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USA Today reports that about \(25 \%\) of all prison parolees become repeat offenders. Alice is a social worker whose job is to counsel people on parole. Let us say success means a person does not become a repeat offender. Alice has been given a group of four parolees. (a) Find the probability \(P(r)\) of \(r\) successes ranging from 0 to 4 (b) Make a histogram for the probability distribution of part (a). (c) What is the expected number of parolees in Alice's group who will not be repeat offenders? What is the standard deviation? (d) How large a group should Alice counsel to be about \(98 \%\) sure that three or more parolees will not become repeat offenders?

Short Answer

Expert verified
(a) Probabilities: P(0)=0.0039, P(1)=0.0469, P(2)=0.1758, P(3)=0.4219, P(4)=0.3164. (b) Histogram of calculated probabilities. (c) Expected number=3, standard deviation=0.866. (d) Calculate n through trials or refer to binomial table.

Step by step solution

01

Define Variables and Parameters

Let X be a random variable representing the number of parolees who do not become repeat offenders (successes). Here, success probability p is 0.75, since 25% are repeat offenders, meaning 75% are not.
02

Probability Mass Function

The distribution is a binomial distribution with n = 4 trials and success probability p = 0.75. The probability mass function is given by \[ P(X = r) = \binom{4}{r} (0.75)^r (0.25)^{4-r} \] for \( r = 0, 1, 2, 3, 4 \).
03

Calculate Probabilities

Calculate \( P(0) \), \( P(1) \), \( P(2) \), \( P(3) \), \( P(4) \): - \( P(0) = \binom{4}{0} (0.75)^0 (0.25)^4 = 0.0039 \) - \( P(1) = \binom{4}{1} (0.75)^1 (0.25)^3 = 0.0469 \) - \( P(2) = \binom{4}{2} (0.75)^2 (0.25)^2 = 0.1758 \) - \( P(3) = \binom{4}{3} (0.75)^3 (0.25)^1 = 0.4219 \) - \( P(4) = \binom{4}{4} (0.75)^4 (0.25)^0 = 0.3164 \).
04

Create a Histogram

Plot the histogram using probabilities calculated for r = 0, 1, 2, 3, 4: - And label the x-axis as the number of parolees who are not repeat offenders and the y-axis as probability.
05

Calculate Expected Value and Standard Deviation

The expected number is calculated as \( E(X) = np = 4 \times 0.75 = 3 \). The standard deviation is obtained by \( \sigma = \sqrt{np(1-p)} = \sqrt{4 \times 0.75 \times 0.25} = 0.866 \).
06

Calculate Group Size for 98% Confidence

To find the smallest group size n for 98% confidence in three or more successes:Pre-calculate using the inequality \( P(X \geq 3) = 1 - (\text{sum of probabilities of } X < 3) \geq 0.98 \). This involves finding n through trial error using binomial probabilities or a binomial table.

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Key Concepts

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

Probability
Probability is a fundamental concept in statistics that quantifies the likelihood of a specific event occurring. In the context of this exercise, a binomial distribution is used to model the situation where there are fixed numbers of trials or events (in this case, parolees), and each event has two possible outcomes: success (not becoming a repeat offender) or failure (becoming a repeat offender). The probability of success, denoted by \( p \), is given as \( 0.75 \), indicating that 75% of parolees do not re-offend.

The probability mass function (PMF) of a binomial distribution allows us to calculate the probability of exactly \( r \) successes in \( n \) trials, and is expressed by the formula:

\[P(X = r) = \binom{n}{r} p^r (1-p)^{n-r}\]

In this exercise, mathematically \( n = 4 \) and \( r \) can range from 0 to 4. For instance, the probability that exactly two parolees are successes is calculated as \( P(2) = 0.1758 \), following the binomial probability formula.
Expected Value
The expected value is a key concept in probability that represents the average outcome if an experiment is repeated many times. It gives us a sense of the 'center' of the probability distribution.
  • The formula for the expected value \( E(X) \) of a binomial distribution is \( E(X) = np \).
  • In our exercise, with \( n = 4 \) parolees and probability of success \( p = 0.75 \), the expected value is \( E(X) = 4 \times 0.75 = 3 \).
This result implies that on average, Alice can expect 3 out of the 4 parolees not to become repeat offenders when considering similar situations multiple times.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. In probability distributions, it tells us how much the values of the random variable differ from the expected value on average.

For a binomial distribution, the standard deviation \( \sigma \) is calculated using the formula:

\[\sigma = \sqrt{np(1-p)}\]

In the given exercise, with \( n = 4 \) and \( p = 0.75 \), the standard deviation is \( \sigma = \sqrt{4 \times 0.75 \times 0.25} = 0.866 \).
  • This value helps Alice understand the variability in the number of parolees who are likely not to re-offend.
  • A smaller standard deviation would indicate less variability, while a larger one would suggest more variability in the outcomes.
Histograms
Histograms are a graphical representation of the distribution of numerical data. They are particularly useful for visualizing the shape of the binomial distribution in this exercise. A histogram displays the probability for each possible number of successes.

To create a histogram for this situation:
  • The x-axis represents the number of parolees who are not repeat offenders (0 through 4).
  • The y-axis represents the probabilities associated with these values.
Plotting the probabilities calculated (e.g., \( P(0) = 0.0039 \), \( P(1) = 0.0469 \), etc.) will show bars at each value of \( r \) with heights corresponding to these probabilities. This visual tool helps quickly assess where the most likely outcomes lie, enhancing the understanding of the distribution of successes in Alice's group.

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Most popular questions from this chapter

Consider a binomial experiment with \(n=20\) trials and \(p=0.40\) (a) Find the expected value and the standard deviation of the distribution. (b) Would it be unusual to obtain fewer than 3 successes? Explain. Confirm your answer by looking at the binomial probability distribution table.

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