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

n=194
P(r >/= 50)
p= .12 (converted percent to decimal)
q= 1p = 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 zchart and there's your answer.