Respuesta :
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
Step-by-step 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.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
9% of all drivers arrested for DUI (Driving Under the Influence) are repeat offenders
This means that [tex]p = 0.09[/tex]
41 people arrested for DUI in Illinois are selected at random.
This means that [tex]n = 41[/tex]
a. What is the probability that exactly 3 people are repeat offenders?
This is P(X = 3).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 3) = C_{41,3}.(0.09)^{3}.(0.91)^{38} = 0.2158[/tex]
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
[tex]P(X = 0) + P(X \geq 1) = 1[/tex]
We want [tex]P(X \geq 1)[/tex].
Then
[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]
In which
[tex]P(X = 0) = C_{41,0}.(0.09)^{0}.(0.91)^{41} = 0.0209[/tex]
[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0209 = 0.9791[/tex]
97.91% probability that at least one person is a repeat offender
c. What is the mean number of repeat offenders?
[tex]E(X) = np = 41*0.09 = 3.69[/tex]
d. What is the standard deviation of the number of repeat offenders?
[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{41*0.09*0.91} = 1.83[/tex]
