Respuesta :
This is a binomial problem. Binompdf(10,0.2,3)=0.201, about 20% chance
Using the binomial distribution, it is found that there is a 0.2013 = 20.13% probability that exactly 3 vehicles were exceeding the limit.
For each vehicle, there are only two possible outcomes. Either they exceed the limit, or they do not. The probability of a vehicle exceeding the limit is independent of any other vehicle, which means that the binomial distribution is used to solve this question.
Binomial probability distribution
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
In this problem:
- Sample of 10 vehicles, thus [tex]n = 10[/tex].
- 20% exceed the speed limit, thus [tex]p = 0.2[/tex].
The probability of exactly 3 exceeding the speed limit is P(X = 3), thus:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 3) = C_{10,3}.(0.2)^{3}.(0.8)^{7} = 0.2013[/tex]
0.2013 = 20.13% probability that exactly 3 vehicles were exceeding the limit.
A similar problem is given at https://brainly.com/question/24756209
