Question 1037363
From Chebyshev's inequality, 

{{{P(abs(X - mu) <= nd) >= 1-1/n^2)}}}.

==> {{{P(abs(X - 101) <= 14n) >= 1-1/n^2)}}}

Letting n = 2, we get

{{{P(abs(X - 101) <= 28) >= 1-1/2^2 = 1-1/4 = 3/4}}},

which is what is needed by the problem.

==> the interval is {{{abs(X - 101) <= 28}}}, or {{{-28 <= X - 101 <= 28}}}, or {{{73 <= X <= 128}}}.