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) 0.211 0.378 0.216 0.162 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 the complement of the probability of no repeat offenders. These probabilities are not related to each other. This is twice the probability of no repeat offenders. This is five times the probability of no repeat offenders. These probabilities are the same. (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
Accepted Solution
A:
Answer:[tex] \mu =E(X) = 0*0.207+1*0.367+ 2*0.227+ 3*0.162+ 4*0.036+ 5*0.001= 1.456[/tex]For the variance we need to calculate first the second moment given by:[tex] E(X) = \sum_{i=1}^n X^2_i P(X_i)[/tex]And replacing we got:[tex] E(X^2) = 0^2*0.207+1^2*0.367+ 2^2*0.227+ 3^2*0.162+ 4^2*0.036+ 5^2*0.001= 3.33[/tex]And the variance is given by:[tex] \sigma^2 = E(X^2) - [E(X)]^2 = 3.334 -[1.456]^2 =1.21[/tex]And the deviation is:[tex]\sigma = \sqrt{1.214} = 1.10[/tex]Step-by-step explanation:For thi case we have the following distribution given:X 0 1 2 3 4 5 P(X) 0.207 0.367 0.227 0.162 0.036 0.001For this case the expected value is given by:[tex] E(X) = \sum_{i=1}^n X_i P(X_i)[/tex]And replacing we got:[tex] \mu =E(X) = 0*0.207+1*0.367+ 2*0.227+ 3*0.162+ 4*0.036+ 5*0.001= 1.456[/tex]For the variance we need to calculate first the second moment given by:[tex] E(X) = \sum_{i=1}^n X^2_i P(X_i)[/tex]And replacing we got:[tex] E(X^2) = 0^2*0.207+1^2*0.367+ 2^2*0.227+ 3^2*0.162+ 4^2*0.036+ 5^2*0.001= 3.33[/tex]And the variance is given by:[tex] \sigma^2 = E(X^2) - [E(X)]^2 = 3.334 -[1.456]^2 =1.21[/tex]And the deviation is:[tex]\sigma = \sqrt{1.214} = 1.10[/tex]