplying watts by hours gives watt hours, which is a thousand times too big. Divide by $1000$ first.

Per day or in total?

Decide whether the question wants one day, one quarter or a year, before you reach for the price.

Round money at the end

Keep full accuracy through the working and round to the nearest cent only in the final line.

In your book. Work the two heaters right through to a monthly cost at $\$0.29$ per kWh, and write one sentence on why the smaller heater wins.

Fill the gaps. A $2\text{ kW}$ heater runs $3$ hours a day for $20$ days, and electricity costs $\$0.30$ per kWh. Energy per day $=$ kWh, total energy $=$ kWh, total cost $= \$$.

Show what you can do

Answer the exam-style questions, then compare your working with the model answers.

Quick-check drill

work before revealing answers

Choose an option for each fixed drill question, then reveal the concept-labelled explanations below.

Drill 1. $60$ fish are tagged and released. Later $80$ are caught and $12$ are tagged. Estimate the population.

$400$
$300$
$480$
$144$

Drill 2. An appliance is rated at $3500\text{ W}$. What is its power in kilowatts?

$3.5\text{ kW}$
$35\text{ kW}$
$0.35\text{ kW}$
$350\text{ kW}$

Drill 3. A $1.5\text{ kW}$ appliance runs for $4$ hours. How much energy does it use?

$6\text{ kWh}$
$600\text{ kWh}$
$0.375\text{ kWh}$
$60\text{ kWh}$

Drill 4. A $2\text{ kW}$ heater runs for $3$ hours. Electricity costs $\$0.28$ per kWh. What is the cost?

$\$1.68$
$\$0.84$
$\$16.80$
$\$0.56$

Drill 5. Which assumption does the capture-recapture method depend on?

The marked animals mix evenly back into the population and the population does not change between samples
Every animal in the population is caught at least once
The second sample is larger than the first
The marked animals stay together in one group

Short answer

ApplyBand 43 marks

SA 1. A ranger tags $45$ koalas in a reserve. Three months later $60$ koalas are surveyed and $9$ of them carry a tag. (a) Estimate the koala population of the reserve. (2 marks) (b) State one assumption the estimate depends on. (1 mark)

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ApplyBand 44 marks

SA 2. An air conditioner is rated at $2200\text{ W}$. (a) Write its power in kilowatts. (1 mark) (b) It runs for $6$ hours a day for $30$ days. Find the energy used, in kilowatt hours. (2 marks) (c) Electricity costs $\$0.31$ per kWh. Find the cost of running it for those $30$ days. (1 mark)

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AnalyseBand 54 marks

SA 3. Heater A is rated at $2.4\text{ kW}$ and runs for $5$ hours a day. Heater B is rated at $1.8\text{ kW}$ and runs for $7$ hours a day. Electricity costs $\$0.29$ per kWh. Over $30$ days, which heater costs more to run, and by how much? (4 marks)

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📖 Comprehensive answers (click to reveal)

Capture-recapture estimate. $N \approx \tfrac{60 \times 80}{12} = \tfrac{4800}{12} = 400$. Check: $12$ out of $80$ is $0.15$, and $60$ out of $400$ is also $0.15$. Option D multiplies by $12$ instead of dividing.

Watts to kilowatts. $3500 \div 1000 = 3.5\text{ kW}$. A kilowatt is a thousand watts, so the number gets smaller, which rules out option B.

Energy in kilowatt hours. Energy $=$ power $\times$ time $= 1.5 \times 4 = 6\text{ kWh}$. Option C divides instead of multiplying.

Electricity running cost. Energy $= 2 \times 3 = 6\text{ kWh}$, so the cost is $6 \times 0.28 = \$1.68$. Option C is the answer you get if you forget that the price is per kilowatt hour and treat it as cents.

Capture-recapture assumptions. The whole method rests on the sample being representative, which needs the marked animals mixed evenly back through the population and the population unchanged by births, deaths or migration between the two samples. Option B describes a full census, which is exactly what the method exists to avoid.

SA 1 (3 marks): (a) $N \approx \tfrac{45 \times 60}{9} = \tfrac{2700}{9}$ [1] $= 300$ koalas [1]. Check: $9$ of $60$ is $0.15$ and $45$ of $300$ is $0.15$. (b) Any one of: the tagged koalas mixed evenly back through the reserve; tagging did not change how easily a koala is spotted; the population did not change through births, deaths or movement in or out over the three months [1].

SA 2 (4 marks): (a) $2200 \div 1000 = 2.2\text{ kW}$ [1]. (b) $2.2 \times 6 = 13.2\text{ kWh}$ per day [1], so over $30$ days that is $13.2 \times 30 = 396\text{ kWh}$ [1]. (c) $396 \times 0.31 = \$122.76$ [1]. Check: $122.76 \div 0.31 = 396$.

SA 3 (4 marks): Heater A uses $2.4 \times 5 = 12\text{ kWh}$ a day, so $12 \times 30 = 360\text{ kWh}$, costing $360 \times 0.29 = \$104.40$ [2]. Heater B uses $1.8 \times 7 = 12.6\text{ kWh}$ a day, so $12.6 \times 30 = 378\text{ kWh}$, costing $378 \times 0.29 = \$109.62$ [1]. Heater B costs more, by $109.62 - 104.40 = \$5.22$ [1]. Note that the less powerful heater is the dearer one to run, because it is switched on for longer.

Retrieve, reflect and finish

Revisit your opening idea, then use the topic challenge and mark the lesson complete.

Revisit your thinking

+5 XP

Go back to what you wrote in section 01. Because $8$ of the $50$ caught were tagged, tagged fish are about $16\%$ of the lake. The $40$ tagged fish are therefore about $16\%$ of the total, giving $N \approx 250$ fish.

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Focus area overview · Year 12 · Maths Standard