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)	0.227	0.380	0.209	0.165	0.018	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.)


How does this number relate to the probability that none of the parolees will be repeat offenders?

This is twice the probability of no repeat offenders.These probabilities are the same.     This is the complement of the probability of no repeat offenders.These probabilities are not related to each other.This is five times the probability of no repeat offenders.

(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 =  prisoners

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

Answer 1

Solution(a)

We need to calculate P(X>=1) = P(X=1)+ P(X=2) + P(X=3) + P(X=4) + P(X=5) = 0.380+ 0.209 + 0.165+ 0.018+ 0.001 = 0.773

this is also complement of zero offenders P(X>=1) = 1-P(X=0) = 1 - 0.227 = 0.773

Solution(b)

P(X>=2) = P(X=2) + P(X=3) + P(X=4) + P(X=5) = 0.209 + 0.165+ 0.018+ 0.001 =0.393

Solution(c)

P(X>=4) = P(X=4) +P(X=5) = 0.018+0.001 = 0.019

Solution(d)

Expected mean = summation(Xi*P(Xi))

X	P(X)	Xi*P(Xi)
0	0.227	0
1	0.38	0.38
2	0.209	0.418
3	0.165	0.495
4	0.018	0.072
5	0.001	0.005

Expected mean = 0+0.38+0.418+0.495+0.072+0.005 = 1.37

Solution(e)

Standard deviation = sqrt(summation(Xi-mean)^2 *P(Xi)))

X	P(X)	Xi*P(Xi)	(Xi-mean)	(Xi-mean)^2	(Xi-mean)^2 *P(Xi)
0	0.227	0	-1.37	1.8769	0.426056
1	0.38	0.38	-0.37	0.1369	0.052022
2	0.209	0.418	0.63	0.3969	0.082952
3	0.165	0.495	1.63	2.6569	0.438389
4	0.018	0.072	2.63	6.9169	0.124504
5	0.001	0.005	3.63	13.1769	0.013177

Standard deviation = sqrt(0.426056+0.052022+0.082952+0.438389+0.438389+0.124504+0.013177) = sqrt(1.1371) = 1.07