Solution:
Given,
sensitivity = 96%, specificity = 98% , false positive, P(FR) = 2%, chances of MNR-HIV-1, P(MNR) = 0.1%,
median cost per positive case = $100,000,
total first year for treating 100,000 = ?
\[\begin{equation} P(MNR|+) = \frac {P(+|MNR).P(MNR)}{P(+|MNR).P(MNR) + P(+|FR).P(FR)} \\ \\ = \frac {(0.96)*(0.001)} {(0.96*0.001) + 0.02*0.98} \\ = \frac {0.00096}{0.00096+0.0196} \\\\ = \frac {0.00096}{0.02056} \\\\ \approx 0.047 \end{equation}\]
Out of 100,000 individuals, the rate of being positive is 0.1% i.e. approximately 100 people will test positive
So, \[\text {The cost of testing 100,000 individuals costs is } 100,000*1000\] \[=100,000,000\] test for positive case \[\begin{equation} = 100 * 100,000 \\\\ = 10,000,000 \\\\ \end{equation}\]
So, \[\text {total cost} = 100,000,000+10,000,000\] \[ = \$ 110,000,000\] 2. (Binomial). 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?
Solution:
Given,
Probability of joint commission inspection P(inspection) = 0.05 probability after 24 months = ?
n = 24
for exactly 2 inspection, \[\begin{equation} P(X=2) = ^nC_x P^x(1-P)^{n-2} \\\\ = ^{24}C_2(0.05)^2*(1-0.05)^{24-2} \\\\ =(0.05)^2*(0.95)^22 \\\\ \approx 0.246 \end{equation}\]
for 2 or more inspection \[\begin{equation} P(X \geq 2) = 1 - P(X \leq 2) \\\\ = 1 - P(X = 0) - P(X=1) \\\\ = 1 - (0.005)^0*0.95^{24} - (0.005)^1 * (0.95)^{23} \\ \approx 0.0317 \end{equation}\]
for less than 2 inpection, \[\begin{equation} P(x<2) = P(X=0) + P(X=1) \\\\ = (0.005)^0 * (0.95)^{24} + (0.005)^1 * (0.95)^{23} \\\\ \approx 0.754 \end{equation}\]
Standard Deviation \[\begin{equation} \sqrt{n*p*(1-p)} \\\\ \sqrt{24*0.05*0.95} \\\\ \approx 0.974 \end{equation}\]
Solution:
Given,
\[\mu = 10/hr\] \[ x_1 = 3 \]
So, \[\begin{equation} P(X = x_1) = \frac {\mu^x*e^{-\mu}}{x!} \\\\ P(X = 3) = \frac {10^3*e^{-10}}{3!} \\\\ = \frac {0.0454}{3!} \approx 0.00757 \end{equation}\]
For more than 10 patient in 1 hr,
\[\begin{equation} P(X>10) = 1 - P(X<10) \\\\ = 1 - e^\mu(\sum_{1-9} \frac {\mu^x}{x!}) \\\\ = 1 - e^{-10}(\frac {10^1}{1!} .... \frac {10^9}{9!}) \\\\ = 1 - e^{-10}(10085.57319) \\\\ = 1-0.4578 \approx 0.5421 \end{equation}\]
Again,
in 8 hrs,
\[\mu_2 = ? \\\\ \mu_1 = 10\\\\ t_2 = 8hrs\\\\ t_1 = 1 \] \[\begin{equation} \frac {\mu_2} {\mu1} = 8 \\\\ \frac {\mu_2}{10} = 8 \\\\ \mu_2 = 80 \end{equation}\]
So, we expect 80 patients in 8 hrs
\[\begin{equation} \sigma = \sqrt{\mu_2} \\\\ = \sqrt {80} \approx 8.94 \end{equation}\]
If there are 3 family practice provider care and see 24 template patient, the utilization is **3*24 = 72**
Currently there are only 10 patient / hr
so,
\[\frac {72}{10} = 7.2 hrs \text{ of patient}\] this means \[\frac {7.2}{24} = 30 \text{ % utilization}\]
Suggestion: increase number of patient that can be seen in each hour. Some new services can be added to attract more patient.
Solution:
Given,
total population (N) = 30 total nurse (A) = 15 total # of employee picked for Disney vacation (n) = 6 total nurses picked out of 6 (k) = 5
we know that, \[\begin{equation} P(K=k) = \frac {{A \choose k} {N-A \choose n-k}} {N \choose n} \\\\ P(K = 5) = \frac {{15 \choose 5} {15 \choose 1}} {30 \choose 6} \\\\ = \frac {3003.15}{593775} \\\\ \approx 0.0758 \end{equation}\]
To find the expected # of nurses selected for a tryp, we can use fact that the expected value of a hypergeometric distribution is given by:
\[\begin{equation} E(X) = n * \frac {k} {N} \\\\ = 6 * \frac {15}{30} \\\\ = 3 \end{equation}\]
for non-nurse, we know that there are 15 staffs, so, \[\begin{equation} E(X_2) = n * \frac {(N-k)}{N} \\\\ = 6 * \frac {15}{30} = 3 \end{equation}\]
therefore, we would have expected the supervisor to send 3 non-nurses to the trip.
