Question 95603
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.
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Comment: This is a binary problem with n=6 and p =0.82
Hopefully you have a TI calculator with Statistics functions.
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A. Determine the probability that the number rearrested within 3 years will be exacly four
P(x=4) = 6C4(0.82)^4 = binompdf(6,0,82,4)= 0.2197...
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at least four
P(4<=x<=6) = 1-binomcdf(6,0.82,3) = 0.9241....
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at most five
P(0<=x<=5) = binomcdf(6,0.82,5) = 0.69599...
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 between two and five; inclusive
P(2<=x<=5) = binomcdf(6,0.82,5)-binomcdf(6.0.82,1) = 0.6950...
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B. Determine the probability distribution of the random variable Y, the number of released prisoners of the six who are arrested within 3 years.
Use binomcdf(6,0.82,Y) and keep changing the Y; let it range from 0 up to 6.
The seven answers constitute the probability distribution.
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C. Determine and interpret the mean of the random variable Y
Use the fact that the mean for a binomial distribution
is np. In you case that would be 6*0.82 = 4.92
On the average close to 5 of 6 prisoners in the stated age category
are arrested again within 3 years of being released.
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D. Obtain the standard deviation of Y.
standard deviation for a binomial distribution is sqrt(npq)
= sqrt(6*0.82*0.18) = 0.9411
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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.
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I'm not sure if you are looking for a Hypothesis test here, or just an opinion
based on the probabilities, or a judgement based on the normal curve.
Anyway, here is the probilty of that happening if you assume that p = 0.82:
P(x=2) = binompdf(6,0.82,2) = 0.0105... or about one out of 100 instead of
82 out of 100.
I'll leave the conjecturing to you.
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Cheers,
Stan H.