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.
