A new test for multinucleoside-resistant (MNR) human immunodeficiency virus type 1 (HIV-1) variants was recently developed. The test maintains 96% sensitivity, meaning that, for those with the disease, it will correctly report “positive” for 96% of them. The test is also 98% specific, meaning that, for those without the disease, 98% will be correctly reported as “negative.” MNR HIV-1 is considered to be rare (albeit emerging), with about a .1% or .001 prevalence rate. Given the prevalence rate, sensitivity, and specificity estimates, what is the probability that an individual who is reported as positive by the new test actually has the disease? If the median cost (consider this the best point estimate) is about $100,000 per positive case total and the test itself costs $1000 per administration, what is the total first-year cost for treating 100,000 individuals?
Event A1 -> Actual positive
Event A2 -> Actual negative
Event B -> Positive test
P(A1) = .001
P(B | A1) = 0.96
P(B | A2) = 0.98
P(A1 | B) = ?
Using Bayes
P(A1 | B) = P (B|A1)P(A1) / P(B|A1)P(A1)+P(B|A2)P(A1)
paste("Probability that an individual who is reported as positive by the new test actually has the disease is ",round(0.96*0.001/(0.96*0.001+0.98*0.001),4)*100,"%")## [1] "Probability that an individual who is reported as positive by the new test actually has the disease is 49.48 %"cost_per_case = 100000+1000=101000
Out of 100000 individuals based on .1% prevalence rate 100 will get HIV positive. With sensitivity 96% actual HIV+ cases will be 96 and 4 will be false positive.
100000 - 100 = 99900 are true HIV-. With 98% specificity, 99900*0.98= 97902 are true -ve leaving (99900-97902=1998) as false -ve.
So actual positive cases will be 96 + 1998.
paste("Total first year cost would be ", (96+1998)*101000)## [1] "Total first year cost would be 211494000"The probability of your organization receiving a Joint Commission inspection in any given month is .05. What is the probability that, after 24 months, you received exactly 2 inspections? What is the probability that, after 24 months, you received 2 or more inspections? What is the probability that your received fewer than 2 inspections? What is the expected number of inspections you should have received? What is the standard deviation?
What is the probability that, after 24 months, you received exactly 2 inspections?
size <- 24
prob <-0.05
x<-2
dbinom(x,size,prob)## [1] 0.2232381What is the probability that, after 24 months, you received 2 or more inspections?
# 1- (P(1) +P(0))
1- (dbinom(1,size,prob) + dbinom(0,size,prob))## [1] 0.3391827What is the probability that your received fewer than 2 inspections?
dbinom(1,size,prob) + dbinom(0,size,prob)## [1] 0.6608173What is the expected number of inspections you should have received?
size*prob## [1] 1.2What is the standard deviation?
sqrt(size*prob*(1-prob))## [1] 1.067708You are modeling the family practice clinic and notice that patients arrive at a rate of 10 per hour. What is the probability that exactly 3 arrive in one hour? What is the probability that more than 10 arrive in one hour? How many would you expect to arrive in 8 hours? What is the standard deviation of the appropriate probability distribution? If there are three family practice providers that can see 24 templated patients each day, what is the percent utilization and what are your recommendations?
What is the probability that exactly 3 arrive in one hour?
dpois(3,10)## [1] 0.007566655What is the probability that more than 10 arrive in one hour?
1-ppois(10,10)## [1] 0.4169602How many would you expect to arrive in 8 hours?
per_hour <- 10
8*per_hour## [1] 80What is the standard deviation of the appropriate probability distribution?
sqrt(10)## [1] 3.162278If there are three family practice providers that can see 24 templated patients each day, what is the percent utilization and what are your recommendations?
3*24 = 72 patients can be served with 8 hours working. But from above we expect 80 to be expected patients. 11.11% over utilization, so either hire more staff or extend work hours
80/72*100## [1] 111.1111Your subordinate with 30 supervisors was recently accused of favoring nurses. 15 of the subordinate’s workers are nurses and 15 are other than nurses. As evidence of malfeasance, the accuser stated that there were 6 company-paid trips to Disney World for which everyone was eligible. The supervisor sent 5 nurses and 1 non-nurse. If your subordinate acted innocently, what was the probability he/she would have selected five nurses for the trips? How many nurses would we have expected your subordinate to send? How many non-nurses would we have expected your subordinate to send?
What was the probability he/she would have selected five nurses for the trips?
x<-5
m<-15
n<-15
k<-6
dhyper(x,m,n,k,log=FALSE)## [1] 0.07586207How many nurses would we have expected your subordinate to send?
15 * 6 / (15 + 15)## [1] 3How many non-nurses would we have expected your subordinate to send?
15 * 6 / (15 + 15)## [1] 3The probability of being seriously injured in a car crash in an unspecified location is about .1% per hour. A driver is required to traverse this area for 1200 hours in the course of a year. What is the probability that the driver will be seriously injured during the course of the year? In the course of 15 months? What is the expected number of hours that a driver will drive before being seriously injured? Given that a driver has driven 1200 hours, what is the probability that he or she will be injured in the next 100 hours?
What is the probability that the driver will be seriously injured during the course of the year?
(1-(1-0.001)^1200) ## [1] 0.6989866In the course of 15 months?
(1-(1-0.001)^1500) ## [1] 0.7770372What is the expected number of hours that a driver will drive before being seriously injured?
1/0.001## [1] 1000Given that a driver has driven 1200 hours, what is the probability that he or she will be injured in the next 100 hours?
pgeom(100, 0.001)## [1] 0.09611265You are working in a hospital that is running off of a primary generator which fails about once in 1000 hours. What is the probability that the generator will fail more than twice in 1000 hours? What is the expected value?
What is the probability that the generator will fail more than twice in 1000 hours?
1 - ppois(2,1)## [1] 0.0803014What is the expected value?
\(\lambda = 1\)
A surgical patient arrives for surgery precisely at a given time. Based on previous analysis (or a lack of knowledge assumption), you know that the waiting time is uniformly distributed from 0 to 30 minutes. What is the probability that this patient will wait more than 10 minutes? If the patient has already waited 10 minutes, what is the probability that he/she will wait at least another 5 minutes prior to being seen? What is the expected waiting time?
What is the probability that this patient will wait more than 10 minutes?
q <- 10
min <- 0
max <- 30
1- punif(q,min,max)## [1] 0.6666667If the patient has already waited 10 minutes, what is the probability that he/she will wait at least another 5 minutes prior to being seen?
PA<-1-punif(15,10,30)
PB<-1-punif(10,10,30)
PA/PB## [1] 0.75What is the expected waiting time?
1/2*(0+30)## [1] 15Your hospital owns an old MRI, which has a manufacturer’s lifetime of about 10 years (expected value). Based on previous studies, we know that the failure of most MRIs obeys an exponential distribution. What is the expected failure time? What is the standard deviation? What is the probability that your MRI will fail after 8 years? Now assume that you have owned the machine for 8 years. Given that you already owned the machine 8 years, what is the probability that it will fail in the next two years?
What is the expected failure time?
10
What is the standard deviation?
10
What is the probability that your MRI will fail after 8 years?
lambda <- 1/10
1 - pexp(8, lambda)## [1] 0.449329Now assume that you have owned the machine for 8 years. Given that you already owned the machine 8 years, what is the probability that it will fail in the next two years?
pexp((10-8),lambda)## [1] 0.1812692