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22) Smokers According to Information Please Almanac, 80% of adult smokers starte

ID: 3159611 • Letter: 2

Question

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

Explanation / Answer

Let X denote the number of smokers among 100 randomly selected smokers who started smoking before they were 18 years old. Clearly X follows a Binomial distribution with parameters n = 100 and p = 0.80.

Since n is large, an excellent approximation to the above Binomial distribution is a Normal distribution with mean np = 100*0.80 = 80 and standard deviation Sqrt[np(1-p)] = Sqrt[100*0.80*0.20] = Sqrt[16] = 4.

Therefore, Z = (X – 80)/4 follows a standard normal distribution N(0,1).

a) The probability that exactly 80 of the smokers started smoking before they were 18 years old is given by,

P[ X = 80] P[ 79.5 < X < 80.5]

                    = P[ (79.5-80)/4 < (X-80)/4 < (80.5-80)/4]

                 = P[- 0.125 < Z < 0.125]

                    = P[ Z < 0.125] - P[Z < - 0.125]

                    = 0.5497 – 0.4503

                 = 0.0994

b) The probability that at least 80 of the smokers started smoking before they were 18 years old is given by

P[X 80] P[ X 79.5]

                    = P[(X-80)/4 (79.5-80)/4]

                 = P[Z -0.125]

                    = 1 - P[Z < -0.125]

                    = 1 – 0.4503

                 = 0.5497

c)The probability that fewer than 70 of the smokers started smoking before they were 18 years old is given by,

P[ X < 70] P[ X < 70.5]

                    = P[(X-80)/4 < (70.5-80)/4]

                 = P[Z < -2.375]

                    = 0.0088

d) The probability that between 70 and 90 of the smokers, inclusive, started smoking before they were 18 years old is given by,

P[70 X 90] P[ 69.5 X 90.5]

                    = P[ (69.5-80)/4 (X-80)/4 (90.5-80)/4]

                 = P[- 2.625 Z 2.625]

                    = P[ Z 2.625] - P[Z - 2.625]

                    = 0.9957 – 0.0043

                 = 0.9914

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