Question

2. In a survey of a large company, 30 percent of all employees are said that they are dissatisfied with their working conditions. Find the probabilities that among 12 randomly selected employee who were surveyed; a) None were dissatisfied with working conditions ( 4 marks) b) All were dissatisfied with working conditions ( 4marks) c) EIGHT were dissatisfied with working conditions ( 4 marks)

Answer 1

For a large population with a known (30%) probability of success with relatively small samples, binomial distribution may be used to model the situation.

p=0.3
n=12
P(X=x)=C(n,x)(p^x)(1-p)^(n-x)
where
C(n,x)=n!/(x!(n-x)!) combination of x objects from n.

(A) x=12
P(X=12)=C(12,12)(0.3^12)(0.7^0)
=5.3*10^(-7)

(B) x=0
P(X=0)=C(12,0)(0.3^0)(0.7^12)
=...

(C) x=exactly 8
P(X=8)=C(12,8)(0.3^8)(0.7^4)
=...