It is claimed that 15% of the ducks in a particular region have patent schistosome infection. suppose that seven ducks are selected at random. let x equal the number of ducks that are infected. (a) assuming independence, how is x distributed? (b) find (i) p(x ≥ 2), (ii) p(x = 1), and (iii) p(x ≤ 3).

(a) X follows a Binomial distribution

(b) (i) P(X ≥ 2) = 0.28348

P(X = 1) = 0.39601

P(X ≤ 3) = 0.98800

Step-by-step explanation:

(a) In this situation, the variable X equal to the number of ducks that are infected follows a Binomial distribution because we have:

(b) So, the probability that X ducks are infected is calculated as:

Then, Probability P(X = 1) is equal to:

At the same way, probability P(X ≥ 2) is equal to:

P(X ≥ 2) = P(2) + P(3) + P(4) + P(5) + P(6) + P(7)

P(X ≥ 2) = 0.2097 + 0.0617 + 0.0109 + 0.0011 + 0.00006 + 0.00002

P(X ≥ 2) = 0.28348

And probability P(X ≤ 3) is equal to:

P(X ≤ 3) = P(0) + P(1) + P(2) + P(3)

P(X ≤ 3) = 0.3206 + 0.3960 + 0.2097 + 0.0617

P(X ≤ 3) = 0.988

Answers:

b)

i) 0.2834

ii) 0.3960

iii) 0.9880

Solution:

To solve this we need to use the binomial probability

P(X=k)=

a)

X= number of ducks infected

n=7

p=15%=0.15

P(X)= ; x=0,1,2,...,7

b)

First we need to calculate by the definition of binomial probability at k=0,1,2,3

P(X=0)= = 0.3206 ;

P(X=1)= = 0.3960 ;

P(X=2)= = 0.2097  ;

P(X=3)= = 0.0617  ;

(i) Find P(X≥2)

Using the complement rule, we have that : P(A´ )= 1-P(A)

P(X≥2)= 1- P(X<2)

= 1- ((P(X=0)+P(X=1))

= 1- 0.3206-0.3960

=0.2834

(ii) Find P(X=1)

We have to evaluate the definition of binomial probaility at k=1

Then we have that

P(X=1)= = 0.3960

(iii) Find P(X ≤ 3)

We have to use the addition tule for mutually exclusive events

Addition rule: P(A∪B)=P(A)+P(B)

P(≤3)= P(X=0)+ P(X=1) + P(X=2) + P(X=3)

=0.3206 + 0.3960+ 0.2097 + 0.0617

= 0.9880

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step-by-step explanation: