How can I use Chebyshev's Inequality to show that P[0 < X < 3] >= 0.75?
note: X ~ Exp (lamba = 1)
and does the Cheby. inequality improve as k --> infinity?
it's a little confusing. can someone help?
There are things you need to make it your business to know:
1. The statement of Chebyshev's Inequality.
2. The mean of an exponential distribution.
3. The standard deviation of an exponential distribution.
Once you know these things you need to see that:
$\displaystyle \Pr(0 < X < 3) = \Pr(0 < X < \mu + 2 \sigma)$
$\displaystyle = \Pr(\mu - 2 \sigma < X < \mu + 2 \sigma)$
since $\displaystyle \mu - 2 \sigma < 0$ and $\displaystyle \Pr(X < 0) = 0$
$\displaystyle = \Pr( - 2 \sigma < X - \mu < 2 \sigma) = \Pr(|X - \mu| < 2 \sigma) $
where $\displaystyle \mu$ and $\displaystyle \sigma$ are respectively the mean and standard deviation of X.