This distribution is clearly non-normal. Skewed to the high values and has a standard deviation (250) larger than the mean (1000), which would predict negative values for the distribution.
Use the fact that means with many samples are nearly normal.
The mean (250) that is being compared against a limit is m=2,700,000/10,000 = 270.
The standard error of the nearly normal distribution of means would be
SE = sigma/sqrt(N) = 1000/sqrt(10000) =1000/100 = 10,
and the question becomes the percentile represented by
z = (270-250)/10 = 20/10 =2.0
P(z <= 2.0) = 0.977. Thus, only 0.023 = 2.3% probability of going over the limit of 270.