Given:
Prevalence of MNR HIV-1, \(P(\text{Disease})\): 0.1% or 0.001
Sensitivity, \(P(\text{Positive} | \text{Disease})\): 96%
Specificity, \(P(\text{Negative} | \text{No Disease})\): 98%
The probability that an individual who tests positive actually has the disease, \(P(\text{Disease} | \text{Positive})\), is calculated using Bayes’ theorem:
\[P(\text{Disease} | \text{Positive}) = \frac{P(\text{Positive} | \text{Disease}) \cdot P(\text{Disease})}{P(\text{Positive})}\]
Where:
\[P(\text{Positive}) = P(\text{Positive} | \text{Disease}) \cdot P(\text{Disease}) + P(\text{Positive} | \text{No Disease}) \cdot P(\text{No Disease})\]
To find \(P(\text{Disease} | \text{Positive})\), we first calculate \(P(\text{Positive})\); Then it yields a \(P(\text{Disease} | \text{Positive})\) of approximately 4.58% (Check below).
# Given values
prevalence <- 0.001
sensitivity <- 0.96
specificity <- 0.98
# Calculating P(No Disease)
p_no_disease <- 1 - prevalence
# Calculating P(Positive | No Disease)
p_positive_no_disease <- 1 - specificity
# Calculating P(Positive)
p_positive <- (sensitivity * prevalence) + (p_positive_no_disease * p_no_disease)
# Calculating P(Disease | Positive)
p_disease_given_positive <- (sensitivity * prevalence) / p_positive
p_disease_given_positive## [1] 0.04584527We calculate the total first-year cost for treating 100,000 individuals, taking into account the cost of testing and the treatment cost for true positive cases. The total first-year cost is approximately $109.6 million (as gotten in the result below).
population <- 100000
test_cost <- 1000
treatment_cost_per_positive_case <- 100000
# Calculating expected number of positive tests
expected_positives <- p_positive * population
# Calculating expected number of true positives
true_positives <- p_disease_given_positive * expected_positives
# Calculating total first-year cost
total_cost <- (true_positives * treatment_cost_per_positive_case) + (population * test_cost)
total_cost## [1] 109600000\(n = 24\): the number of trials (months).
\(p = 0.05\): the probability of success on a single trial (receiving an inspection).
\(k\): the number of successes (number of inspections received).
The probability mass function (PMF) of a binomial distribution, which gives the probability of receiving exactly \(k\) successes (inspections) out of \(n\) trials (months), is given by:
\[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]
Where \(\binom{n}{k}\) is the binomial coefficient, representing the number of ways to choose \(k\) successes out of \(n\) trials.
The expected number of inspections (\(E[X]\)) and the standard deviation (\(\sigma\)) for a binomial distribution are calculated as follows:
\(E[X] = np\)
\(\sigma = \sqrt{np(1-p)}\)
n <- 24
p <- 0.05
k <- 2
prob_exact_2 <- dbinom(k, n, p)
prob_exact_2## [1] 0.2232381prob_2_or_more <- 1 - pbinom(1, n, p)
prob_2_or_more## [1] 0.3391827prob_fewer_than_2 <- pbinom(1, n, p)
prob_fewer_than_2## [1] 0.6608173# Expected number of inspections
expected_inspections <- n * p
# Standard deviation
sd_inspections <- sqrt(n * p * (1 - p))
list(Expected = expected_inspections, Standard_deviation = sd_inspections)## $Expected
## [1] 1.2
##
## $Standard_deviation
## [1] 1.067708The arrival of patients at the family practice clinic follows a Poisson distribution with the following parameters:
Let \(\lambda = 10\) be the average rate of arrival (patients per hour).
Let \(k\) be the number of arrivals (patients) we’re interested in.
The probability mass function (PMF) of the Poisson distribution, giving the probability of observing exactly \(k\) arrivals in a given period, is:
\[P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}\]
For a Poisson distribution, the expected number of arrivals (\(E[X]\)) is \(\lambda\), and the standard deviation (\(\sigma\)) is \(\sqrt{\lambda}\).
