Assume the statistic is (265±40)kg. Assume also that we are looking for the top 5%.
If X is the required weight, then Z=(X-265)/40, where Z is the Z-score for 95%=1.645.
This tells us Z for a deviation up to 1.645 times the standard deviation from the mean. Beyond this (N(Z)>95%) we can find X, the weight required to be in the top 5%.
X-265=40×1.645, X=265+65.8=330.8kg.