Question

A manufacturer of aspirin claims that the proportion of headache sufferers who get relief with just two aspirins is 64%. What is the probability that in a random sample of 480 headache sufferers, less than 59.1% obtain relief?

Answer 1

$P(p<0.591)= P(Z<\dfrac{0.591-p}{\sqrt{\frac{p(1-p)}{n}}})\\=P(Z<\dfrac{0.591-0.64}{\sqrt{\frac{0.64(1-0.64)}{480}}})\\\ \\=P(Z<-\dfrac{0.049}{0.021909})=P(Z<-2.2365)$

Using Standard Normal Distribution Table:

$P(Z<-2.2365)=0.0127$