IIRC, the correct solution to this is in sqrt(N) time; for a 100-floor building, you drop at floors 10, 20, 30, 40, 50, 60, 70, 80, 90, and 100, in that order. In general, this is sqrt(N) floors, spaced sqrt(N) floors apart. The bulb will break at floor sqrt(N) * X [10X in this example]. Now, drop the bulb from the approximately sqrt(N) [10] floors between floor sqrt(N) * (X-1) [10 * (X-1)] and sqrt(N)* X - 1 [10 * X - 1], inclusive. When it breaks, you have the exact tolerance rounded down to the nearest floor, and you've done it with at worst 2 * sqrt(N) drops.
By the way, this is a great example of a dumb interview question; it's not particularly obvious that there is a solution in sqrt(N) worst case time, but once you're told that there is one (and, perhaps, that binary search + linear search is linear search in the worst case), it's not too difficult to get it in sqrt(N).
Also, note that average case running time analysis is meaningless if you don't have a distribution for the input; simply assuming that the tolerance is distributed uniformly at random from 1 to 100 just because you have a 100-floor building makes little sense.