(a) Solution:

During second catch and release, 10 trout were tagged out of 100 trout that were caught. Assuming that for every 100 trout caught 10% will always have tags, total trout in the lake approximately = \(10 * 100 = 1000\).

(b) Solution:

Using hypergeometric distribution, the following is the probability function of \(X\).

\(P(X=x) = h(N,k,n,x) = \frac{{k \choose x}{{N-k} \choose {n-x}}}{N \choose n}\)

where \(N\) is population estimate

\(k\) is sample size of trout caught first time\((k = 100)\)

\(n\) is sample size of trout caught second time\((n = 100)\)

\(x\) is tagged trout in caught second time\((x = 10)\)

\(P(N) = h(N,100,100,10) = \frac{{100 \choose 10}{{N-100} \choose {100-10}}}{N \choose 100}\) = \(\frac{{100 \choose 10}{{N-100} \choose 90}}{N \choose 100}\).

(c) Solution:

Using growth quotient of hypergeometric model, _Maximum likelihood estimate(\(MLE\))

\(Q(N) = \frac{P(N)}{P(N-1)} > 1\)

where \(P(N) = \frac{{k \choose x}{{N-k} \choose {n-x}}}{N \choose n}\)

\(P(N - 1) = \frac{{k \choose x}{{N-1-k} \choose {n-x}}}{N-1 \choose n}\)

\(\frac{P(N)}{P(N-1)} > 1\)

\(P(N) > P(N-1)\)

\(= \frac{{k \choose x}{{N-k} \choose {n-x}}}{N \choose n} > \frac{{k \choose x}{{N-1-k} \choose {n-x}}}{N-1 \choose n}\)

Solving the equation will result in

\(= N(N - k -n + x) < (N - x)(N-n)\)

\(= N < \frac{kn}{x}\)

Substituting values for \(k, n\ and\ x\)

\(= N < \frac{100 * 100}{10}\) = \(1000\)

Let’s use R function to get Maximum likelihood estimate(\(MLE\))

library(knitr)
#initiate values
N1<- 100*100/10
k<- 100
n<- 100
x<- 10

PN<- matrix(NA, nrow=10, ncol=2)

#calculate upper values
for (i in 0:4){
  N<- N1 + i
  #hypergeometric distribution
  output<- (choose(k,x) * choose(N-k, n-x))/choose(N,n)
  PN[6+i,1]<- N
  PN[6+i,2]<- output
}

#calculate lower values
for (i in 1:5){
  N<- N1 - i
  #hypergeometric distribution
  output<- (choose(k,x) * choose(N-k, n-x))/choose(N,n)
  PN[6-i,1]<- N
  PN[6-i,2]<- output
}

kable(PN, format="pandoc", align="l", col.names = c("N", "P(N)"), caption = "Hypergeometric distribution")
Hypergeometric distribution
N P(N)
995 0.1389680
996 0.1389749
997 0.1389801
998 0.1389835
999 0.1389853
1000 0.1389853
1001 0.1389836
1002 0.1389801
1003 0.1389750
1004 0.1389682

At \(N = 1000\) and \(N = 999, P(N) = 0.1389853\), hence it is maximum value, for all other values of \(N, P(N)\) is lower.

Therefore, Maximum likelihood estimate(\(MLE\)) = \(1000\) Or \(999\).

References: