the form of a normal distribution with a mean of 265 and a standard deviation of 45, if an athlete wants to be in the top five of the distribution, what weight must he lift

Question

In a weightlifting championship that takes the form of a normal distribution with a mean of 265 and a standard deviation of 45, if an athlete wants to be in the top five of the distribution, what weight must he lift to achieve his desire?

Comments

Do you mean top 5% or in the top 5 of some unspecified number of athletes?

What are the units of weight: lbs, kg? I assume the statistic is the weight.

Answer 1

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.