I recently got interested in this topic of standard deviation. My TA did not have any time to go over this topic so I was trying to teach myself it recently. My TA said if he had more time he would teach us a problem just like this although it was not part of the ciriculum. He said you can look over this particular question if I wanted to but he will not go over it. Can someone help me with this? He said this would be a two stage type problem.
$X$ is an exponential random variable. Let $P(X<2.16)=.4215$. Find $x_0$ such that $P(X>x_0)=.2184$
My answer:
We can let $x_1$ be $2.16$. My TA said this. Now solving we get:
$P(X<2.16)=.4215$
$=.5000-.4215=.0785$
$z_1=.202$ After using interpolation and the $z$ value table (the one that only has positive values).
Now use $Z=\frac{x_0-\mu}\sigma$
$.202=\frac{2.16-0}\sigma$
$\sigma=10.69$
Now, $P(X>x_0)=.2184$
$.5000-.2184=.2816$
$z_0=.782$ using interpolation again.
Now use $Z=\frac{x_0-\mu}\sigma$
$.782=\frac{x_0-0}{10.69}$
$8.36$ which is my answer.
Can someone please tell me if this is correct? If so, is there another way of doing it? I know this topic was not part of the ciriculum but, I really wanted to know how to solve this. It seems like an interesting problem to know for later on.
See the exponential distribution.
Assume $X\sim \text{Exp}(\lambda)$, so $P(X\leq x) = 1-exp(-\lambda x)\,,\quad x>0$.
$P(X< 2.16) = 1-\exp(-2.16 \lambda ) = .4215$, so
$\log(1-.4215 ) = -2.16\lambda$.
$ \lambda=\log(1-0.4215 )/ -2.16 \approx 0.253387 $
$P(X>x_0) = 0.2184$
$\therefore \exp(-\lambda x_0) = 0.2184$
$ x_0 = 2.16 \frac{\log(0.2184)}{\log(1-0.4215)} \approx 6.0044$