In a clinical trial of a drug used to help subjects stop smoking. 797 subjects w
ID: 3230478 • Letter: I
Question
In a clinical trial of a drug used to help subjects stop smoking. 797 subjects were treated with 1 mg doses of the drug. That group consisted of 47 subjects who experienced nausea. The probability of nausea for subjects not receiving the treatment was 0.0125. Complete parts (a) through (c). a. Assuming that the drug has no effect, so that the probability of nausea was 0.0125. find the mean and standard deviation for the numbers of people in groups of 797 that can be expected to experience nausea. The mean is people. (Round to one decimal place as needed.) The standard deviation is people. (Round to one decimal place as needed.) b. Based on the result from part (a). is it unusual to find that among 797 people, there are 47 who experience nausea? Why or why not? A. It is unusual because 47 is outside the range of usual values. B. It is not unusual because 47 is within the range of usual values. C. It is not unusual because 47 is outside the range of usual values. D. It is unusual because 47 is within the range of usual values. c. Based on the preceding results. does nausea appear to be an adverse reaction that should be of concern to those who use the drug A. The drug does not appear to be the cause of any nausea. B. The drug does appear to be the cause of some nausea. Since the nausea rate is still quite low (about 6%), it appears to be an adverse reaction that does not occur very often. C. The drug does appear to be the cause of some nausea. Since the nausea rate is quite high (about 6%), it appears to be an adverse reaction that occurs very often.Explanation / Answer
a) mean =np=797*0.0125=9.9625
std deviation =|(np(1-p))1/2 =3.137
b)option A is correct
c)option C is correct