Amanufacturer of aspirin claims that the proportion of headache sufferers who get relief with just two aspirins is 63.8%. what is the probability that in a random sample of 520 headache sufferers, less than 58.2% obtain relief? probability =

n=410 (sample of headache sufferers) p = 0.637 (proportion of those who get relief with just two aspirins) Mean = np = (410)(0.637) = 261.17 Standard deviation = sqrt [ np(1-p)] = sqrt [ (410)(0.637)(0.363)] = 9.7368 We want the probability that more than 66.2% obtain relief. (410)(0.662) = 271.42 Using the normal distribution with mean 261.17 and standard deviation 9.7368: ÎĽ = 261.17 Ď = 9.7368 standardize x to z = (x - ÎĽ) / Ď P(x > 271.42) = P( z > (271.42-261.17) / 9.7368) = P(z > 1.0527) = 0.1469 (probability that in a random sample of 410 headache sufferers, more than 66.2% obtain relief)

Ithink it’s b, hope this .