5 it claimed that 26 of the ducks in a particular region hav

(5) it claimed that 26% of the ducks in a particular region have patent schistosome infection. Suppose that 10 ducks are selected at random. Let X equals the number of ducks that are infected (a)at most are infected. (b) exactly five are infected. (c)the expected value(Mu) and variance of X (sigma^2).

Solution

This is binomial.

Number of trials, n = 10.

Probability of succes, p = .25

The probability formula for binomial is:

P(X = x) = nCx*P^x*(1-p)^(n-x)

a. At most three:

This x equals 0, 1, 2, or 3.

P(X = 0) = 10C0*.25^0*(1-.25)^(10-0) = .056314

P(X = 1) = 10C1*.25^1*(1-.25)^(10-1) = .187712

P(X = 2) = 10C2*.25^2*(1-.25)^(10-2) = .281568

P(X = 3) = 10C3*.25^3*(1-.25)^(10-3) = .250282

Then add them up: answer is 0.7759

alternatively, you can use a calculator with a binomcdf function:

binomcdf(10, .25, 3) = 0.7759

b. Exactly 5 are infected:

P(X = 5) = 10C5*.25^5*(1-.25)^(10-5) = 0.0584

answer: 0.0584

alternatively, you can use a calculator with a binompdf function:

binompdf(10, .25, 5) = 0.0584

c. Expected value

Expected value = n*p = 10*.25 = 2.5 (answer)

Variance = n*p*(1-p) = 10*.25*(1-.25) = 1.875 (answer)