1. Complete problem #31 on page 163 (business venture).

You are involved in a venture.

Option 1: Loses $8000, x1= -8000 Option 2: Loses $0, x2= 0 Option 3: Makes $10000, x3=10000

To figure out the probabilities, just write out the fact that x1 is .5x2 and x3 is .3x2. Since this is a probability distribution function, the probabilities must sum to one.

.5x2 + x2 + .333x2 = 1= (11/6)x2 = x2=6/11 x2 = .545454 x1 = .272727 x3=.181818

We can then write out our probability mass function (PMF).

x = c(-8000,0, 10000)
px=c(6/22, 6/11, 6/33)

#verify pmf
sum(px)  #equals 1
## [1] 1
#calculate EX
EX=sum(x*px)
EX  #-363.6364
## [1] -363.6364
#Calculate Var(X)

VarX=sum(px*(x-EX)^2)  #35504132
VarX
## [1] 35504132
#Calculate SD(X)
SDX=sqrt(VarX)  # $5,958.534

2. The national 30-day case fatality rate for intracranial hemorrhage (ICH) surgical intervention is .16. The neurosurgical team performed 20 of these operations last year and experienced 6 deaths. Does sufficient evidence exist to suggest that the neurosurgical practice pattern requires review? In other words, what is the probability that this team would experience something this or more severe, i.e., 6 or more deaths in 20 trials if they were operating at the national rate of .16? Use an evidentiary standard that investigates if the event is rare (routinely defined as less than .05 probability.)

n=20
pi=.16
#P(Y>= 6 | n=20, pi = .16)
EY=n*pi
VY=n*pi*(1-pi)
SDY=sqrt(VY)
PY=1-pbinom(5,20,.16)
mylist=c(EY, VY, SDY, PY)
names(mylist)=c("E(Y)", "V(Y)", "SD(Y)", "P(Y)")
mylist
##       E(Y)       V(Y)      SD(Y)       P(Y) 
## 3.20000000 2.68800000 1.63951212 0.08699685

3. The rate of medication errors across the United States is 2 per 1000 orders. (Volume). A sample from your local hospital finds 5 errors in a sample of 2000. What is the probability that this hospital is within the US standard? In other words, what is the probability of having 5 or more errors in such a sample size if the hospital is operating at the national rate?

#Binomial, P(R>=5 | N = 2000, pi = 2/1000)
n2=2000
pi2=2/1000
PR=1-pbinom(4,n2, pi2)
ER=n2*pi2
VR=n2*pi2*(1-pi2)

#Poisson,  P(S>=5 | lambda x t = 4)
lambdat=4
PS=1-ppois(4,lambdat)
ES=lambdat
VS=ES

mylist2=c(PR, ER, VR, PS, ES, VS)
names(mylist2)=c("P(R)", "E(R)", "V(R)", "P(S)", "E(S)", "V(S)")

mylist2
##      P(R)      E(R)      V(R)      P(S)      E(S)      V(S) 
## 0.3711630 4.0000000 3.9920000 0.3711631 4.0000000 4.0000000

Complete problem #41 on page 212 (industrial accidents).

The annual number of industrial accidents occurring in a particular manufacturing plant is known to follow a Poisson distribution with mean 12.

#P(X=12 | Lambda = 12)
PXEQ12=dpois(12, 12)

#P(X<=12 | ...)
PXLTE12 = ppois(12,12)

#P(X>=15 | ...)
PXGTE15=1-ppois(14,12)

#P(10<=X<=15 | ...)
PRange=ppois(15, 12)-ppois(9,12)

#P(X<=k)=.99
Q=qpois(.99,12) 

mylist3=c(PXEQ12, PXLTE12, PXGTE15, PRange, Q)

names(mylist3)=c("P(X=12)", "P(X<=12)", "P(X>=15)", "P(10<=X=15)", "P(X<=k)=.99")

mylist3
##     P(X=12)    P(X<=12)    P(X>=15) P(10<=X=15) P(X<=k)=.99 
##   0.1143679   0.5759652   0.2279755   0.6020235  21.0000000