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This is probably very simple, but I could not find the answer. How to solve? I tried Reduce as well to no avail.

dist[j_] := CDF[BinomialDistribution[1001, 1/2], j]
Solve[dist[j] <= 1/2]
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    $\begingroup$ InverseCDF[BinomialDistribution[1001, 1/2], 1/2]. Use the function intended for the purpose. (The situation is similar to this answer.) $\endgroup$
    – J. M.'s missing motivation ♦
    Commented Oct 20, 2018 at 13:34
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    $\begingroup$ Or Median[BinomialDistribution[1001, 1/2]] $\endgroup$
    – Bob Hanlon
    Commented Oct 20, 2018 at 16:12

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