[tex]\mathbb P(X>79)=\mathbb P\left(\dfrac{X-70}3>\dfrac{79-70}3\right)=\mathbb P(Z>3)[/tex]
Recall that the empirical rule for normal distributions says that approximately 99.7% of the distribution falls within three standard deviations of the mean. Therefore 0.3% must lie outside that range, with 0.15% on either side.
So, [tex]\mathbb P(Z>3)\approx0.15\%[/tex].
