I am not very sure where to go or what to do with thisquestion. I\'ll gladly pro
ID: 2950681 • Letter: I
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I am not very sure where to go or what to do with thisquestion. I'll gladly provide Lifesaver points. Pleaseshow me how to work through this 2 part problem. Q: In testing a new drug, researchers found that 5% of allpatients using it will have a mild side effect. A randomsample of 11 patients using the drug is selected. 1- Find the probability that exactly two will this mild sideeffect. Answers to choosefrom: .078, .019, .099 or .087 2- Find the probability that at least one will have this mildside effect. Answers to choosefrom: .431, .134, .314 or .341 Please explain how you came to get the answer, I am sittinghere punching numbers and can not come to the correctsolution. Thanks I am not very sure where to go or what to do with thisquestion. I'll gladly provide Lifesaver points. Pleaseshow me how to work through this 2 part problem. Q: In testing a new drug, researchers found that 5% of allpatients using it will have a mild side effect. A randomsample of 11 patients using the drug is selected. 1- Find the probability that exactly two will this mild sideeffect. Answers to choosefrom: .078, .019, .099 or .087 2- Find the probability that at least one will have this mildside effect. Answers to choosefrom: .431, .134, .314 or .341 Please explain how you came to get the answer, I am sittinghere punching numbers and can not come to the correctsolution. ThanksExplanation / Answer
Apply binomial probability. 1- Find the probability that exactly two will this mild sideeffect. Answers to choosefrom: .078, .019, .099 or .087 P(x)=nCxx(1-)n-x P(2) =11C2(.05)2(1-.05)11-2=(55)(.0025)(.630249409) = .87 Aslo if you look at your probability distribution in the backof your book you will find n=11; x=.05 which gives you .87 2- Find the probability that at least one will have this mildside effect. Answers to choosefrom: .431, .134, .314 or .341 Look at your probability distribution for n=11; x=.05. To findat least 1 = add all the numbers starting at 1; therefore itis .329+.087+.014+.001= .431 Look at your probability distribution for n=11; x=.05. To findat least 1 = add all the numbers starting at 1; therefore itis .329+.087+.014+.001= .431 Please explain how you came to get the answer, I am sittinghere punching numbers and can not come to the correct solution. ThanksRelated Questions
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