Question

A pond contains 110 fish, of which 20 are carp. If 10 fish are caught from the lake, what are the mean and variance of the number of carp among those 10?

Answer 1

Answer:

Mean = 1.82

Variance = 1.365

Step-by-step explanation:

We define X as the random variable that counts the number of the carp fishes that has been caught. Also we assume that all the fishes are equally likely to be caught. In that case we have that X as hypergeometric distribution.

The of the Hypergeometric Distribution and Its variance is given below.

Mean of the distribution = $n*\frac{k}{N}$

Variance = $\frac{n*k*(N-k)(N-n)}{N^2(N-1)}$

N= 110

k = 20

n = 10

Upon calculation we get

Mean = $10*\frac{20}{110}$

          = 1.82

Variance = $\frac{10*20*(110-20)(110-10)}{110*110*(110-1)}$ = 2*90*100/11*11*109 = 1.365

Therefore the mean is 1.82 and variance is 1.365

Answer 2

The mean and the variance of the number of carp are 20/11 and 180/121, respectively

The given parameters are:

Fishes = 110

Carp = 20

The probability of selecting a carp is:

p = 20/110

Simplify

p = 2/11

In a selection of 10 fishes, we have:

n = 10

The mean is:

E(x) = np

This gives

E(x) = 10 * 2/11

E(x) = 20/11

The variance is

Var(x) = E(x) * (1 - p)

So, we have:

Var(x) = 20/11 * (1 - 2/11)

Simplify

Var(x) = 20/11 * 9/11

Evaluate the product

Var(x) = 180/121

Hence, the mean and the variance of the number of carp are 20/11 and 180/121, respectively

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