Given that each provider can see 24 patients a day, and assuming an 8-hour workday, the total capacity per day for all providers combined is 3 providers \(\times\) 24 patients/provider = 72 patients.
lambda <- 10
k <- 3
prob_exact_3 <- dpois(k, lambda)
prob_exact_3## [1] 0.007566655prob_more_than_10 <- 1 - ppois(10, lambda)
prob_more_than_10## [1] 0.4169602# Expected number of arrivals in one hour is lambda
expected_arrivals_1hr <- lambda
# Standard deviation for one hour
sd_arrivals_1hr <- sqrt(lambda)
# Expected number of arrivals in 8 hours
expected_arrivals_8hr <- lambda * 8
list(Expected_in_1hour = expected_arrivals_1hr, Standard_deviation = sd_arrivals_1hr, Expected_in_8hours = expected_arrivals_8hr )## $Expected_in_1hour
## [1] 10
##
## $Standard_deviation
## [1] 3.162278
##
## $Expected_in_8hours
## [1] 80# Total patient capacity for all providers
total_capacity <- 3 * 24
expected_demand_8hr <- expected_arrivals_8hr # Is also expected demand in 8 hours
# Percent utilization
percent_utilization <- (expected_demand_8hr / total_capacity) * 100
percent_utilization## [1] 111.1111Recommendations: adjusting the number of providers; scheduling to optimize operations; ensuring that the clinic can handle the expected patient volume efficiently.
Employing the hypergeometric distribution, which is suitable for modeling the probability of a given number of successes (nurse selections) without replacement from a finite population. In this scenario:
Total number of workers, \(N = 30\)
Number of nurses (successes in the population), \(K = 15\)
Number of draws (people sent on trips), \(n = 6\)
Number of observed successes (nurses sent on trips), \(k = 5\)
The probability mass function (PMF) of the hypergeometric distribution is:
\[P(X = k) = \frac{{\binom{K}{k} \binom{N-K}{n-k}}}{{\binom{N}{n}}}\]
where:
\(\binom{K}{k}\) is the number of ways to choose \(k\) successes out of \(K\) possible successes,
\(\binom{N-K}{n-k}\) is the number of ways to choose \(n-k\) failures out of \(N-K\) possible failures,
\(\binom{N}{n}\) is the total number of ways to choose \(n\) workers out of \(N\).
For the hypergeometric distribution: - The expected number of nurses sent on the trips is calculated as \(E[\text{Nurses}] = n \frac{K}{N}\).
N <- 30
K <- 15
n <- 6
k <- 5
# Probability of selecting 5 nurses
prob_select_5_nurses <- dhyper(k, K, N-K, n)
prob_select_5_nurses## [1] 0.07586207# Expected number of nurses sent on trips
expected_nurses <- n * (K / N)
# Expected number of non-nurses sent on trips
expected_non_nurses <- n * ((N - K) / N)
list(Expected_nurses = expected_nurses, Expected_non_nurses = expected_non_nurses)## $Expected_nurses
## [1] 3
##
## $Expected_non_nurses
## [1] 3In analyzing the risk associated with prolonged driving, we consider the probability of being seriously injured in a car crash, which is about 0.1% per hour (\(p = 0.001\)). This scenario is well-modeled by the geometric distribution, which describes the probability of observing the first success (in this case, a serious injury) on the \(k\)-th trial (hour of driving).
The probability mass function (PMF) for the geometric distribution is given by:
\[P(X = k) = (1 - p)^{k-1}p\]
where:
\(k\) is the trial number at which the first success occurs.
\(p\) is the probability of success on any given trial.
For the geometric distribution, the expected number of trials (\(E[X]\)) until the first success is \(\frac{1}{p}\), and this gives us the average time until an injury in hours.
What is the probability that the driver will be seriously injured during the course of the year?
What is the probability that the driver will be seriously injured 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?
