Question 465416
Chebyshev's rule {{{P(abs(X - mu) <= k*sigma) >= 1 - 1/k^2)}}}.

==>{{{P(abs(X - 180) <= 32k) >= 1 - 1/k^2 = 0.8889)}}}.

Solving the equation {{{1 - 1/k^2 = 0.8889)}}}, we get

{{{1/k^2 = 0.1111}}} ==> {{{k^2 = 9}}} ==> k = 3.

==> {{{P(abs(X - 180) <= 32*3 = 96) >= 0.8889)}}}, or 
the interval in question is 

{{{abs(X - 180) <= 96}}}

<==> {{{-96 <= X - 180 <= 96}}}, or {{{84 <= X <= 276}}}.