Question

please the question. need R code to solve the question

Assume that the literature on this topic suggests that the distribution of days of hospital stay are normally distributed with a mean of 5 and a standard deviation of 3. Recent publications have indicated that hypervirulent strains of C. Difficile are on the rise. Such strains are associated with poor outcomes, including extended hospital stays.An investigator is interested in showing that the average hospital stay durations have increased versus published literature. He has a sample of 10 patients from his hospital. If the published data are consistent with the truth, what is the probability that the sample mean in his sample will be greater than 7 days?

(the question before this question is:  Assume that the literature on this topic suggests that the distribution of days of hospital stay are normally distributed with a mean of 5 and a standard deviation of 3. What percentage of patients are in the hospital for less than a week? (Note: in percent, it is a number between 0 and 100)

and I know this answer is 74.75075% by pnorm(7,5,3) )

Answer 1

Given that, mean $(\mu)$ = 5 and standard deviation $(\sigma)$ = 3

sample size ( n ) = 10

Therefore, mean and standard deviation of the sampling diatribution of sample mean are,

$\mu_{\bar x}= \mu = 5$

$\sigma_{\bar x}= \frac { \sigma}{\sqrt n } = \frac { 3}{\sqrt { 10}}$

a) P(X < 7 days) = 0.7475

R-command: pnorm(7, 5, 3) = 0.7475

Therefore, 74.75% of patients are in the hospital for less than a week.

b) We want to find, $P(\bar x > 7)$

$P(\bar x > 7) = 1 - P(\bar x < 7) = 0.0175$

R-command: pnorm(7, 5, (3/sqrt (10))) = 0.0175

Therefore, the probability that the sample mean in his sample will be greater than 7 days is 0.0175

( In percentage = 1.75%)