Respuesta :
Answer:
a) [tex] P(X \leq 2) = \frac{2+1}{3+1}= 0.75[/tex]
b) [tex] P(X \geq 1) = 1- P(X <1) =1-\frac{1+1}{3+1}= 1-0.5=0.5[/tex]
c) [tex]P(-0.5 <X <1.5)= P(X<1.5)- P(X<-0.5)= \frac{1.5+1}{4} -\frac{-0.5+1}{4}=0.625-0.125= 0.5[/tex]
d) [tex] f(x) = \frac{1}{3+1}= 0.25[/tex]
Step-by-step explanation:
Let X the random variable of interest and we know that the distribution is given by:
[tex]X \sim Unif (a= -1, b=3)[/tex]
And for this problem we can use the cumulative distribution function in order to solve the items:
[tex] F(x) =\frac{x-a}{b-a}, a\leq x \leq b[/tex]
Part a
We want to find this probability:
[tex] P(X \leq 2) = \frac{2+1}{3+1}= 0.75[/tex]
Part b
[tex] P(X \geq 1) = 1- P(X <1) =1-\frac{1+1}{3+1}= 1-0.5=0.5[/tex]
Part c
[tex]P(-0.5 <X <1.5)[/tex]
And we can calculate the probability with this difference:
[tex]P(-0.5 <X <1.5)= P(X<1.5)- P(X<-0.5)= \frac{1.5+1}{4} -\frac{-0.5+1}{4}=0.625-0.125= 0.5[/tex]
Part d
Since we have a continuous distribution the the probability for an unique value would be:
[tex] f(x) = \frac{1}{3+1}= 0.25[/tex]
