More than a decade ago, high levels of lead in the blood put 88% of children at risk. A concerted effort was made to remove lead from the environment. Now, suppose only 12% of children in the United States are at risk of high blood-lead levels.

Question

(a) In a random sample of 194 children taken more than a decade ago, what is the probability that 50 or more had high blood-lead levels? (Round your answer to three decimal places.)

Answer 1

n=194
P(r >/= 50)
p= .12 (converted percent to decimal)
q= 1-p = 1-.12 = .88
P(x) = r-.5 if r is to the left(greater than) therefore,

P(x>/= 49.5)

You need P(z>/= ?)

So:
mean (mu) = np = (194)(.12) = 23.28

stan. dev. (sigma) = sqrt(npq) = (194)(.12)(.88) = 4.526

z= (x - mu)/sigma = (49.5 - 23.28)/4.526 = 5.79

So: P(z >/= 5.79), now you look up the probability on your z-chart and there's your answer.