SOLUTION: 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 sel

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Question 846156: 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 70 and 90 of them, inclusive, started smoking before they were 18 years old.

Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!
 

Hi,

.0993 vs .1034, .5595 vs .5517, .0061 vs .0043, .9916 vs .9914 ???

book may find few who would agree with those answers given

for ex. long hand: P(x=80) = 100C80*  = .0993

 p = .80, n = 100

Using TI Calculator

P(x = 80)= binompdf(n, p, x-value) = binompdf(100, .80, 80)= .0993  0r  9.93%

P(x ≥ 80)= 1 – binomcdf(n, p, largest x-value)= 1 - binomcdf(100, .80,79)= .5595  0r 55.95%

P( x < 70)=  binomcdf(n, p, largest x-value)= binomcdf(100, .80, 69) = .0061 0r  .66%

P( 70≤ x ≤90) = binomcdf(100, .80, 90) - binomcdf(100, .80, 70)

   .9977- .0061 = .9916   0r 99.16% 

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