1. Human temperature is normally distributed with a mean of 98.6 and a known standard deviation of 1 degree. What is the probability that a single person drawn at random would have a temperature of greater than 99.8? Write the probability statement and solve.

\(P(X\ge 99.8 | \mu = 98.6, \sigma = 1)\)

1-pnorm(99.8,98.6,1)
## [1] 0.1150697

2. Human temperature is normally distributed with a mean of 98.6 and a known standard deviation of 1 degree. What is the probability that a random sample of 10 would have a mean temperature of greater than 99.8? Write the probability statement and solve.

\(P(X\ge 99.8 | \mu = 98.6, \sigma \bar{_x} = \frac{1}{\sqrt{10}})\)

#P(Xbar < 99.8 | Mu=98.6, Se= 1/sqrt(10))
Xbar=99.8
Mu=98.6
Se=1/sqrt(10)
1-pnorm(Xbar,Mu,Se)
## [1] 7.390116e-05

3. What is the 95% confidence interval for temperature in the above problem? Interpret it.

\(\bar{x}\pm Z_.975 \frac{\sigma}{\sqrt{n}}\)

Z=qnorm(.975)
lower = Xbar-Z*Se
upper = Xbar+Z*Se
mylist=c(lower,upper)
names(mylist)=c("Lower", "Upper")
mylist
##    Lower    Upper 
##  99.1802 100.4198

4. An administrator knows takes a random sample of 100 patients and asks them if they were pleased with the intake process. Exactly 47 patients were pleased. What is the 95% confidence interval satisfaction with the process?

\(\hat{p}\pm Z_.975 \frac{\hat{p}(1-\hat{p})}{\sqrt{n}}\)

phat=47/100
Z2=qnorm(.975)
N=100
Se2=sqrt(phat*(1-phat)/N)

lower2=phat-Z2*Se2
upper2=phat+Z2*Se2
mylist2=c(lower2,upper2)
names(mylist2)=c("Lower", "Upper")
mylist2
##     Lower     Upper 
## 0.3721784 0.5678216

5. An administrator wants a survey that will determine patient’s satisfaction with intake (yes or no) within 3% at a 95% confidence level. What size of sample should he/she take?

n=\(\lceil\frac{{Z^2}\sigma^2}{\epsilon^2}\rceil\)= \(\lceil\frac{{Z^2}\pi(1-\pi)}{\epsilon^2}\rceil\)

You should generally assume that \(\pi\) is .5, as it maximizes the required sample size.

MOE=.03
Z3=qnorm(.975)
Sigma2=sqrt(.5*.5)

n=ceiling(Z3^2*Sigma2^2/MOE^2)
n
## [1] 1068