Solution:
In this case, p = 0.001 and X is the number of hours the driver must drive before being seriously injured. The distribution is geometric because each hour is independent of the others and has the same probability of a serious injury occurring.
The probability of the driver not being seriously injured in one hour is \[1 - p = 0.999 \]
Therefore, the probability of the driver not being seriously injured in 1200 hours is \[(0.999)^{1200} \approx 0.818\] This means that the probability of being seriously injured at some point during the year is approximately \[1 - 0.818 = 0.182 \text { or 18.2%} \]
To find the probability of being seriously injured during the course of 15 months, we need to calculate the probability of not being seriously injured in the first 1080 hours (which is 9 months) and being seriously injured in the next 360 hours (which is 3 months). The probability of not being seriously injured in 1080 hours is \[(0.999)^{1080} \approx 0.361\]
Therefore, the probability of being seriously injured in the next 360 hours is \[1 - 0.361 = 0.639 \text { or about 63.9%}\]
The expected number of hours a driver will drive before being seriously injured is 1/p, which is 1000 hours.
Given that a driver has driven 1200 hours, the probability of being seriously injured in the next 100 hours is the probability that the first serious injury occurs between hour 1201 and hour 1300. This probability can be calculated as:
\[\begin{equation} P(1201 \leq X \leq 1300 | X > 1200) = [1 - (1 - p)^{100}] / (1 - (1 - p)^{1200}) \\\\ \approx 0.009 \end{equation}\]
Therefore, the probability of being seriously injured in the next 100 hours, given that the driver has driven 1200 hours, is approximately 0.009, or about 0.9%.
This situation can be modeled using a Poisson distribution with \[\lambda = 1/1000 \text{ failures/hour.}\] The probability of the generator failing more than twice in 1000 hours is: \[\begin{equation} P(X > 2) = 1 - P(X \leq 2) \\\\ = 1 - (P(X=0) + P(X=1) + P(X=2)) = 1 - (e^{(-\lambda)} * (\frac {\lambda^0} {0!}) + e^{(-\lambda)} * (\frac {\lambda^1}{1!}) + e^{(-\lambda)} * (\frac {\lambda^2} {2!})) \\\\ = 1 - (e^{(-1)} * (\frac {1^0} {0!}) + e^{(-1)} * (\frac {1^1}{1!}) + e^{(-1)} * (\frac {1^2}{2!})) \\\\ = 1 - (0.3679 + 0.3679 + 0.1839) \\\\ \approx 0.0803 \end{equation}\]
So the probability of the generator failing more than twice in 1000 hours is 0.0803.
The expected value for the number of failures in 1000 hours is:
\[\begin{equation} E(X) = \lambda * t = (1/1000) * 1000 = 1 \end{equation}\]
So we can expect 1 failure in 1000 hours.
Solution:
The waiting time for the surgical patient is uniformly distributed from 0 to 30 minutes. Let X denote the waiting time in minutes.
The probability that the patient will wait more than 10 minutes is:
\[\begin{equation} P(X > 10) = \frac {(30 - 10)} {30} = \frac {2}{3} = 0.6667 \end{equation}\]
So the probability that the patient will wait more than 10 minutes is 0.6667.
If the patient has already waited 10 minutes, the remaining waiting time is uniformly distributed from 0 to 20 minutes. Let Y denote the remaining waiting time in minutes.
The probability that the patient will wait at least another 5 minutes prior to being seen is:
\[\begin{equation} P(Y \geq 5 | X = 10) = P(X + Y \geq 15 | X = 10) = P(Y \geq 5) = \frac {(30 - 5)}{30} = \frac {5}{6} \approx 0.8333 \end{equation}\]
So the probability that the patient will wait at least another 5 minutes prior to being seen, given that they have already waited 10 minutes, is 0.8333.
The expected waiting time is:
\[E(X) = \frac{(0 + 30)} {2} = 15\]
So the expected waiting time is 15 minutes.
Solution:
Since the failure of MRIs obeys an exponential distribution, the probability density function (PDF) for the lifetime of the MRI is given by:
\[f(x) = \lambda*e^{-λx}\]
where λ is the failure rate (i.e., the number of failures per unit time) and x is the time.
Since the expected lifetime of the MRI is 10 years, we know that λ = 1/10 = 0.1 per year.
The expected failure time (i.e., the mean) is given by:
E(X) = 1/λ = 10 years
The standard deviation is given by:
SD(X) = 1/λ = 10 years
The probability that the MRI will fail after 8 years is given by:
\[P(X > 8) = \int_8^\infty \lambda e^{-\lambda x} dx = e^{(-0.8)} \approx 0.4493\]
Now assume that you have owned the machine for 8 years. Given that you already owned the machine for 8 years, the probability that it will fail in the next two years is given by:
\[\begin{equation} P(X \leq 2 | X > 8) = \frac {P(8 < X \leq 10)} {P(X > 8)} \\\\ = \frac {\int_8^{10} \lambda e^{-\lambda x} dx} {\int_8^\infty \lambda e^{-\lambda x} dx} \\\\ = \frac{ [e^{(-0.8)} - e^{(-1)}]} {e^{(-0.8)}} \\\\ \approx 0.2587 \end{equation}\]