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?
P_A <- 0.001 # Prevalence rate
P_B_given_A <- 0.96 # Sensitivity
P_notB_given_notA <- 0.98 # Specificity
# Calculating P(A|B) using Bayes' theorem
P_B_given_notA <- 1 - P_notB_given_notA
P_notA <- 1 - P_A
P_A_given_B <- (P_B_given_A * P_A) / (P_B_given_A * P_A + P_B_given_notA * P_notA)
print(P_A_given_B)## [1] 0.04584527If 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?
total_people <- 100000
test_cost <- 1000
positive_people <- 2094
treatment_cost <- 100000
total_cost <- (total_people * test_cost) + (positive_people * treatment_cost)
total_cost## [1] 309400000total_people * (test_cost + treatment_cost)## [1] 1.01e+10The total cost will be $309,400,000 under the conditions of the prevalence, sensitivity, and specificity. Furthermore, if all 100,000 people test positive, the cost will amount to $10,100,000,000.
What is the probability that, after 24 months, you received exactly 2 inspections? - After 24 months, the possibility of exactly 2 inspections is 22.32%
months <- 24
probability <- 0.05
inspect <- 2
dbinom(inspect,months,probability)## [1] 0.2232381What is the probability that, after 24 months, you received 2 or more inspections? - The probability is 39.92%
months <- 24
probability <- 0.05
1-(dbinom(0,months,probability)+dbinom(1,months,probability))## [1] 0.3391827What is the probability that your received fewer than 2 inspections? - The possibility is 66.08%
(dbinom(0,months,probability)+dbinom(1,months,probability))## [1] 0.6608173What is the expected number of inspections you should have received? - 1.2 inspections
expect <- months*probability
expect## [1] 1.2What is the standard deviation? - 1.068 is the standard deviation.
std <- sqrt(probability*months*(1-probability))
std## [1] 1.067708What is the probability that exactly 3 arrive in one hour? - 0.757%
dpois(3, 10)## [1] 0.007566655What is the probability that more than 10 arrive in one hour? - The probability is 41.69%
1-ppois(10, 10)## [1] 0.4169602How many would you expect to arrive in 8 hours? - 80 people
10 * 8## [1] 80What is the standard deviation of the appropriate probability distribution? - 3.162 is the standard deviation
std <- sqrt(10)
std## [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?
(80/72*100%=111.11%). The percent utilization is 111%. This means that there are not enough providers for the expected number of patients, as the utilization exceeds 100%. To address this, they should have more doctors, or could refrain from accepting new patients.
If your subordinate acted innocently, what was the probability he/she would have selected five nurses for the trips? - 7.59%
#dhyper(k, K, N - K, n)
dhyper(5, 15, 15, 6)## [1] 0.07586207How many nurses would we have expected your subordinate to send? How many non-nurses would we have expected your subordinate to send? - 3 nurses
6 * (15/30)## [1] 3What is the probability that the driver will be seriously injured during the course of the year? - 69.93%
injured_1200 <- pgeom(q=1200 ,prob=0.001)
injured_1500 <- pgeom(q=1500 ,prob=0.001)
injured_1200## [1] 0.6992876In the course of 15 months? What is the expected number of hours that a driver will drive before being seriously injured?
injured_1500## [1] 0.7772602Given that a driver has driven 1200 hours, what is the probability that he or she will be injured in the next 100 hours?
injured_1300 <- pgeom(q=1300 ,prob=0.001)
injured_1300 - injured_1200## [1] 0.02863018The likelihood of the generator failing more than twice within a span of 1000 hours is 8%, while the expected value is 1.
p_fail_once <- 1 / 1000
lambda <- 1000 * p_fail_once
prob_more_than_twice <- 1 - ppois(2, lambda)
prob_more_than_twice## [1] 0.0803014expected_value <- lambda
expected_value## [1] 1What is the probability that this patient will wait more than 10 minutes? The probablity is 66.67%
min=0
max=30
1-punif(10, 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? - 0.5
1 - punif(15, 0, 30)## [1] 0.5What is the expected waiting time? - The expected wait time is 15 minutes
(0 + 30) / 2## [1] 15What is the expected failure time? - 10 years
lifetime <- 10
expected_failure_time <- lifetime
expected_failure_time## [1] 10What is the standard deviation? - 10
standard_deviation <- lifetime
standard_deviation## [1] 10What is the probability that your MRI will fail after 8 years? - 44.93%
time_passed <- 8
probability_failure_after_8_years <- pexp(time_passed, rate = 1/lifetime, lower.tail = FALSE)
probability_failure_after_8_years## [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? - 18%
time_remaining <- 2
probability_failure_next_two_years <- pexp(time_remaining, rate = 1/lifetime, lower.tail = TRUE)
probability_failure_next_two_years## [1] 0.1812692