Question 1011770: If the probability of being a smoker among a group of cases with lung cancer is .6, what’s the probability that in a group of 8 cases you have less than 2 smokers? More than 5? What are the expected value and variance of the number of smokers?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! If the probability of being a smoker among a group of cases with lung cancer is .6, what’s the probability that in a group of 8 cases you have less than 2 smokers? If the probability of being a smoker among a group of cases with lung cancer is .6, what’s the probability that in a group of 8 cases you have less than 2 smokers?
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Binomial Problem with n = 8 and p(smoke) = 0.6
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P(0<= x <=1) = binomcdf(8,0.6,1) =
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More than 5?
P(6<= x <=8) = 1 - binomcdf(8,0.6,5) = 1- 0.6846 = 0.3154
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What are the expected value and variance of the number of smokers?
E(x) = np = 8*0.6 = 4.8
Var(x) = sqrt(npq) = sqrt(4.8*0.4) = 1.3856
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Cheers,
Stan H.
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