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. x 0 1 2 3 4 5 P(x) 0.211 0.378 0.216 0.162 0.032 0.001 (a) Find the probability that one or more of the five parolees will be repeat offenders. (Round your answer to three decimal places.) How does this number relate to the probability that none of the parolees will be repeat offenders? This is the complement of the probability of no repeat offenders. These probabilities are not related to each other. This is twice the probability of no repeat offenders. This is five times the probability of no repeat offenders. These probabilities are the same. (b) Find the probability that two or more of the five parolees will be repeat offenders. (Round your answer to three decimal places.) (c) Find the probability that four or more of the five parolees will be repeat offenders. (Round your answer to three decimal places.) (d) Compute μ, the expected number of repeat offenders out of five. (Round your answer to three decimal places.) μ = prisoners (e) Compute σ, the standard deviation of the number of repeat offenders out of five. (Round your answer to two decimal places.) σ = prisoners

The probability that no parolees are repeat offenders is 0.211. This means the probability of at least one is a repeat offender is the complement of this event. To find this probability, subtract from 1:

1-0.211 = 0.789.

For part b,

To find the probability that 2 or more are repeat offenders, add together the probability that 2, 3, 4 or 5 parolees are repeat offenders:

0.216+0.162+0.032+0.001 = 0.411.

For part c,

To find the probability that 4 or more are repeat offenders, add together the probabilities that 4 or 5 parolees are repeat offenders:

0.032+0.001 = 0.033.

For part d,

To find the mean, we multiply each number of parolees by their probability and add them together:

0(0.211)+1(0.378)+2(0.216)+3(0.162)+4(0.032)+5(0.001)

= 0 + 0.378 + 0.432 + 0.486 + 0.128 + 0.005 = 1.429

For part e,

To find the mean, we first subtract each number of parolees and the mean to find the amount of deviation. We then square it and multiply it by its probability. Then we add these values together and find the square root.

First the differences between each value and the mean:

0-1.429 = -1.429;

1-1.429 = -0.429;

2-1.429 = 0.571;

3-1.429 = 1.571;

4-1.429 = 2.571;

5-1.429 = 3.571

Next the differences squared:

(-1.429)^2 = 2.0420

(-0.429)^2 = 0.1840

(0.571)^2 = 0.3260

(1.571)^2 = 2.4680

(2.571)^2 = 6.6100

(3.571)^2 = 12.7520

Next the squares multiplied by the probabilities:

0(2.0420) = 0

1(0.1840) = 0.1840

2(0.3260) = 0.652

3(2.4680) = 7.404

4(6.6100) = 26.44

5(12.7520) = 63.76

Next the sum of these products:

0+0.1840+0.652+0.7404+26.44+63.76 = 91.7764

Lastly the square root:

√(91.7764) = 9.58

MrRoyal MrRoyal · Answer 2 · 2021-09-30T18:50:57+02:00

Probabilities are used to determine the outcomes of events.

The probability that one or more are repeat offenders is 0.789
The probability that two or more are repeat offenders is 0.411
The probability that four or more are repeat offenders is 0.033
The standard deviation of repeat offenders is 1.093
The expected number of repeat offenders is 1.429

The table is given as:

[tex]\left[\begin{array}{ccccccc}x &0 &1 &2 &3 &4 &5 &P(x) &0.211 &0.378 &0.216& 0.162 &0.032 &0.001\end{array}\right][/tex]

(a) Probability that one or more are repeat offenders

This is represented as: [tex]P(x \ge 1)[/tex]

Using the complement rule, we have:

[tex]P(x \ge 1) = 1 - P(x = 0)[/tex]

So, we have:

[tex]P(x \ge 1) = 1 - 0.211[/tex]

[tex]P(x \ge 1) = 0.789[/tex]

The probability that one or more are repeat offenders is 0.789

(b) Probability that two or more are repeat offenders

This is represented as: [tex]P(x \ge 2)[/tex]

Using the complement rule, we have:

[tex]P(x \ge 2) = 1 - P(x = 0) - P(x = 1)[/tex]

So, we have:

[tex]P(x \ge 2) = 1 - 0.211 - 0.378[/tex]

[tex]P(x \ge 2) = 0.411[/tex]

The probability that two or more are repeat offenders is 0.411

(c) Probability that four or more are repeat offenders

This is represented as: [tex]P(x \ge 4)[/tex]

So, we have:

[tex]P(x \ge 4) = P(x = 4) + P(x = 5)[/tex]

[tex]P(x \ge 4) = 0.032 + 0.001[/tex]

[tex]P(x \ge 4) = 0.033[/tex]

The probability that four or more are repeat offenders is 0.033

(d) The expected number of repeat offenders

This is calculated as:

[tex]\mu = \sum x \times P(x)[/tex]

So, we have:

[tex]\mu = 0 \times 0.211+ 1\times 0.378 + 2 \times 0.216 + 3 \times 0.162 + 4 \times 0.032 + 5 \times 0.001[/tex]

[tex]\mu = 1.429[/tex]

The expected number of repeat offenders is 1.429

(e) The standard deviation

This is calculated as:

[tex]\sigma= \sqrt{\sum (x^2 \times P(x)) - \mu^2}[/tex]

[tex]\sum (x^2 \times P(x))[/tex] is calculated as:

[tex]\sum (x^2 \times P(x)) = 0^2 \times 0.211+ 1^2 \times 0.378 + 2^2 \times 0.216 + 3^2 \times 0.162 + 4^2 \times 0.032 + 5^2 \times 0.001[/tex]

[tex]\sum (x^2 \times P(x)) = 3.237[/tex]

So, we have:

[tex]\sigma= \sqrt{\sum (x^2 \times P(x)) - \mu^2}[/tex]

[tex]\sigma = \sqrt{3.237 - 1.429^2}[/tex]

[tex]\sigma = \sqrt{1.194959}[/tex]

[tex]\sigma = 1.093[/tex]

The standard deviation of repeat offenders is 1.093

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