document.write( "Question 95603This question is from textbook Introductory Statistics
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document.write( ": Recidivism. In May 2003 issue of Scientific American, R. Doyle examined rehabilitation of felons in the article, \"Reducing Crime: Rehabilitation is making a Comeback\". \r
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document.write( " One aspect of the article discussed recidivism of juvenile prisoners between 14 and 17 years old, indicating the 82% of those released in 1994 were rearrested within 3 years. Assuming that recidivism rate still applies today, solve the following problems for six newly released juvinelle prisoners between 14 and 17 years old. \r
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document.write( "A. Determine the probability that the number rearrested within 3 years will be exacly four; at least four;at most five; between two and five; inclusive?\r
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document.write( "B. Determine the probability distribution of the random variable Y, the number of released prisoners of the six who are arrested within 3 years.\r
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document.write( "C. Determine and interpret the mean of the random variable Y.\r
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document.write( "D. Obtain the standard deviation of Y.\r
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document.write( "E. If in fact, exactly two of the six released are rearrested within 3 years, would you be inclined to conclude that recidivism rate today has decreased from the 82% rate in 1994? Explain your reasoning. \n" );
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Algebra.Com's Answer #69671 by stanbon(75887) ![]() You can put this solution on YOUR website! 82% of those released in 1994 were rearrested within 3 years. Assuming that recidivism rate still applies today, solve the following problems for six newly released juvinelle prisoners between 14 and 17 years old. \n" ); document.write( "------------- \n" ); document.write( "Comment: This is a binary problem with n=6 and p =0.82 \n" ); document.write( "Hopefully you have a TI calculator with Statistics functions. \n" ); document.write( "------------------------------------------------ \n" ); document.write( "A. Determine the probability that the number rearrested within 3 years will be exacly four \n" ); document.write( "P(x=4) = 6C4(0.82)^4 = binompdf(6,0,82,4)= 0.2197... \n" ); document.write( "------------ \n" ); document.write( "at least four \n" ); document.write( "P(4<=x<=6) = 1-binomcdf(6,0.82,3) = 0.9241.... \n" ); document.write( "-------------- \n" ); document.write( "at most five \n" ); document.write( "P(0<=x<=5) = binomcdf(6,0.82,5) = 0.69599... \n" ); document.write( "-------------- \n" ); document.write( " between two and five; inclusive \n" ); document.write( "P(2<=x<=5) = binomcdf(6,0.82,5)-binomcdf(6.0.82,1) = 0.6950... \n" ); document.write( "---------------------- \n" ); document.write( "B. Determine the probability distribution of the random variable Y, the number of released prisoners of the six who are arrested within 3 years. \n" ); document.write( "Use binomcdf(6,0.82,Y) and keep changing the Y; let it range from 0 up to 6. \n" ); document.write( "The seven answers constitute the probability distribution. \n" ); document.write( "------------------- \n" ); document.write( "C. Determine and interpret the mean of the random variable Y \n" ); document.write( "Use the fact that the mean for a binomial distribution \n" ); document.write( "is np. In you case that would be 6*0.82 = 4.92 \n" ); document.write( "On the average close to 5 of 6 prisoners in the stated age category \n" ); document.write( "are arrested again within 3 years of being released. \n" ); document.write( "--------------------. \n" ); document.write( "D. Obtain the standard deviation of Y. \n" ); document.write( "standard deviation for a binomial distribution is sqrt(npq) \n" ); document.write( "= sqrt(6*0.82*0.18) = 0.9411 \n" ); document.write( "------------------- \n" ); document.write( "E. If in fact, exactly two of the six released are rearrested within 3 years, would you be inclined to conclude that recidivism rate today has decreased from the 82% rate in 1994? Explain your reasoning. \n" ); document.write( "-------- \n" ); document.write( "I'm not sure if you are looking for a Hypothesis test here, or just an opinion \n" ); document.write( "based on the probabilities, or a judgement based on the normal curve. \n" ); document.write( "Anyway, here is the probilty of that happening if you assume that p = 0.82: \n" ); document.write( "P(x=2) = binompdf(6,0.82,2) = 0.0105... or about one out of 100 instead of \n" ); document.write( "82 out of 100. \n" ); document.write( "I'll leave the conjecturing to you. \n" ); document.write( "================ \n" ); document.write( "Cheers, \n" ); document.write( "Stan H.\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( " |