Nicotine Absent Nicotine Present
Self-Reported Non-Smoker 82 14
Self-Reported Smoker 12 52
A. What is the probability that at least 3 are hyperlipidemic?
P(X=0) = [12!/0(12-0)!]0.30(1-0.3)12-0 = 0.0138
P(X=1) = [12!/1(12-1)!]0.31(1-0.3)12-1 = 0.0712
P(X=2) = [12!/2(12-2)1]0.32(1-0.3)12-2 = 0.1678
P = [0.0138 + 0.0712 + 0.1678] = 1 - 0.2528 = 0.7472 = 75%
B. What is the probability that 3 are hyperlipidemic?
P(X=3) = [12!/{3!(12-3)}](0.3)3 (1-0.3)12-3 = 0.2397
C. How many would be expected to meet the criteria for hyperlipidemia?
4
Is this right?