SOLUTION: 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? Wh

Question 1011720: 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 mathmate(429) (Show Source): You can put this solution on YOUR website!

Question:
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?

Solution:
The problem can be solved using the binomial distribution, with p=0.6.

For n=8,
P(X<2)=P(X=0)+P(X=1)
=C(8,0)*p^0*(1-p)^8 + C(8,1)*p^1*(1-p)^(8-1)
=0.0065536+0.0078643
=0.008520

P(X>5)=P(X=6)+P(X=7)+P(X=8)
=0.2090+0.0896+0.0168
=0.3154

Expected value = np=8*0.6=4.8
Variance = npq=8*0.6*(1-0.6)=1.92
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