More than a decade ago, high levels of lead in the blood put 87% of children at risk. A concerted effort was made to remove lead from the environment. Now, suppose only 17% of children in the United States are at risk of high blood-lead levels.
(a) In a random sample of 188 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.)
(b) In a random sample of 188 children taken now, what is the probability that 50 or more have high blood-lead levels? (Round your answer to three decimal places.)
a)
X ~ Bin ( n , p)
Where n = 188 , p = 0.87
Mean = n * p = 188 * 0.87 = 163.56
Standard deviation = sqrt [ n p ( 1 - p) ] = sqrt ( 188 * 0.87 * ( 1 - 0.87) ) = 4.6112
Using normal approximation,
P(X < x) = P(Z < ( x - Mean) / SD)
With continuity correction
P(X >= 50) = P(Z > 49.5 - 163.56) / 4.6112)
= P(Z > -24.74)
= P(Z < 24.74)
= 1
b)
X ~ Bin ( n , p)
Where n = 188 , p = 0.17
Mean = n * p = 188 * 0.17 = 31.96
Standard deviation = sqrt [ n p ( 1 - p) ] = sqrt ( 188 * 0.17 * ( 1 - 0.17) ) = 5.1504
Using normal approximation,
P(X < x) = P(Z < ( x - Mean) / SD)
With continuity correction
P(X >= 50) = P(Z > 49.5 - 31.96) / 5.1504)
= P(Z > 3.41)
= 1 - P(Z < 3.41)
= 1 - 0.9997 (From Z table)
= 0.000
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