please show work! thank you! 34. childhood lead poisoning is a public health con
ID: 2953636 • Letter: P
Question
please show work! thank you!34. childhood lead poisoning is a public health concern in theUnited States. in a certain population, one child in 8 has a highblood lead level (defined as 30 ug/dLi or more). in a randomlychosen group of 16 children from the population, what is theprobability that (a) none has high blood lead? (b) one has high blood lead? (c) 2 have high blood lead? (d) 3 or more have high blood lead? [hint: use parts (a)-(c)to answer part (d)]
34. childhood lead poisoning is a public health concern in theUnited States. in a certain population, one child in 8 has a highblood lead level (defined as 30 ug/dLi or more). in a randomlychosen group of 16 children from the population, what is theprobability that (a) none has high blood lead? (b) one has high blood lead? (c) 2 have high blood lead? (d) 3 or more have high blood lead? [hint: use parts (a)-(c)to answer part (d)]
Explanation / Answer
This is a binomial distribution with PMF - P(X=x) =nCxpx(1-p)n-x Where X: observed value n - sample size p - probability ofoccuring (1/8 in this example) a)P(X=0) = 16C0(1/8)0(7/8)16 = 0.118 b)P(X=1) =16C1(1/8)1(7/8)15 =0.269 c)P(X=2) =16C2(1/8)2(7/8)14 =0.289 d)P(X=>3) = 1 - P(X<=2) = 1- (P(X=0)+P(X=1)+P(X=2)) = 1 - (0.118+0.269+0.289) =0.324 Where X: observed value n - sample size p - probability ofoccuring (1/8 in this example) a)P(X=0) = 16C0(1/8)0(7/8)16 = 0.118 b)P(X=1) =16C1(1/8)1(7/8)15 =0.269 c)P(X=2) =16C2(1/8)2(7/8)14 =0.289 d)P(X=>3) = 1 - P(X<=2) = 1- (P(X=0)+P(X=1)+P(X=2)) = 1 - (0.118+0.269+0.289) =0.324Related Questions
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