Question

Reconsider problem 1 but suppose that the sample is selected with replacement. A small pond contains 30 fish, 10 of which have been tagged. Suppose that a fisherman’s catches four fish by catching a fish, noting whether it is tagged, releasing it, and repeating until four fish have been caught. (Assume that these four fish form a random sample of size four selected with replacement.) Let X denote the number of tagged fish that are caught.

a) What is the name of the distribution of X and what is the p.m.f. of X for this example? Provide an expression for p_X(x).

b) Find the probability that the fisherman’s catch will contain at most 3 tagged fish. c) Use the appropriate expressions from section 7.9 to compute expected value of X and the variance of X.

Answer 1

(a)

Since out of 30 fishes, 10 are tagged so the probability that a fish is tagged is

p = P(tagged) = 10/30 = 1/3 = 0.3333

Since fish are drawn with replacement so each time probability of getting tagged fish remain same.

Here X has binomial distribution with parameter n= 4 and p = 0.3333.

The expression of pmf is

$p_{X}(x)=\binom{4}{x}(0.3333)^{x}(1-0.3333)^{30-x},x=0,1,2,3,4$

Following table shows the pmf:

X	p(x)
0	0.19757
1	0.39508
2	0.29627
3	0.09874
4	0.01234

(b)

$P(X\leq 3)=1-P(X>3)=1-p_{X}(4)=0.98766$

(c)

Following table shows the calculations:

X	p(x)	x*p(x)	x^2*p(x)
0	0.19757	0	0
1	0.39508	0.39508	0.39508
2	0.29627	0.59254	1.18508
3	0.09874	0.29622	0.88866
4	0.01234	0.04936	0.19744
Total		1.3332	2.66626

The mean is

$\mu=E(X)=\sum xp_{X}(x)=1.3332$

The variance is

$\sigma^{2}=E(X^{2})-[E(X)]^{2}=\sum x^{2}p_{X}(x)-\left [\sum xp_{X}(x) \right ]^{2}=0.8888$

Another way to find mean and variance:

$\mu=np=4\cdot 0.3333=1.3332$

$\sigma^{2}=np(1-p)=4\cdot 0.3333\cdot (1-0.3333)=0.88884444\approx 0.8888$