91Ó°ÊÓ

Criminal Justice: Parole 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. The methods of Section \(5.2\) can be used to compute the probability assignments for the \(x\) distribution. \begin{tabular}{l|cccccc} \hline\(x\) & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline\(P(x)\) & \(0.237\) & \(0.396\) & \(0.264\) & \(0.088\) & \(0.015\) & \(0.001\) \\ \hline \end{tabular} (a) Find the probability that one or more of the five parolees will be repeat offenders. How does this number relate to the probability that none of the parolees will be repeat offenders? (b) Find the probability that two or more of the five parolees will be repeat offenders. (c) Find the probability that four or more of the five parolees will be repeat offenders. (d) Compute \(\mu\), the expected number of repeat offenders out of five. (e) Compute \(\sigma\), the standard deviation of the number of repeat offenders out of five.

Step by step solution

01

Determining Probability of At Least One Repeat Offender

To find the probability of one or more repeat offenders, we use the complement rule: \[ P(x \geq 1) = 1 - P(x = 0) \]Given that \(P(x = 0) = 0.237\), \[ P(x \geq 1) = 1 - 0.237 = 0.763 \]So, the probability that at least one parolee is a repeat offender is 0.763.

02

Finding Probability of Two or More Repeat Offenders

To find the probability of two or more repeat offenders, sum the probabilities for \(x = 2, 3, 4,\) and \(5\):\[ P(x \geq 2) = P(x = 2) + P(x = 3) + P(x = 4) + P(x = 5) \]Plugging in values: \[ P(x \geq 2) = 0.264 + 0.088 + 0.015 + 0.001 = 0.368 \]

03

Determining Probability of Four or More Repeat Offenders

To find the probability of four or more repeat offenders, sum the probabilities for \(x = 4\) and \(x = 5\):\[ P(x \geq 4) = P(x = 4) + P(x = 5) \]Substitute the given probabilities:\[ P(x \geq 4) = 0.015 + 0.001 = 0.016 \]

04

Calculating Expected Number of Repeat Offenders

The expected number \(\mu\) of repeat offenders is given by:\[ \mu = E[x] = \sum (x * P(x)) \]Calculate using the table values:\[ \mu = 0*0.237 + 1*0.396 + 2*0.264 + 3*0.088 + 4*0.015 + 5*0.001 \]\[ \mu = 0 + 0.396 + 0.528 + 0.264 + 0.06 + 0.005 = 1.253 \]

05

Computing Standard Deviation of Repeat Offenders

The standard deviation \(\sigma\) is given by:\[ \sigma = \sqrt{\sum(x^2 * P(x)) - \mu^2} \]Calculate the value:\[ x^2 * P(x) = 0^2*0.237 + 1^2*0.396 + 2^2*0.264 + 3^2*0.088 + 4^2*0.015 + 5^2*0.001 \]\[ = 0 + 0.396 + 1.056 + 0.792 + 0.24 + 0.025 = 2.509 \]\[ \sigma = \sqrt{2.509 - (1.253)^2} = \sqrt{2.509 - 1.570} = \sqrt{0.939} \approx 0.97 \]

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Binomial Distribution

When dealing with statistical probability regarding events that have two potential outcomes – like success or failure – the binomial distribution comes into play. It's like flipping a coin where heads and tails are the two possible outcomes.
In our exercise, the prisoners can either become repeat offenders or not while on parole. Key characteristics of a binomial distribution include:

• Number of trials (n): determines the number of experiments, which is 5 in our case for the five parolees.
• Probability of success (p): refers to the likelihood of a prisoner being a repeat offender, known to be 25% or 0.25.
• Probability of failure (1-p): naturally, this is the probability that a prisoner does not become a repeat offender, here 0.75.

Each prisoner on parole represents a trial with the same probability of success and failure. This constant probability and fixed number of trials define the binomial distribution. By using this distribution, we can easily calculate the probabilities of different numbers of prisoners becoming repeat offenders. For example, the probability of only one offender is found using the formula \[ P(x=k) = \binom{n}{k} \, p^k \, (1-p)^{n-k} \], where \( \binom{n}{k} \) represents the binomial coefficient.

Expected Value

The expected value, often denoted as \( \mu \), provides an average outcome for a random variable in a binomial experiment. This is like the center of data in the event's probability distribution. It predicts the average number of prisoners expected to reoffend if the parole decision is repeated multiple times. To compute the expected value for a binomial distribution, multiply the number of trials by the probability of success: \[ \mu = n \times p \] For the prisoners' scenario, this becomes: \[ \mu = 5 \times 0.25 = 1.25 \] So, if parole boards often examined groups of five prisoners, they could expect about 1.25 of them to reoffend on average. This concept helps in predicting future outcomes based on probability, essential for planning and decision-making, especially in justice systems aiming to reduce recidivism.

Standard Deviation

Standard deviation, symbolized by \( \sigma \), measures how spread out the values of a probability distribution are from the mean or expected value. In simpler terms, it's an indicator of the variability or consistency in the outcomes. Did the number of repeat offenders stay close to the expected average of 1.25, or did it vary widely? For a binomial distribution, calculating the standard deviation involves taking the square root of the variance (denoted as \( \sigma^2 \)). The formula for variance is: \[ \sigma^2 = n \times p \times (1-p) \] Then, standard deviation is found by: \[ \sigma = \sqrt{ \sigma^2 } \] In our case, substituting the known values: \[ \sigma^2 = 5 \times 0.25 \times 0.75 = 0.9375 \] \[ \sigma = \sqrt{0.9375} \approx 0.97 \] This result tells us that the number of repeat offenders typically falls within a range, fluctuating around the average by about 0.97 prisoners.