Question

USA Today reported that approximately 25% of all state prison inmates released on parole become repeat offenders while on parole. Suppose the parole board is examining five prisoners up for parole. Let x = number of prisoners out of five on parole who become repeat offenders. x 0 1 2 3 4 5 P(x) 0.217 0.388 0.210 0.155 0.029 0.001

(a) Find the probability that one or more of the five parolees will be repeat offenders. (Round your answer to three decimal places.)

(b) Find the probability that two or more of the five parolees will be repeat offenders. (Round your answer to three decimal places.)

(c) Find the probability that four or more of the five parolees will be repeat offenders. (Round your answer to three decimal places.)

(d) Compute μ, the expected number of repeat offenders out of five. (Round your answer to three decimal places.)
μ =

(e) Compute σ, the standard deviation of the number of repeat offenders out of five. (Round your answer to two decimal places.)
σ =

Answer 1

a)P(one or more of the five parolees will be repeat offender)=P(X≥1)=1-P(x=0) = 0.783

b) P(two or more of the five parolees will be repeat offenders)=P(X≥2)=1-P(X=0)-P(X=1) = 0.395

c) P(X≥4)=P(X=4)+P(x=5)=0.030

d)

X	P(X)	X*P(X)	X² * P(X)
0	0.217	0	0
1	0.388	0.388	0.388
2	0.21	0.42	0.84
3	0.155	0.465	1.395
4	0.029	0.116	0.464
5	0.001	0.005	0.025

	P(X)	X*P(X)	X² * P(X)
total sum =	1	1.394	3.112

µ= E[X] = Σx*P(X) =           1.394

e)

E [ X² ] = ΣX² * P(X) =            3.112
          
variance = E[ X² ] - (E[ X ])² =            1.168764
          
std dev ,σ = √(variance) =            1.08