Since [tex]145+2\times11=167[/tex], which means the corresponding z-score for Seif's test score is [tex]z=2[/tex]. The empirical rule asserts that for any normal distribution, approximately 95% of it lies within two standard deviations of the mean, i.e. [tex]\mathbb P(|Z|<2)=0.95[/tex], which leaves 5% outside of that range, and specifically 2.5% to either side.
This means
[tex]\mathbb P(Z<2)=\mathbb P(Z<-2)+\mathbb P(|Z|<2)=0.025+0.95=0.975[/tex]
In other words, a test score of 167, which corresponds to a z-score of 2, marks the 97.5th percentile.
