Question

Hi,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.

Answer 1

The question covers the concept of binomial probability in Mathematics to calculate the likelihood of released prisoners becoming repeat offenders. Utilizing the given recidivism rate, we can apply the binomial formula to determine the probability distribution for the number of repeat offenders out of five prisoners.

Step-by-step explanation:

The question pertains to the field of probability, specifically regarding the subject of repeat offenses among parolees. We are asked to understand the probability of a specific number of released inmates becoming repeat offenders while on parole. Given that USA Today reported a 25% recidivism rate among parolees, we can treat the scenario as a binomial probability problem, where x represents the random variable for the number of released prisoners who become repeat offenders out of the five prisoners considered.

To find the probability distribution for x, we would use the binomial probability formula:

P(X = k) = (n choose k) * p^k * (1-p)^(n-k),

where 'n' is the total number of trials (in this case, prisoners), 'k' is the number of successes (repeat offenders), and 'p' is the probability of success (the 25% recidivism rate).

The fact that the United States has the highest incarceration rate in the world and that recidivism is a significant issue within the corrections system substantiates the relevance of such probability calculations. They can help in understanding the dynamics of the criminal justice system, assist in planning for resources, and suggest areas for potential reform.

Answer 2

a)0.783

b)0.409

c) 0.024

d)1.422

e) σ = 1.09

Step - by - Step Explanation

What to find?

• Find the probability that one or more of the five parolees will be repeat offenders.

• Find the probability that two or more of the five parolees will be repeat offenders.

• Find the probability that four or more of the five parolees will be repeat offenders.

• Compute μ, the expected number of repeat offenders out of five

,

• Compute σ, the standard deviation of the number of repeat offenders out of five

Given Parameters:

x 0 1 2 3 4 5

P(x) 0.217 0.374 0.204 0.181 0.023 0.001

a) Probability that one or more of the five parolees will be repeat offenders.

P(x≥ 1) = 1 - p(x < 1)

= 1 - P(x=0)

= 1 - 0.217

= 0.783

OR

P(x≥1) = p(x=1) + p(x=2) + p(x=3) + p(x=4) + p(x=5)

= 0.374 + 0.204 + 0.181 + 0.023 + 0.001

=0.783

Hence, Probability that one or more of the five parolees will be repeat offenders is 0.783

This is the complement of the probability of no repeat offenders as shown above.

Hence, the first option is the correct answer.

b) The probability that two or more of the five parolees will be repeat offenders.

P(x≥ 2) = p(x=2) + p(x=3) + p(x=4) + p(x=5)

= 0.204 + 0.181 + 0.023 + 0.001

=0.409

Hence, probability that two or more of the five parolees will be repeat offenders is 0.409

c)The probability that four or more of the five parolees will be repeat offenders.

P(x≥4) = p(x=4) + p(x=5)

= 0.023 + 0.001

= 0.024

Hence, the probability that four or more of the five parolees will be repeat offenders is 0.024

d)Compute μ, the expected number of repeat offenders out of five

To compute μ, multiply each x by its corresponding P(x) and sum it up.

That is;

Mean = E(X) = Σ x * P(x)

=0(0.217) + 1(0.374) + 2(0.204) + 3(0.181) + 4(0.023) + 5(0.001)

= 0 + 0.374 + 0.408 + 0.543 + 0.092 + 0.005

= 1.422

Hence, the expected number of repeat offenders out of five is 1.422

e)Compute σ, the standard deviation of the number of repeat offenders out of five

To compute σ,

compute E(x²) = Σ x² * p(x)

= 0²(0.217) + 1²(0.374) + 2²(0.204) + 3²(0.181) + 4²(0.023) + 5²(0.001)

= 0(0.217) + 1(0.374) + 4(0.204) + 9(0.181) + 16(0.023) + 25(0.001)

= 0 + 0.374 + 0.816 + 1.629 + 0.368 + 0.025

= 3.212

Variance = Σ x² *p(x) - [Σ x *p(x) ] ²

= 3.212 - 1.422²

= 3.212 - 2.022084

=1.189916

Standard deviation (σ) = √ 1.189916 = 1.09

Hence, σ = 1.09