I have a question about the given solution of this problem.
Problem: A farmer uses $1000$ gallons of water on average per day. The variance of the water rate does not exceed $40000$ gallons. What's the probability that on any given day the water rate will not exceed $2000$ gallons? Use Chebyshev's inequality.
Given solution: Let $X$ be the daily water rate in gallons.
Because the interval $0 \le X \le 2000$ is symmetrical in terms of $EX=1000$, we can apply Chebyshev's inequality $$P\left(X\le2000\right)=P\left(0\le X\le2000\right)=P\left(\left|X-1000\right|\le1000\right)\ge1-\frac{40000}{1000^{2}}=0.96$$
My questions:
- Why do we need symmetry here to apply Chebyshev's inequality?
- What if we need to find the probability that on any given day the water rate will not exceed $1500$ or $2500$ gallons?
I appreciate any help.
Chebishev inequality gives an approximated bound for the probability of the following event
$$|X-\mu|<\epsilon$$
that is
$$\mu-\epsilon\leq X\leq \mu+\epsilon$$
which is a symmetric interval w.r.t. the mean
Are you sure they want you to calculate $P(X<1500)$ using Chebischev inequality? I think only $P(X<2500)$ can be calculated using this inequality.
EDIT: if you want to calculate $P(X<1500)$ given that the mean is 1000 it is clear that Chebishev inequality can give an approx of the probability that
$$|X-1000|<500$$
that is
$$500<X<1500$$
and this is not what you want...
On the contrary, if you are interested in $P(X<2500)$ you can set
$$|X-1000|<1500$$ that is
$$-500<X<2500$$
and this is good for you as it is the same as
$$0<X<2500$$
given that $X$ cannot be negative.