In: Statistics and Probability
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) | 0.218 | 0.361 | 0.223 | 0.165 | 0.032 | 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 five times the probability of no repeat offenders.These probabilities are the same. This is twice the probability of no repeat offenders.These probabilities are not related to each other.This is the complement of 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 = prisoners
a)
probability that one or more of the five parolees will be repeat offenders =1-P(X=0)
=1-0.218 =0.782
This is the complement of the probability of no repeat offenders.
b)
probability that two or more of the five parolees will be repeat offenders=1-P(X=0)-P(X=1)
=0.421
c)
probability that four or more of the five parolees will be repeat offenders=0.032+0.001 =0.033
d)
x | P(X=x) | xP(x) | x2P(x) |
0 | 0.218 | 0.00000 | 0.00000 |
1 | 0.361 | 0.36100 | 0.36100 |
2 | 0.223 | 0.44600 | 0.89200 |
3 | 0.165 | 0.49500 | 1.48500 |
4 | 0.032 | 0.12800 | 0.51200 |
5 | 0.001 | 0.00500 | 0.02500 |
total | 1.4350 | 3.2750 | |
E(x) =μ= | ΣxP(x) = | 1.4350 | |
E(x2) = | Σx2P(x) = | 3.2750 | |
Var(x)=σ2 = | E(x2)-(E(x))2= | 1.215775 | |
std deviation= | σ= √σ2 = | 1.10262 |
μ =1.435
e)
σ =1.10