I want to solve the following exercise:
Given is an expected value of random variable X, E(X) = 200 at time point t and a standard deviation of 50.
(a) Find the 95% confidence interval of the random variable at this time point t.
(b) What is the probability that the value of the Random Variable is at least 300?
There is no probability distribution and no further data given.
My approach: As one of the comments points out, I am using Chebyshev's theorem:
$$P(|X-\mu|\ge k\sigma)\le\frac{1}{k^2}$$
I find the following results - are these correct?
(a) For $1 - \frac{1}{k^2}$ = 0.95 I find k = 4.4721 and hence with $k\sigma-\mu \leq X \leq k\sigma+\mu$ I find the lower bound of 23.6 and the upper bound of 423.6068 for the 95% confidence interval.
(b) Not quite sure here, but my approach would be using $$|X-200|\ge 100$$ with $\sigma = 50$ we get $\frac{1}{2^2}$ and hence a probability of 25% for a value of X being at least 300. Is that correct?