So I am having problems with this exercise. Can someone help me?
The waiting time $T$ in hours for the order in the pizzeria DaFranco has the following density function:
$f_T(X):=\left\{\begin{matrix} -\frac{3}{4} x(x-2)& 0\leq x\leq 2\\ 0& else \end{matrix}\right.$
Calculate the expected value and the standard deviation of the waiting time $T$ for your order. Also, using Chebyshev's inequality, give a lower bound for the probability that you have to wait between $30$ and $90$ minutes for your order.
So I have found $E[X]=1$ and $\sigma(x)=1/\sqrt{5}$
Now I am having problems with the second part of the problem.
What I thought:
$P(30 \leq X \leq 90)=P(30-E[X]\leq X-E[X]\leq 90-E[X])$
So what I thought is $P(30-E[X]\leq X-E[X]\leq 90-E[X])=P(X-E[X]\leq 90 -E[X])-P(X-E[X]\leq 30-E[X] )$
und we know that $P(|X-E[X]|\geq 90 -E[X])=P[|X-E[X]|\geq 90-E[X]]-P[X-E[X] \leq -(90-E[X])]$
and therefore
$P(|X-E[X]|\geq 90-E[X] )=P[|X-E[X]|\geq 90-E[X]]$ and therefore
$P(|X-E[X]|\leq 90-E[X] )=P[|X-E[X]|\leq 90-E[X]]$
So back to the problem
$P(X-E[X]\leq 90-E[X] )-P(X-E[X]\leq 30-E[X] )= \\ P(|X-E[X]|\leq 90-E[X] )-P(|X-E[X]|\leq 30-E[X] )\geq (1-\frac{Var(X)}{89})-P(|X-E[X]|\leq 30 )$
We have that
$P(|X-E[X]|\leq 30-E[X] )=1-P(|X-E[X]|\geq 30-E[X] ) \leq 1- (1-\frac{Var(x)}{29})=\frac{Var(x)}{29}$
and so $P(30 \leq X \leq 90) \\ = P(X-E[X]\leq 90-E[X] )-P(X-E[X]\leq 30-E[X] )\\ \geq (1-\frac{Var(X)}{89})-(\frac{Var(x)}{29})\\ =1-\frac{Var(x)}{29}-\frac{Var(X)}{89}$
Have I done any mistakes? Thanks for the help