Assignment
keniajin
Question 1
Let ET denote existing treatment, NT denote new treatment, S denote surviving and D denote dying. The probability tree will look as
We want \(P(NT|S)\) by Bayes Theorem we have that
\[ P(NT|S x) = \frac{P(NT|S) P(NT)}{P(S)} = \frac{P(S|NT) P(NT)}{P(S|NT) P(NT) + P(S|ET) P(ET)} \]
\[ P(NT|S x) = \frac{0.9*05}{(0.9*0.5)+(0.7*0.5)} = \frac{0.45}{0.8)} = 0.5625 \]
Therefore 56.25 % of the randomly selected patients survived having received the new treatment.
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Question 2
\(X ~ Poisson(3)\) Therefore the PDF is given by
\[ P(X=x) = \frac{3^xe-3}{x!} for x=1,2,3,... \]
- We need to find \(P(x>4)=1-P(x>4)\)
\[ P(x>4) = 1-[P(x=0) + P(x=2) +P(x=2) +P(x=3) +P(x=4) ] \]
\[ P(x>4) = 1-[\frac{3^0e-3}{0!} + \frac{3^1e-3}{1!} + \frac{3^2e-3}{2!} + \frac{3^2e-3}{2!} + \frac{3^4e-3}{4!} ] \]
\[ P(x>4) = 1-0.8152=0.1848 \]
- The table output
| Number of patients | \(P(X=x)\) | \(F(x\) |
|---|---|---|
| 1 | \(0.0498\) | \(0.0498\) |
| 2 | \(0.1494\) | \(0.1992\) |
| 3 | \(0.2240\) | \(0.4232\) |
| 4 | \(0.2240\) | \(0.4232\) |
| 5 | \(0.2240\) | \(0.4232\) |
Or you can decide ans use r
| Number of patients | P(X=x) | F(X) |
|---|---|---|
| 0 | 0.0498 | 0.0498 |
| 1 | 0.1494 | 0.1494 |
| 2 | 0.2240 | 0.2240 |
| 3 | 0.2240 | 0.2240 |
| 4 | 0.1680 | 0.1680 |
| 5 | 0.1008 | 0.1008 |
| 6 | 0.0504 | 0.0504 |
| 7 | 0.0216 | 0.0216 |
At 6 patients, the facility can handle patients arriving on at least 95% of the Saturdays. Therefore the present facility should be increased by at least 2 more patients.
\(E(X) = \mu\) where \(\mu\) is the parameter of the Poisson distribution. Therefore the expected number of patients arriving each Saturday is 3.
The most probable number would be 2 or 3 patients as both values have the highest probability of occurrence 0.224 as shown in the table above.
Main assumptions; 1. Patients arriving at a hospital during the peak period occur randomly in time. 2. The events occur at a constant rate per unit time. 3. The events occur independently of each other