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 55 and 1010 parts per million

a. Find E(x) and interpret its value
b. Compute P(2.875 x35)
c. Computn Plx<4.125)

Answer 1

a)
`E(A)=(1+6)/(2)=3.5 ppm`

b)
`P(2.875 <X < 3.5) = F(3.5) -F(2.875) = (3.5-1)/(5)- (2.875-1)/(5)= (1)/(8)= 0.125`

c)
`P(X<4.125) = F(4.125) = (4.125-1)/(5)= 0.625`

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
`E(A) = (5+10)/(2)= 7.5`

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:

`A\sim Uniform(1 ,6)`

Part a

For this uniform distribution the expected value is given by
`E(X) =(a+b)/(2)` where X is the random variable, and a,b represent the limits for the distribution. If we apply this for our case we got:

`E(A)=(1+6)/(2)=3.5 ppm`

Part b

For this case we can use the cumulative distribution function for the uniform distribution given by:

`F(X=x)= (x-a)/(b-a) = (x)/(6-1) =(x-1)/(5) , 1 \leq X \leq 6`

And we want this probability:
`P(2.875 <X < 3.5) = F(3.5) -F(2.875) = (3.5-1)/(5)- (2.875-1)/(5)= (1)/(8)= 0.125`Part c

For this case we want this probability:

`P(X<4.125) = F(4.125) = (4.125-1)/(5)= 0.625`