More than a decade ago, high levels of lead in the blood put 88%of children at r

Question

More than a decade ago, high levels of lead in the blood put 88%of children at risk. A concerted effort was made to removelead from the environment. Now, according to the Centers forDisease Control, only 9% of children in the U.S. are at risk ofhigh blood lead levels.

a) In a random sample of 200 children taken more than adecade ago, what is the probability that 50 or more had high bloodlead levels?

b) In a random sample of 200 children taken now, what is theprobability that 50 or more have high blood lead levels?

Explanation / Answer

a) More than a decade ago,probability that children had highblood lead levels = p= 0.88 More than a decade ago,probability that children did not have high blood lead levels = q= 1-0.88 = 0.12 n = 200 mean, = np = 200x0.88 = 176 standard deviation, = npq = 4.6 standard error, e = / n =4.6 / 200 = 0.33 hence probability that 50 or more had high blood lead levels,P(X>50) P(Z>(X-)/e ) =P(Z>(50-176)/0.33) =P(Z>-381.8) ˜ 1-0 ˜1 b) Now, according to the Centers for Disease Control,probability that children in the U.S. are at risk of highblood lead levels = p = 0.09
Now, according to the Centers for Disease Control, probabilitythat children in the U.S. are not at risk of high blood leadlevels = q = 1-0.09 = 0.91 n = 200 mean, = np = 200x0.09 = 18 standard deviation, = npq = 4.05 standard error, e = / n =4.05 /200 = 0.29 hence probability that 50 or more had high blood lead levels,P(X>50) P(Z>(X-)/e ) =P(Z>(50-18)/0.29) =P(Z>110.35) ˜ 1-0.99999999 ˜0 n = 200 mean, = np = 200x0.09 = 18 standard deviation, = npq = 4.05 standard error, e = / n =4.05 /200 = 0.29 hence probability that 50 or more had high blood lead levels,P(X>50) P(Z>(X-)/e ) =P(Z>(50-18)/0.29) =P(Z>110.35) ˜ 1-0.99999999 ˜0

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