(12)Question
To calculate the values requested, we can use the following formulas:
a) Sensitivity = (True Positives / (True Positives + False Negatives)) * 100
b) Specificity = (True Negatives / (True Negatives + False Positives)) *
100
c) Predictive Value Positive = (True Positives / (True Positives + False
Positives)) * 100
d) Negative Predictive Value = (True Negatives / (True Negatives + False
Negatives)) * 100
e) Prevalence = (True Positives + False Negatives) / Total Population
Let's calculate each value step by step:
a) Sensitivity:
Sensitivity = (480 / (480 + 20)) * 100
Sensitivity = 96%
b) Specificity:
Specificity = (1880 / (1880 + 120)) * 100
Specificity = 94%
c) Predictive Value Positive:
Predictive Value Positive = (480 / (480 + 120)) * 100
Predictive Value Positive = 80%
d) Negative Predictive Value:
Negative Predictive Value = (1880 / (1880 + 20)) * 100
Negative Predictive Value = 98.9%
e) Prevalence:
Prevalence = (480 + 20) / 1500
Prevalence = 33.3%
Therefore, the calculated values are:
a) Sensitivity = 96%
b) Specificity = 94%
c) Predictive Value Positive = 80%
d) Negative Predictive Value = 98.9%
e) Prevalence = 33.3
(13)Question
13. To calculate the predictive value negative (PVN) of the new test for
hepatitis C in a group of IV drug users with a known prevalence of
hepatitis C of 20%, we need to use the following formula
PVN = (1 - Prevalence) × Specificity / [(1 - Prevalence) × Specificity +
Prevalence × (1 - Sensitivity)]
Given information:
- Sensitivity of the new test = 50%
- Specificity of the new test = 75%
- Prevalence of hepatitis C in the group of IV drug users = 20%
Plugging in the values, we get:
PVN = (1 - 0.20) × 0.75 / [(1 - 0.20) × 0.75 + 0.20 × (1 - 0.50)]
PVN = 0.80 × 0.75 / (0.80 × 0.75 + 0.20 × 0.50)
PVN = 0.60 / (0.60 + 0.10)
PVN = 0.60 / 0.70
PVN = 0.857 or 85.7%
Therefore, the predictive value negative of the new test for hepatitis C
in the group of IV drug users is 85.7%.
(14)Question
To solve this problem, we need to use the standard normal distribution
and the given parameters (mean and standard deviation) to calculate
the probabilities.
a) Probability of having a cholesterol value between 180 and 200
mg/100 ml:
Mean (μ) = 200 mg/100 ml
Standard deviation (σ) = 20 mg/100 ml
Z-score for 180 mg/100 ml = (180 - 200) / 20 = -1
Z-score for 200 mg/100 ml = (200 - 200) / 20 = 0
Probability = P(Z between -1 and 0) = P(Z ≤ 0) - P(Z ≤ -1) = 0.5 - 0.1587 =
0.3413 or 34.13%
b) Probability of having a cholesterol value greater than 225 mg/100
ml:
Z-score for 225 mg/100 ml = (225 - 200) / 20 = 1.25
Probability = P(Z > 1.25) = 1 - P(Z ≤ 1.25) = 1 - 0.8944 = 0.1056 or
10.56%
c) Probability of having a cholesterol value less than 150 mg/100 ml:
Z-score for 150 mg/100 ml = (150 - 200) / 20 = -2.5
Probability = P(Z < -2.5) = 0.0062 or 0.62%
d) Probability of having a cholesterol value between 190 and 210
mg/100 ml:
Z-score for 190 mg/100 ml = (190 - 200) / 20 = -0.5
Z-score for 210 mg/100 ml = (210 - 200) / 20 = 0.5
Probability = P(Z between -0.5 and 0.5) = P(Z ≤ 0.5) - P(Z ≤ -0.5) = 0.6915
- 0.3085 = 0.383 or 38.3%
(15)Qestion
let's solve this problem step-by-step:
a) Probability that in the next year, among patients receiving
rocuronium, exactly three will experience anaphylaxis:
The Poisson probability mass function is given by:
P(X = x) = (e^(-λ) × λ^x) / x!
Where:
- e is the base of the natural logarithm (approximately 2.718)
- λ is the average number of incidents per year
- x is the number of events (in this case, 3 anaphylaxis incidents)
Since the question does not provide the value of λ, let's assume it is 2
incidents per year.
Plugging in the values:
P(X = 3) = (e^(-2) × 2^3) / 3!
P(X = 3) = (e^(-2) × 8) / 6
P(X = 3) = 0.1839 or 18.39%
b) Probability that at least three patients in the next year will
experience anaphylaxis if rocuronium is administered with anesthesia:
The probability of at least three anaphylaxis incidents is the
complement of the probability of less than three incidents:
P(X ≥ 3) = 1 - P(X < 3)
To calculate P(X < 3), we need to find the probability of 0 incidents, 1
incident, and 2 incidents, and then add them together:
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
Plugging in the values:
P(X = 0) = e^(-2) = 0.1353
P(X = 1) = 2e^(-2) = 0.2707
P(X = 2) = 2^2e^(-2) / 2! = 0.2707
P(X < 3) = 0.1353 + 0.2707 + 0.2707 = 0.6767
P(X ≥ 3) = 1 - 0.6767 = 0.3233 or 32.33%
Therefore, the probability that at least three patients in the next year
will experience anaphylaxis if rocuronium is administered with
anesthesia is 32.33%.
