[tex]\mathbb P(X>60)=\mathbb P\left(\dfrac{X-50}{10}>\dfrac{60-50}{10}\right)=\mathbb P(Z>1)[/tex]
Since approximately 68% of a (roughly) normal distribution falls within one standard deviation of the mean, that leaves 32% that falls outside, and by symmetry 16% should lie above one standard deviation of the mean.
