Hi, can anyone help with the last 2 answers? Thank you! Today reported that appr

Question

Hi, can anyone help with the last 2 answers? Thank you! Today reported that approximately 25% of all state prison inmates released on parole become repeat offenders while on parole. Suppose the (a) Find the probability that one or more of the five parolees will be repeat offenders. () 0.781 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. This is five times the probability of no repeat offenders. This is twice the probability of no repeat offenders. These probabilities are not related to each other. These probabilities are the same. (b) Find the probability that two or more of the five parolees will be repeat offenders () 0.409 (c) Find the probability four or more of the five parolees will be repeat offenders () 0.025 (d) Compute mu. the expected number of repeat offenders out of five. () mu = _____ prisoners (e) compute sigma, the standard deviation of the number of repeat offenders out of five ()

Explanation / Answer

D) Mean = sum(x P(x)) = 0 + 1x0.372 + 2x0.206 + 3x0.178 + 4x0.024 + 5x0.001

= 1.419

E) variance = sum of (x - mean)2*P(x)

= (0-1.419)2 x 0.219 + (1-1.419)2 x 0.372 + (2-1.419)2x 0.206 + (3-1.419)2 x 0.178 + (4-1.419)2x0.024 + (5-1.419)2x0.001

= 1.193

So, standard deviation = sq root(1.193) = 1.092

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