The 70th percentile is the same as the cutoff recovery time [tex]t[/tex] such that every point above this time falls in the longest 30%, i.e.
[tex]\mathbb P(X>t)=0.30[/tex]
Transform to the standard normal distribution:
[tex]\mathbb P\left(\dfrac{X-5.9}{2.5}>\dfrac{t-5.9}{2.5}\right)=\mathbb P(Z>t^*)[/tex]
where [tex]t^*[/tex] is the z-score corresponding to the cutoff time [tex]t[/tex], which is approximately [tex]t^*\approx0.5244[/tex]. Solve for [tex]t[/tex]:
[tex]t^*\approx0.5244=\dfrac{t-5.9}{2.5}\implies t\approx7.21\text{ days}[/tex]
