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((X-5.9)/(2.5)>(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=(t-5.9)/(2.5)\implies t\approx7.21\text{ days}`