Question 1041007
Chebyshev's Theorem:  {{{P(abs(X - mu) <= k*sigma) >= 1-1/k^2}}}

==> {{{P(abs(X - 5.02) <= 0.09k) >= 1-1/k^2}}}

Letting k = 2,

==> {{{P(abs(X - 5.02) <= 0.18) >= 1-1/2^2 = 3/4}}}
==> {{{P(-0.18 <= X - 5.02 <= 0.18) >= 3/4}}}
==>  {{{P(4.84 <= X <= 5.20) >= 3/4 = 0.75}}}

==> the range in which at least 75% of the values fall is the interval 
[4.84, 5.20].