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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.)

Verified Answer :

LammettHash

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 \(t\):

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

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