Assuming the required randomness (reasonable), this is a standard statistics problem. For 70 jobs average intake is $700 with standard (sd) deviation of $15. But is this really true? It is not. It overstates the sd of the 700. You should square the sd (gives the variance), add them, then take the sq root. In your case 1.5^2 = 2.25. Variance of work = 22.5, sq root of work = 4.74. You now have the usual bell curve with average 700 and sd of 4.7. You want the probability of exceeding the average by about 20 x sd. Look up the result in a table. It is VERY low, essentially zero.
I get a sd of $12.55, but 7.968 standard deviations from the mean is still essentially impossible.
It would be far more likely that the tail of your distribution is actually thicker then Normal or that the pay for the various jobs are not really independent.
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