Question 846128: 22. 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. 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.
(b) Approximate the probability that at least 80 of them started smoking before they were 18 years old.
(c) Approximate the probability that fewer than 70 of them started smoking before they were 18 years old.
(d) Approximate the probability that between 7- and 90 of them, inclusive, started smoking before they were 18 years old.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 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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