At the Brigham and Women’s Hospital approximately 25% of all births are by caesarian section. You plan to record details of the next 100 births at that hospital. 

(a) What is the sampling distribution of the proportion of those 100 births that are by caesarian section?

 

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The sampling distribution is b is approximately N (0.25, 0.0433) The sample is random. 100*0.25 = 25 > 10 n(1-p) = 100*(1-0.25) = 75 > 10: success/ failure is met. SD (b) = vp (1-p)/n = v0.25 (1-0.25)/100 = v0.1875/100 = v0.001875 = 0.0433 np = %3D %3D (b) What is the probability that more than 30% of the births are by caesanan section? Draw a picture with appropriate labeling that shows your answer. 11 pts P (b > 0.30) = P (Z > The sampling distribution is b is approximately N (0.25, 0.0433) The sample is random. 100*0.25 = 25 > 10 n(1-p) = 100*(1-0.25) = 75 > 10: success/ failure is met. SD (b) = vp (1-p)/n = v0.25 (1-0.25)/100 = v0.1875/100 = v0.001875 = 0.0433 np = %3D %3D (b) What is the probability that more than 30% of the births are by caesanan section? Draw a picture with appropriate labeling that shows your answer. 11 pts P (b > 0.30) = P (Z >

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Solution let X denotes the number of births out of n100 births which are by caesarian secti... View the full answer