(uniform) suppose x follows a continuous uniform distribution from 3 to 9. (a) write down the pdf of x. (b) find p(x ≤ 6). round your answer to 4 decimals. (c) find p(4 < x ≤ 6). round your answer to 4 decimals. (d) determine the conditional probability p(x > 4 | x ≤ 6). round your answer to 4 decimals

a)  P(x) = {0.1667 for x in (3,9) ,   0 otherwise}

b) P(x≤6) = 0.5

c) P(4 < x ≤ 6) = 0.3333

d) P(X > 4 | X ≤ 6) = 0.3333 or 1 *see below*

Step-by-step explanation:

a)

Since it is a uniform distribution you can imagine it as a rectangle which extends from a = 3 to b = 9 and whose total area is 1.

Therefore the height must be equal to the inverse of the side:

P(X) = 1/(b-a) = 1/(9-3)= 1/6       for every x in (3,9)  and 0 otherwise

P(X) = 0.1667

b)

To find this probability we must find the area of the rectangle which extends from 3 to 6 and whose height is equal to P(x)

That is:

P(X ≤ 6) = (6-3)/(9-3) = 3/6 = 1/2 = 0.5

That make sense since 6 is half way between 3 and 9

c)

Similar to b) but now we consider tha base of the rectangle from 4 to 6:

P(4 < X ≤ 6) = (6-4)/(9-3) = 2/6 = 1/3 = 0.3333

d)

If  '' | '' means " or ", the probability is one, since every number in the range (3,9) is either higher than 4 or less than 6.

If " | " means " and " the answer is the same as in c , that is because 4<x≤6 is the set of numbers which are higher than 4 and at most 6 which can be expressed as x>4 and x≤6

y=x+9

?

step-by-step explanation:

c) yes, because of asa or aas