Question

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 1

Answer: 0.9713

Step-by-step explanation:

Given : Mean : $\mu = 5.2\text{ day}$

Standard deviation : $\sigma = 1.7\text{ days}$

The formula of z -score :-

$z=\dfrac{X-\mu}{\sigma}$

At X = 2 days

$z=\dfrac{2-5.2}{1.7}=-1.88235294118\approx-1.9$

Now, $P(X>2)=1-P(X\leq2)$

$=1-P(z<-1.9)=1- 0.0287166=0.9712834\approx0.9713$

Hence, the probability of spending more than 2 days in recovery = 0.9713

Answer 2

Answer:

There is a 98.54% probability of spending more than 2 days in recovery.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 5.7, \sigma = 1.7$

What is the probability of spending more than 2 days in recovery?

This probability is 1 subtracted by the pvalue of Z when X = 2. So:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{2 - 5.7}{1.7}$

$Z = -2.18$

$Z = -2.18$ has a pvalue of 0.0146.

This means that there is a 1-0.0146 = 0.9854 = 98.54% probability of spending more than 2 days in recovery.