The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.2 days and a standard deviation of 1.7 days. What is the probability of spending more than 2 days in recovery? (Round your answer to four decimal places.)

Answer: 0.9713Step-by-step explanation:Given : Mean : [tex]\mu = 5.2\text{ day}[/tex]Standard deviation : [tex]\sigma = 1.7\text{ days}[/tex]The formula of z -score :-[tex]z=\dfrac{X-\mu}{\sigma}[/tex]At X = 2 days[tex]z=\dfrac{2-5.2}{1.7}=-1.88235294118\approx-1.9[/tex]Now, [tex]P(X>2)=1-P(X\leq2)[/tex][tex]=1-P(z<-1.9)=1- 0.0287166=0.9712834\approx0.9713[/tex]Hence, the probability of spending more than 2 days in recovery = 0.9713