Question

5. Recidivism According to the Scientific American article "Reducing Crime: Rehabilitation is Making a Comeback," R. Doyle examined rehabilitation of convicted felons. One aspect of the article discussed recidivism of juvenile prisoners between 14 and 17 years of age, indicating that 82% of those released in 1994 were rearrested within 3 years. Suppose that today, 25 newly released prisoners between 14 and 17 years old are selected at random for monitoring over the next 3 years. a. Assuming that the recidivism rate is the same today as it was in 1994, determine a reasonable probability distribution for the random variable, Y = the number (of these 25 individuals) who are rearrested within the 3 year monitoring period. b. Determine the mean of the random variable Y having the distribution identified in part a. C. What does this mean value suggest might be likely to occur over the next 3 years? d. If, in fact, exactly 10 of the 25 newly released juvenile prisoners are rearrested within the 3 year monitoring period, would you be inclined to conclude that the recidivism rate today has decreased from the 82% rate observed in 1994? Why or why not? (Hint: Consider the probability P[Y S 10].) e. If, in fact, exactly 16 of the 25 newly released juvenile prisoners are rearrested within the 3 year monitoring period, would you be inclined to conclude that the recidivism rate today has decreased from the 82% rate observed in 1994? Why or why not?

Answer 1

According to the question, n=25 p=0.82 Where, n and p are the sample size and probability of prisoner being rearrested within 3 years (probability of success). Let Y be the number of prisoners rearrested within 3 years. (a) There are only two possible outcomes, that is, a prisoner will be rearrested or not and each prisoner is independent of other. This satisfies both the conditions of binomial distribution. Therefore, we can say that Y follows binomial distribution with parameter n and p. Y - B(25,0.82) (b) Compute the mean of binomial distribution as follows: Mean = np = 25x0.82 = 20.5 +21 (c) The mean values suggest that approximately 21 of the released 25 prisoners will be rearrested within 3 years. (d) Consider the null and alternate hypotheses as follows:

Ho: p = 0.82 H:P <0.82 We will consider 5% level of significance (a = 0.05). Now, compute the test statistic as follows: P(Y <10)=P(0)+ P(1) +...+ P(9)+ P(10) = [C, (0.82)°(1–0.82)"]+-+[%C0 (0.82)"º (1–0.82)"S] = 0.000003 Conclusion: Since p-value (0.000003) < a(0.05), we reject the null hypothesis. Therefore, we can conclude that recidivism rate has decreased from 82% rate observed in year 1994. Consider the null and alternate hypotheses as follows: H:p=0.82 H:p <0.82 We will consider 5% level of significance (a = 0.05). Now, compute the test statistic as follows: P(Y <16)=P(0)+P(1) +...+ P(15) + P(16) = [ 35C, (0.82)°(1–0.82)%]+...+[ >C16 (0.82)“ (1– 0.82)'] = 0.025 Conclusion: Since p-value (0.025) < a(0.05), we reject the null hypothesis. Therefore, we can conclude that recidivism rate has decreased from 82% rate observed in year 1994.