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Measurement, STD2 M7 2022 HSC 4 MC

Lily wanted to estimate the number of fish in a lake.

She randomly captured 30 fish, then tagged and released them.

One week later she randomly captured 40 fish from the same lake. She found that 12 of these 40 fish were tagged.

What is the best estimate for the total number of fish in the lake?

  1. 58
  2. 70
  3. 82
  4. 100
Show Answers Only

`D`

Show Worked Solution

`text{Let}\ \ P=\ text{population of fish in lake}`

`text{Capture}\ = 30/P`

`text{Recapture}\ = 12/40`

`30/P` `=12/40`  
`12P` `=30 xx 40`  
`P` `=1200/12`  
  `=100`  

 
`=>D`


♦♦ Mean mark 37%.

Measurement, STD2 M7 2016 HSC 28a

Jacob has a large jar of silver coins. He adds 20 gold coins into the jar. He then seals the jar and shakes it to ensure that the gold coins are mixed in thoroughly with the silver coins. Jacob then opens the jar and takes a handful of coins. In his hand he has 33 silver coins and 4 gold coins. 

  1. Based on Jacob’s handful, if a coin is selected at random from the jar, what is the probability that it is a gold coin?  (1 mark)

  2. Jacob returns the handful of coins to the jar. Estimate the total number of coins in the jar.  (2 marks)

Show Answers Only
  1. `4/37`
  2. `185`
Show Worked Solution
i.    `P(G)` `= 4/(4 + 33)`
    `= 4/37`
♦ Mean mark part (i) 28%.

 

ii.  `text(Let)\ \ X =\ text(total coins in jar.)`

`20/X` `=4/37`
`:.X` `=(20 xx 37)/4`
  `=185`

Measurement, STD2 M7 2007 HSC 23c

A scientific study uses the ‘capture-recapture’ technique.

In the first stage of the study, 24 crocodiles were caught, tagged and released.

Later, in the second stage of the study, some crocodiles were captured from the same area. Eighteen of these were found to be tagged, which was 40% of the total captured during the second stage. 

  1. How many crocodiles were captured in total during the second stage of the study?  (1 mark)

  2. Calculate the estimate for the total population of crocodiles in this area.   (2 marks)

Show Answers Only
  1. `text(45 crocodiles)`
  2. `text(60 crocodiles)`
Show Worked Solution
i.    `text(Let)\ C_1` `=\ text(crocodiles captured in stage 1)`
  `C_2` `=\ text(crocodiles captured in stage 2)`
`C_1` `=\ text(40%)\ xx C_2`
`18` `=\ text(40%)\ xx C_2`
`:.\ C_2` `= 18/0.4 = 45\ text(crocodiles)`

COMMENT: Std2 sample exam questions from NESA included capture/recapture as examinable content within M7 Rates and Ratios.

 

ii.    `text(Capture) = 24/text(Population)`

`text(Recapture) = 18/45`

`24/text(Population)` `= 18/45`
`:.\ text(Population)` `= (24 xx 45)/18`
  `= 60`

Measurement, STD2 M7 2012 HSC 26f

The capture-recapture technique was used to estimate a population of seals in 2012.

•   60 seals were caught, tagged and released.

•   Later, 120 seals were caught at random.

•   30 of these 120 seals had been tagged.

The estimated population of seals in 2012 was 11% less than the estimated population for 2008. 

What was the estimated population for 2008?   (2 marks)

Show Answers Only

`text{270 seals (nearest whole)}`

Show Worked Solution

`text(Let population in 2012 =)\ P(2012)`

COMMENT: Std2 sample exam questions from NESA included capture/recapture as examinable content within M7 Rates and Ratios.

`text(Capture)`

`⇒ 60/{P(2012)} `

`text(Recapture)`

`=> 30/120 = 1/4`

`60/{P(2012)}` `=1/4`
`:. P(2012)` `= 60 xx 4 =240`

 

 

♦♦ Mean mark 26%
MARKER’S COMMENT: The most successful approach was to use the ‘unitary method’ (i.e. calculate what 1% is worth, then multiply it by 89), as shown in the Worked Solution.

`text(We know)\ P(2008)\ text(less 11% = 240)`

`text{(100% – 11%)} xx  P(2008)` `= 240`
`text(89%) xxP(2008)` `=240`
`:. text(1%) xxP(2008)` `=240/89=2.6966…`
`P(2008)` `= 100xx2.6966…`
  `= 269.6629…`
  `=270\ \ text{(nearest whole)}`

 

`:.\ text{2008 population estimate = 270 seals}`