Answer:
a) 21.58% probability that exactly 3 people are repeat offenders
b) 97.91% probability that at least one person is a repeat offender
c) 3.69
d) 1.83
Explanation:
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
![P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)](https://img.qammunity.org/2021/formulas/mathematics/college/mj488d1yx012m85w10rpw59rwq0s5qv1dq.png)
In which
is the number of different combinations of x objects from a set of n elements, given by the following formula.
![C_(n,x) = (n!)/(x!(n-x)!)](https://img.qammunity.org/2021/formulas/mathematics/college/qaowm9lzn4vyb0kbgc2ooqh7fbldb6dkwq.png)
And p is the probability of X happening.
The expected value of the binomial distribution is:
![E(X) = np](https://img.qammunity.org/2021/formulas/mathematics/college/66n16kmn896qth698tyf6rfu48vhaipkmv.png)
The standard deviation of the binomial distribution is:
![√(V(X)) = √(np(1-p))](https://img.qammunity.org/2021/formulas/mathematics/college/50rvo6hmelacol69fy9pzbmom4zmpsvsnd.png)
9% of all drivers arrested for DUI (Driving Under the Influence) are repeat offenders
This means that
![p = 0.09](https://img.qammunity.org/2021/formulas/mathematics/college/f9rp9n2grjbvg9enwecqebp5afiuy87k9l.png)
41 people arrested for DUI in Illinois are selected at random.
This means that
![n = 41](https://img.qammunity.org/2021/formulas/mathematics/college/4smqy5xx6sv89rqvzoq1hbvarjpf03220h.png)
a. What is the probability that exactly 3 people are repeat offenders?
This is P(X = 3).
![P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)](https://img.qammunity.org/2021/formulas/mathematics/college/mj488d1yx012m85w10rpw59rwq0s5qv1dq.png)
![P(X = 3) = C_(41,3).(0.09)^(3).(0.91)^(38) = 0.2158](https://img.qammunity.org/2021/formulas/mathematics/college/msulc88uc7ln8vqhtprrvjjsbvcp126a1p.png)
21.58% probability that exactly 3 people are repeat offenders
b. What is the probability that at least one person is a repeat offender?
Either none are repeat offenders, or at least one is. The sum of the probabilities of these outcomes is 1. So
![P(X = 0) + P(X \geq 1) = 1](https://img.qammunity.org/2021/formulas/mathematics/college/3y4i11vw3n4ugfq1uhu4er92ncdwchnt5i.png)
We want
.
Then
![P(X \geq 1) = 1 - P(X = 0)](https://img.qammunity.org/2021/formulas/mathematics/college/mrh0qjcttwa4i58cxv41mpzosdbbpdfl58.png)
In which
![P(X = 0) = C_(41,0).(0.09)^(0).(0.91)^(41) = 0.0209](https://img.qammunity.org/2021/formulas/mathematics/college/ig2i4b8igtd68drntesyfm37jaaqxyuv9s.png)
![P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0209 = 0.9791](https://img.qammunity.org/2021/formulas/mathematics/college/qksgc738xdcqp4yjplwde6ltabdxp9sihm.png)
97.91% probability that at least one person is a repeat offender
c. What is the mean number of repeat offenders?
![E(X) = np = 41*0.09 = 3.69](https://img.qammunity.org/2021/formulas/mathematics/college/1va2jhze9btflvv9gknslu1uyfmqp3nnzf.png)
d. What is the standard deviation of the number of repeat offenders?
![√(V(X)) = √(np(1-p)) = √(41*0.09*0.91) = 1.83](https://img.qammunity.org/2021/formulas/mathematics/college/p722vvno6qc7fk9u6dfvch4jzc52oo9x3m.png)