Respuesta :
Answer:
a) [tex]E(A)=\frac{1+6}{2}=3.5 ppm[/tex]
b) [tex] P(2.875 <X < 3.5) = F(3.5) -F(2.875) = \frac{3.5-1}{5}- \frac{2.875-1}{5}= \frac{1}{8}= 0.125[/tex]
c) [tex] P(X<4.125) = F(4.125) = \frac{4.125-1}{5}= 0.625[/tex]
Step-by-step explanation:
If we work with the limits defined from 5 to 10 then part b and c from this question not makes sense. If we work with the limits 1 and 6 all the parts for the question makes sense because if we work with 5 and 10 the only thing that we can find is the expected value [tex] E(A) = \frac{5+10}{2}= 7.5[/tex]
Assuming the following correct question : "A publication released the results of a study of the evolution of a certain mineral in the Earth's crust. Researchers estimate that the trace amount of this mineral x in reservoirs follows a uniform distribution ranging between 1 and 6 parts per million"
Solution to the problem
Let A the random variable that represent " amount of the mineral x ". And we know that the distribution of A is given by:
[tex]A\sim Uniform(1 ,6)[/tex]
Part a
For this uniform distribution the expected value is given by [tex]E(X) =\frac{a+b}{2}[/tex] where X is the random variable, and a,b represent the limits for the distribution. If we apply this for our case we got:
[tex]E(A)=\frac{1+6}{2}=3.5 ppm[/tex]
Part b
For this case we can use the cumulative distribution function for the uniform distribution given by:
[tex] F(X=x)= \frac{x-a}{b-a} = \frac{x}{6-1} =\frac{x-1}{5} , 1 \leq X \leq 6[/tex]
And we want this probability:[tex] P(2.875 <X < 3.5) = F(3.5) -F(2.875) = \frac{3.5-1}{5}- \frac{2.875-1}{5}= \frac{1}{8}= 0.125[/tex]Part c
For this case we want this probability:
[tex] P(X<4.125) = F(4.125) = \frac{4.125-1}{5}= 0.625[/tex]