Considering a driver is required to traverse an area for 1200 hours in a year:
p <- 0.001
hours_year <- 1200
# Probability of being injured at least once in the year
prob_injury_year <- 1 - (1 - p)^hours_year
prob_injury_year## [1] 0.6989866hours_15_months <- 1500
prob_injury_15_months <- 1 - (1 - p)^hours_15_months
prob_injury_15_months## [1] 0.7770372# Average time until a serious injury
expected_hours_before_injury <- 1 / p
expected_hours_before_injury## [1] 1000Given that a driver has already driven 1200 hours without injury, the probability of being injured in the next 100 hours is the same as being injured in any 100-hour period:
next_100_hours <- 100
# Probability of being injured in the next 100 hours
prob_injury_next_100 <- 1 - (1 - p)^next_100_hours
prob_injury_next_100## [1] 0.09520785Given that a hospital’s primary generator fails on average once every 1000 hours, we apply the Poisson distribution to model the probability of the generator failing more than twice in a 1000-hour period.
To find the probability of more than two failures, we calculate the complement of the probability of 0, 1, or 2 failures:
lambda <- 1
# Probability of 0, 1, or 2 failures (less than or equal to 2 failures)
prob_up_to_two_failures <- ppois(2, lambda)
# Probability of more than two failures
prob_more_than_two_failures <- 1 - prob_up_to_two_failures
prob_more_than_two_failures## [1] 0.0803014As in the poisson distribution, the expected number of generator failures in a 1000-hour period can be directly taken as the average rate of failure:
# Expected number of failures is lambda
expected_failures <- lambda
expected_failures## [1] 1The waiting time for a surgical patient is assumed to be uniformly distributed across a continuous range. This assumption implies that within the specified interval, every moment is equally likely for the patient’s surgery to begin.
The probability density function (PDF) for a uniform distribution is defined as:
\[f(x) = \frac{1}{b - a} \quad \text{for} \quad a \leq x \leq b\]
Probability of Waiting More Than 10 Minutes: This can be found by considering the proportion of the time interval from 10 to 30 minutes relative to the entire interval from 0 to 30 minutes.
Probability After Waiting 10 Minutes: Given the patient has already waited 10 minutes, we reassess the waiting time distribution from 0 to 20 minutes and calculate the probability of waiting at least another 5 minutes.
Expected Waiting Time: The average time a patient can expect to wait, given by the midpoint of the distribution’s range.
To calculate the probability that a patient waits more than 10 minutes:
a <- 0
b <- 30
# Proportion of the distribution above 10 minutes
wait_more_than_10 <- (b - 10) / (b - a)
wait_more_than_10## [1] 0.6666667# New range is 0 to 20 minutes after waiting 10 minutes
remaining_time_after_10 <- b - 10
# Proportion of the new range above 5 more minutes
wait_at_least_5_more <- (remaining_time_after_10 - 5) / remaining_time_after_10
wait_at_least_5_more## [1] 0.75# Midpoint of the distribution's range (in case of uniform distr.)
expected_waiting_time <- (a + b) / 2
expected_waiting_time## [1] 15An old MRI machine’s failure time follows an exponential distribution, with an expected lifetime of 10 years. This section calculates the expected failure time, standard deviation, and probabilities of failure at specific times.
lambda <- 1 / 10
# Expected failure time
expected_failure_time <- 1 / lambda
# Standard deviation
standard_deviation <- 1 / lambda
list(expected_failure_time = expected_failure_time, standard_deviation = standard_deviation)## $expected_failure_time
## [1] 10
##
## $standard_deviation
## [1] 10years <- 8
# Probability of surviving past 8 years
prob_survive_past_8 <- exp(-lambda * years)
# Probability of failing after 8 years is the complement
prob_fail_after_8 <- 1 - prob_survive_past_8 #survival function of the exponential distribution
prob_fail_after_8## [1] 0.550671next_years <- 2
# Probability of failing in the next two years is the same as failing within two years
prob_fail_next_2 <- 1 - exp(-lambda * next_years)
prob_fail_next_2## [1] 0.1812692