Question

The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.5 days and a standard deviation of 2.3 days. What is the probability of spending more than 4 days in recovery? (Round your answer to four decimal places.)

Answer 1

`P(X>4)=P((X-\mu)/(\sigma)>(4-\mu)/(\sigma))=P(Z>(4-5.5)/(2.3))=P(z>-0.652)`

And we can find this probability with the complement rule and with the normal standard table and we got:

`P(z>-0.652)=1-P(z<-0.652)= 1-0.2572= 0.7428`

Explanation:

Let X the random variable that represent the patient recovery time of a population, and for this case we know the distribution for X is given by:

`X \sim N(5.5,2.3)`

Where
`\mu=5.5` and
`\sigma=2.3`

We are interested on this probability

`P(X>4)`

And we can solve this problem using the z score formula given by:

`z=(x-\mu)/(\sigma)`

And using this formula we got:

`P(X>4)=P((X-\mu)/(\sigma)>(4-\mu)/(\sigma))=P(Z>(4-5.5)/(2.3))=P(z>-0.652)`

And we can find this probability with the complement rule and with the normal standard table and we got:

`P(z>-0.652)=1-P(z<-0.652)= 1-0.2572= 0.7428`