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

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

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 = (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 = (X - \mu)/(\sigma)`

`Z = (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.

Answer 2

Answer: 0.9713

Explanation:

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

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

The formula of z -score :-

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

At X = 2 days

`z=(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