Question 1102207: 3. A study estimates that the rate of preventable Adverse Drug Events (ADEs) in hospital is 35%. Preventable ADEs typically result from inappropriate care or medication errors, such as errors in commission and omission. In a hospital, 5 randomly selected patients reported experiencing an ADE. Let X represent the number of preventable ADE, i.e., X ~ b(5, 0.35)
a. Calculate the probability of preventable ADEs of the five patients and present them in tabular form as we did in class; (Hint: There are 6 possible outcomes.)
b. Present the probability distribution of preventable ADEs in a bar-like graph.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! for 0 errors it is 0.65^5=0.1160
for 1 error it is 5C1 (5 ways it can happen once)*0.65^4*0.35=0.3124
for 2 errors it is 5C2 (10 ways it can happen twice)*0.65^3*0.35^2=0.3364
for 3 errors, it is 5C3*0.65^2*0.35^3=0.1811
for 4 errors, it is 5*0.65*0.35^4=0.0488
For all 5, 0.35^5=0.0053
They add to 1.
Table and graph may be made from this. Expected value is np=1.75, so 2 is closest to that and has the highest single probability.
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