```Question 87626
Suppose the national average for medication error is one out of every 1000 patients. A hospital believes that their medication error rate is comparable to the national average. If the hospital randomly selects 500 patients,
P(error) = 0.001 n=500
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a. Find the expected number of patients in the sample that will have had a medication error.
E(x) = np = 0.5
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b. What is the standard deviation of the number of patients in the sample that will have had a medication error?
sigma = sqrt(npq) = sqrt(500*0.001*0.999) = 0.70675...
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c. What is the probability of observing one or more patients who have had medication errors in the same sample?
P(at least one) = 1 - P(none) = 1 - 0.999^500 = 0.393621...
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d. What is the probability of observing two or more patients who have had medication errors in the same sample?
P(1 or 2) = 1-[P(none)+P(one)] = 1 - binomcdf(500,0.001,1) = 0.0901
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e. If you observed three patients who have had medication errors in the sample, would you believe that the hospital's medication error rate was comparable to the national average? Give reasons for your conclusions
Based on national average P(x=3) = binompdf(500,0.001,3) = 0.0125949...
Based on the observed results 3/500 = 0.006
The observed rate is better than the rate predicted by the national average.
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Cheers,
Stan H.

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