SME
Ecology · Problem type
Mark–recapture
Use this for “40 marked, 50 caught later, 8 of them marked. How many are in the pond?”
Diagram
The
marked
share
of
the
second
catch
is
the
marked
share
of
the
population:
R
C
=
M
N
,
8
50
=
40
N
.
N
=
MC
R
=
40
×
50
8
=
250.
This is an example. Type your own values to start a new problem.
N
=
×
R
p
Marked share of the catch
worked out (not typed) · %
Formulas
p
=
100
×
R
C
16
=
100
×
8
50
N
=
M
×
C
R
250
=
40
×
50
8
Assumptions
•
The population is closed: no births, deaths or migration between the two catches.
•
Marks are not lost, and marked animals mix back in and are as easy to catch as the rest.
•
The marked share of the second catch, R ÷ C, estimates the marked share of the population, M ÷ N.
Step-by-step
WE KNOW
M = 40, C = 50, R = 8
FIND
marked share of the catch (p), estimated population (N)
Step 1 · Find marked share of the catch (p)
Use p = 100 × R ÷ C
The marked animals as a percent of the second catch.
p
=
100
×
8
50
p
=
800
50
p
=
16%
Step 2 · Find estimated population (N)
Use N = M × C ÷ R
Set
M
N
=
R
C
,
the
marked
share
in
the
catch
and
in
the
population,
and
solve
for
N
.
N
=
40
×
50
8
N
=
2,000
8
N
=
250
Check
Put the numbers back in. Both sides should match.
16
=
100
×
8
50
✓
250
=
40
×
50
8
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.