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?

rate = 0.001
sense = 0.96
spec = 0.98

prob = sense*rate/((sense*rate)+(1-spec)*(1-rate))
prob
## [1] 0.04584527

There is a 4.58% 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?

test_cost <- 1000*100000
treatment_cost <- 100000*100000*0.001
tc <- test_cost + treatment_cost
tc
## [1] 1.1e+08

There are two ways to calculate this, the first is assuming we had perfect knowledge of the situation, and then it would require the test cost * population + treatment_costpopulationprevalence. This assumes that we know exactly who is sick and who is not. However, if we do not have perfect knowledge:

people <- 100000
tp = sense*people*rate
fp = (1-sense)*people*rate
cost = 1000*people+100000*(tp+fp)
cost
## [1] 1.1e+08

Practically we look at the prevailing rate of infection, then split it up into true positives and false positives, we receive the same answer!

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?

p <- dbinom(2,24,0.05)
p
## [1] 0.2232381

The probability that there will be exactly 2 inspection is 22%

What is the probability that, after 24 months, you received 2 or more inspections?

p_2p <- 1 - (dbinom(1, 24, 0.05) + dbinom(0, 24, 0.05))
p_2p
## [1] 0.3391827

The probability is 33%

What is the probability that your received fewer than 2 inspections?

p_2l <- dbinom(0,24,.05) + dbinom(1,24,.05)
p_2l
## [1] 0.6608173

The probability that there is less than 2 inspections is 66%, which aligns with the prior question.

What is the expected number of inspections you should have received?

n <- 24       
r <- 0.05      
i <- 1 - 0.05  
num_inspect <- n * r * i
num_inspect
## [1] 1.14

The expected number of inspections is 1.

What is the standard deviation?

(n*r*i)^0.5
## [1] 1.067708

The Standard deviation is 1.06

You 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?

(p_exactly_3 <- dpois(3,10))
## [1] 0.007566655

There is a 0.7% chance of exactly 3.

What is the probability that more than 10 arrive in one hour?

P_greater_10 <- 1-sum(dpois(0:10,10))
P_greater_10
## [1] 0.4169602

There is a 41.6% chance of more than 10 visits per hour.

How many would you expect to arrive in 8 hours?

EV is 10/hour * 8 hours yields 80 Visits.

What is the standard deviation of the appropriate probability distribution?

sqrt(10)
## [1] 3.162278

The standard deviation is 3.16

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?

If the three family practice providers can see 24 patients per day, it would equate to a rate of 8 per provider per day, or an effective rate or 1 per hour, which is approximately a 12% utilization rate. My recommendation would be to increase staffing as these providers will have a serious backlog.

Your subordinate with 30 supervisors was recently accused of favoring nurses. 10 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?

phyper(4,15,15,6)
## [1] 0.9157088

There is a about a 90% chance that is valid.

How many nurses would we have expected your subordinate to send?

I would expect my subordinate to send 3 nurses as that is representative of the available population.

How many non-nurses would we have expected your subordinate to send?

I would expect my subordinate to send 3 non-nurses as that is representative of the available population.

The probability of being seriously injured in a car crash in an unspecified location is about 0.1% 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?

(1-(1-0.001)^1200)
## [1] 0.6989866

There is a 69.9% chance that the driver will be injured in the course of a year.

In the course of 15 months?

(1-(1-0.001)^1500)
## [1] 0.7770372
There is a 77.7% chance the driver will be injured in the course of 15 months

What is the expected number of hours that a driver will drive before being seriously injured?

1/0.001
## [1] 1000
It is expected the driver will be injured around the 1000 hour mark.

Given that a driver has driven 1200 hours, what is the probability that he or she will be injured in the next 100 hours?

1-((1-0.001)^100)
## [1] 0.09520785

There is a 9.5% chance

You 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?

(1 - (pbinom (2, size = 1000, prob = 0.001)))
## [1] 0.08020934
There is a 8% chance of that failure condition.

What is the expected value?

The expected value of a binomial distribution is attempts * probability, so I believe the expected value would be 0.001.

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?

Given the uniform distribution, I believe there is a 30-10/30= 66.67% chance the patient will wait more than 10 minutes. But I can guarantee you it will certainly feel a lot more than 10 minutes to them.

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?

This is the same as will the patient have already waited at least 15 minutes, or 50%.

What is the expected waiting time?

The expected wait time is 15 minutes.

Your 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?

The expected failure time is 10 years.

What is the standard deviation?

The standard deviation is 0.1.

What is the probability that your MRI will fail after 8 years?

1-pexp(8,.1)
## [1] 0.449329

There is a 45% chance that the machine 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?

pexp(2,.1)
## [1] 0.1812692

There is an 18% chance that the machine will fail in the next two years.