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

Answer:

$P(X>4)=P(\frac{X-\mu}{\sigma}>\frac{4-\mu}{\sigma})=P(Z>\frac{4-5.5}{2.3})=P(z>-0.652)$

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

$P(z>-0.652)=1-P(z$

Step-by-step 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:

$z=\frac{x-\mu}{\sigma}$

And using this formula we got:

$P(X>4)=P(\frac{X-\mu}{\sigma}>\frac{4-\mu}{\sigma})=P(Z>\frac{4-5.5}{2.3})=P(z>-0.652)$

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

$P(z>-0.652)=1-P(z$