Question 472559
The distribution is not necessarily normal, so we use Chebyshev's inequality, 

{{{P(abs(X - mu) <= k*sigma) >= 1-1/k^2}}}.  Since in this case, k = 2, we get


{{{P(abs(X - 150) <= 50) >= 1-1/4 = 3/4}}}.

Hence at least 75% of the distribution lies with 100 and 200.

Note that if the distribution is normal, then by the empirical rule, around 95% would lie within 2 sd's of the mean, or between 100 and 200.