Question

The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days.

What is the probability of spending more than two days in recovery?

a. 0.0580

b. 0.8447

c. 0.0553

d. 0.9420

Short answer

The correct answer is option dthat is value of$P(X>2)$is$0.9420$.

Step-by-step solution

Given information

Given in the question that

Mean$=5.3$

Standard deviation$=2.1$

Step: Solution

Here we need to find the probability of spending more than two days in recovery

$X~N(5.3,2.1)$

So, the probability can be calculated as

$P(x>2)=1-P(x<2)$

$=1-P\left(\frac{x-\mu }{\sigma }<\frac{2-\mu }{\sigma }\right)$

$=1-P\left(Z<\frac{2-5.3}{2.1}\right)$

$=1-P(Z<-1.5714)$

$=1-0.0580\left\{\begin{array}{l}\text{From standard}\\ \text{normal table}\end{array}\right\}$

$=0.9420$