Researchers claim that a new treatment for chronic bronchitis is 80percentage ef

Question

Researchers claim that a new treatment for chronic bronchitis is 80percentage effective in reducing a symptoms. If the new treatment is given to 12 patients suffering from bronchitis what is the probability that: The treatment is effective in exactly 10 patients? The treatment is effective in at most 10 patients? The treatment is effective in at most 10 patients given that the treatment is effective in at least 4 out or 12 patients? Given the true effectiveness rate of 80percentage, compute the expected number and the standard deviation of patients that would have reduced symptoms? Problem 4: A pilot study is proposed to evaluate whether a new medication is effective in controlling the symptoms of asthma. A sample of 10 patients with asthma is enrolled in the study. Each patient agrees to take the study medication and to report back in 30 days for a clinical evaluation. Similar studies report that 90percentage of all patients complete follow-up visits 1 month from the initial contact. What is the distribution of X = {a number of patients show up for the clinical assessment}? What is the probability that at least 8 patients show up for the clinical assessment? What is the probability that all patients show up for the clinical assessment? What is the probability that no more than 4 patients DO NOT show up for the clinical assessment What is the probability that all show up if the true show rate is 87percentage? What is the expected number of patients (out of 125 patients) who DO NOT show up (if the true show rate is 87percentage)?

Explanation / Answer

3

Answers to the questions are:

p (effective) = .8

n=12

a. P(X=10) = 12C10 (.8^10)(.2^2) = .2834

b.P(X>=10) = 12C10 (.8^10)(.2^2) +12C11 (.8^11)(.2^1) +12C12 (.8^12)(.2^0) = 0.5583

c.P(X<=10)/P(X>=4) =0.725 /0.999 = .726

d. E(X) = np = .8*12 = 9.6

Var(X) = npq = 9.6*.8 = 7.68

Standard dviation = sqrt(7.68) = 2.772

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