Well here are a couple of things: you show 99 records on your excel spreadsheet, with an average of 709 and a standard deviation of 1073.

With that large of a standard deviation and that average you can't assume a normal distribution so you need to use the T distribution.

Using the T distribution at a 95% confidence level I come up with an estimated average of [495, 923] which results in a number of jobs in the top 1000 from 495,000 to 923,000 with a 95% certainty.

Just for kicks I wanted to see what would happen if you sampled 500 times but had that same standard deviation. At that point your spread would be about +/- 100 for the average value of X.

If you are looking for a singular number you'll need to check all 1,000 of them, otherwise it will be an estimated range.