MathCastle
Statistics & DataSuperchallengeThe Normal Distribution

The Normal Distribution · Superchallenge problem

The normal density f(x)=1σ2πe−(x−μ)2/2σ2f(x) = \dfrac{1}{\sigma\sqrt{2\pi}}e^{-(x-\mu)^2/2\sigma^2} has its peak at x=μx = \mu. What fraction of the distribution lies where the density is at least half of its peak value? Round to two decimal places.