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It is claimed that [tex]$46 \%$[/tex] of all violent felons in the prison system are repeat offenders. Based on this, if 35 violent felons are randomly selected, find the probability that:

a. Exactly 15 of them are repeat offenders.
[tex]\[ P(X = 15) = \square \][/tex]

b. At most 16 of them are repeat offenders.
[tex]\[ P(X \leq 16) = \square \][/tex]

c. At least 15 of them are repeat offenders.
[tex]\[ P(X \geq 15) = \square \][/tex]

d. Between 12 and 18 (including 12 and 18) of them are repeat offenders.
[tex]\[ P(12 \leq X \leq 18) = \square \][/tex]

e. The mean of this distribution is
[tex]\[ \mu = \square \][/tex]

f. The standard deviation of this distribution is
[tex]\[ \sigma = \square \][/tex]



Answer :

GinnyAnswer
To solve this problem, let [tex]\( X \)[/tex] be the random variable representing the number of repeat offenders among the 35 violent felons. This follows a binomial distribution with parameters [tex]\( n = 35 \)[/tex] (sample size) and [tex]\( p = 0.46 \)[/tex] (probability of being a repeat offender).

### Parts a to f:

a. Probability that exactly 15 of them are repeat offenders
[tex]\[ P(X = 15) = 0.12613949618246523 \][/tex]

b. Probability that at most 16 of them are repeat offenders
[tex]\[ P(X \leq 16) = 0.555600200320101 \][/tex]

c. Probability that at least 15 of them are repeat offenders
[tex]\[ P(X \geq 15) = 0.7048545001307283 \][/tex]

d. Probability that between 12 and 18 (including 12 and 18) of them are repeat offenders
[tex]\[ P(12 \leq X \leq 18) = 0.7344738288879837 \][/tex]

e. Mean of the distribution
[tex]\[ \mu = n \times p = 35 \times 0.46 = 16.1 \][/tex]

f. Standard deviation of the distribution
[tex]\[ \sigma = \sqrt{n \times p \times (1 - p)} = \sqrt{35 \times 0.46 \times (1 - 0.46)} = 2.9485589700733477 \][/tex]

Each value is determined based on binomial distribution properties using provided probabilities and parameters.

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