91Ó°ÊÓ

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. $$\begin{array}{c|cccccc} \hline x & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline P(x) & 0.237 & 0.396 & 0.264 & 0.088 & 0.015 & 0.001 \\ \hline \end{array}$$ (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

Understanding Probability for One or More

To find the probability that one or more of the parolees become repeat offenders, begin by noting that this is the complement of none becoming repeat offenders. The probability that none of the parolees becomes repeat offenders is given by \( P(0) = 0.237 \). Thus, the probability that one or more do become repeat offenders is \( 1 - P(0) \).

02

Calculate Probability for One or More

Using the complement rule, calculate: \( P(\text{one or more}) = 1 - P(0) = 1 - 0.237 = 0.763 \).

03

Understanding Probability for Two or More

To find the probability that two or more parolees become repeat offenders, consider the probabilities that 0 or 1 become repeat offenders. We are looking for the complement of these events.

04

Calculate Probability for Two or More

Calculate: \( P(x \geq 2) = 1 - (P(0) + P(1)) = 1 - (0.237 + 0.396) = 0.367 \).

05

Understanding Probability for Four or More

To determine the probability that four or more become repeat offenders, you need the sum of the probabilities when \( x = 4 \) and \( x = 5 \).

06

Calculate Probability for Four or More

Sum the probabilities: \( P(x \geq 4) = P(4) + P(5) = 0.015 + 0.001 = 0.016 \).

07

Compute Expected Number of Repeat Offenders

The expected value \( \mu \) is calculated as \( \mu = \sum_{x=0}^{5} x \cdot P(x) \).

08

Calculate Expected Number of Repeat Offenders

Compute \( \mu = (0 \times 0.237) + (1 \times 0.396) + (2 \times 0.264) + (3 \times 0.088) + (4 \times 0.015) + (5 \times 0.001) = 1.25 \).

09

Compute Standard Deviation

The standard deviation \( \sigma \) is given by \( \sqrt{\text{Var}(X)} \) where \( \text{Var}(X) = \sum_{x=0}^{5} (x - \mu)^2 \cdot P(x) \).

10

Calculate Variance and Standard Deviation

Compute variance: \( \text{Var}(X) = (0-1.25)^2 \times 0.237 + (1-1.25)^2 \times 0.396 + (2-1.25)^2 \times 0.264 + (3-1.25)^2 \times 0.088 + (4-1.25)^2 \times 0.015 + (5-1.25)^2 \times 0.001 = 0.9375 \). Then find the standard deviation: \( \sigma = \sqrt{0.9375} \approx 0.968 \).

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Key Concepts

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

Binomial Distribution

Imagine a situation where something can either happen or not happen, like a coin flip resulting in heads or tails. A binomial distribution models such situations. It helps us understand scenarios with two possible outcomes, such as success or failure.
In the exercise, we look at parolees who can either become repeat offenders or not. This aligns with the binomial distribution, where:

• A trial is the parole decision for a prisoner.
• Success is a prisoner becoming a repeat offender.
• The probability of success is given as 25% or 0.25.

The distribution helps compute probabilities for different numbers of successes (repeat offenders) in five trials (parolees). In our case, the probabilities were already calculated in a table, helping answer further questions about expected values and deviations.

Expected Value

Expected value is a key concept in probability and statistics. It tells us the average or "expected" outcome over many repeated trials.
A helpful analogy is thinking of it as the "center of mass" for the probabilities, showing us where most outcomes will cluster.
In the context of our prison exercise, the expected number of repeat offenders ( ) tells us, on average, how many out of the five parolees might become repeat offenders. We calculate the expected value using the formula: \[ \mu = \sum_{x=0}^{5} x \cdot P(x) \] Here, each potential number of repeat offenders is multiplied by its probability, and then all results are added up. So, using the probabilities given, we find:

• When there's 0 offenders: \(0 \times 0.237\)
• For 1 offender: \(1 \times 0.396\).
• And so on, all the way up to 5 offenders: \(5 \times 0.001\)

This calculation gives us an expected value \( \mu = 1.25 \). Therefore, on average, we expect 1.25 out of 5 parolees to become repeat offenders.

Standard Deviation

While the expected value gives us an average, standard deviation tells us how spread out the outcomes might be around this average. It's a measure of the variability or "dispersion" from the expected value.
To calculate standard deviation \(\sigma\), we first find the variance, which involves looking at how each possible outcome (0 through 5 repeat offenders) varies from the expected value. The formula is: \[ \text{Var}(X) = \sum_{x=0}^{5} (x - \mu)^2 \cdot P(x) \]This formula checks how far each possible \( x \) is from the expected value \( \mu = 1.25 \), squares it, and adjusts by \(P(x)\).Finally, the standard deviation is the square root of the variance:\[ \sigma = \sqrt{\text{Var}(X)} \]In our example, the variance calculated was 0.9375, so the standard deviation is approximately 0.968. This tells us the average range from the expected value (1.25) within which most of our data will fall.

Complementary Probability

Complementary probability is a simple but powerful concept in probability, where the probability of something not happening is used to find the probability of it happening.
It's based on the idea that probabilities of all possible outcomes add up to 1.
In our exercise, this is used to find probabilities like "one or more" repeat offenders. If the probability that none are repeat offenders is given as 0.237, then the complementary probability that there is at least one is: \[ P(\text{at least one}) = 1 - P(0) \] This method can make complex probability questions simpler by allowing you to find what's tricky by first finding its opposite. In our examples, this approach quickly answers how many parolees could potentially become repeat offenders through complementary insights.