Question

A federal study reported that 7.5% of the United States workforce has drug problems. An official Indiana state drug investigation will investigate this claim. In a sample of 20 workers: Binomial

a) How many workers are expected to have drug addiction problems?

b) What is the probability that none of the workers in the sample has addiction problems?

c) What is the probability that at least 25% of the workers in the sample have addiction problems?

Please show your procedure so I can understand it.

Answer 1

a)

here this is binomial with parameter n=20 and p=0.075

expected number =np=20*0.075 =1.5

b)

probability that none of the workers in the sample has addiction problems =P(X=0) =(1-0.075)^20 =0.2103

c)

25% of the workers =20*0.25 =5

probability that at least 25% of the workers in the sample have addiction problems

=P(at least 5 have addiction problems)

=P(X>=5) =1-P(X<=4)

=1-(P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4))

=1-(20C0)*(0.075)^0*(1-0.075)^20-(20C1)*(0.075)^1*(1-0.075)^19-(20C2)*(0.075)^2*(1-0.075)^18-(20C3)*(0.075)^3*(1-0.075)^17-(20C4)*(0.075)^4*(1-0.075)^16

=0.0142