I believe that you have not put enough information in the question like what is the mean all you told us was that its with in 2.5 of the mean.
chebs thm
P(M-OK< X < M+Ok)>= 1- 1/k^2
a) Chebyshev's inequality tells us that:
P(|x-mu|>= k sigma) <= 1/k^2
so P(|x-mu|) >= 2.5 sigma) <= 1/2.5^2 = 0.16
So P(|x-mu| < 2.5 sigma) = 1-0.16 = 0.84.
b) here I presume that we are to assume normality, so we look 2.5 up
in a table of the standard normal distribution to get:
So P(|x-mu| < 2.5 sigma) = 0.9876.
RonL