Normal distribution

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?

Answer 1

If written as 5%...

The z-score for the 95th percentile is 1.645.

So any score greater than x=μ+z*σ=265+1.645*45=339.025

One would assume the weight is measured in whole lbs, so 340 lbs is likely the answer they are looking for.

1. Jason