Question 846128
 Smokers, According to Information Please Almanac, 80% of adult smokers started smoking before they were 18 years old. 
Suppose 100 smokers 18 years old or older are randomly selected.
mean = np = 100*0.80 = 80
std = sqrt(npq) = sqrt(80*0.20) = sqrt(16) = 4 
Use the normal approximation to the binomial to:
(a) Approximate the probability that exactly 80 of them started smoking before they were 18 years old.
Find P(79.5 < x < 80.5)
z(79.5) = (79.5-80)/4 = -0.5/4 = -0.125
z(80.5) = (80.5-80)/4 = +0.125
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P(x = 80) = P(-0.125< z < 0.125) = 0.3867 
(b) Approximate the probability that at least 80 of them started smoking before they were 18 years old.
P(x >= 80) = P(z >= 0.125) = 0.45
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(c) Approximate the probability that fewer than 70 of them started smoking before they were 18 years old.
z(69.5) = (69.5-80)/4 = -10.5/4 = -2.625
P(x <= 69.5) = P(z <= -2.625) = 0.004
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(d) Approximate the probability that between 7- and 90 of them, inclusive, started smoking before they were 18 years old.
Find (6.5 < x < 90.5)
I'll leave that to you.
Cheers,
Stan H.
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