Question

x is exponential random variable
f(x) = (1/θ)e^-x/θ
given: P(x < 2.40) = .520
find: P(x > 4.36)

Answer 1

You need to integrate f(x) from 0 to 2.40 and set that equal to .520. You will have an equation you can solve for theta. Once you find theta integrate the pdf again from 4.36 to infinity, or take one minus the integration from 0 to 4.36 whichever you prefer.

Answer 2

sorry i dont know why it posted that many times

Answer 3

i keep getting a negative number for the theta
-1.5694
i end up with a negative answer which i don't think it's correct

Answer 4

make that theta = -3.67

Answer 5

-e^2.40/theta - [-e^-0/theta] = .520
-e^-2.40/theta + 1 = .520
e^2.40/theta = .48
-2.40/theta = ln(.48)
theta = 3.2699

i was trying to ln the limit 0 at first and that's why i was having problem.
yay for perseverance :)

Answer 6

-e^-2.40/theta - [-e^-0/theta] = .520
-e^-2.40/theta + 1 = .520
e^-2.40/theta = .48
-2.40/theta = ln(.48)
theta = 3.2699

-e-x/theta from 4.36 to infinity
-e^-inf/3.269 - [-e^-4.36/3.269]
= 0 + .2635
=.2635

Answer 7

yep I think that's correct