Question

The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.9 days and a standard deviation of 2.5 days.
What is the 70th percentile for recovery times? (Round your answer to two decimal places.)

Answer 1

The 70th percentile is the same as the cutoff recovery time $t$ such that every point above this time falls in the longest 30%, i.e.

$\mathbb P(X>t)=0.30$

Transform to the standard normal distribution:

$\mathbb P\left(\dfrac{X-5.9}{2.5}>\dfrac{t-5.9}{2.5}\right)=\mathbb P(Z>t^*)$

where $t^*$ is the z-score corresponding to the cutoff time $t$, which is approximately $t^*\approx0.5244$. Solve for $t$:

$t^*\approx0.5244=\dfrac{t-5.9}{2.5}\implies t\approx7.21\text{ days}$