1a) 𝑁(8) = 1000 + 200 × 8
= 2600
1b) 45/300=300/N
N=2000
1c) n=300, p = 300/2000 = 0.15
1d) Each fish caught is independent of previous fish caught
1e) Var(X) = np(1-p)
=300 * 300/2000 * 1700/2000
=38.25
1f) Var(X/300 = (Var(X))/3002
=0.000425
1g) 0.15+1.5(0.000425)0.5 = 0.181
And
0.15-1.5(0.000425)0.5 = 0.119
1h) 300/N = 0.181
And
300/N = 0.119
Lower bound: 1658 and upper bound: 2519
1i) Linear model prediction falls outside this range so unlikely to be a good model
1j) 1000= 5000/(1+c)
C=4
1200=5000/(1+4e-k)
e-k= 3800/(4 * 1200)
k= -ln(0.791) = 0.234
1k) N(8) = 5000/(1+ 4e-0.234 * 8)
= 3090
1l) Much higher than the upper bound thus the fish are unlikely to reach the carrying capacity
of 5000