Question

According to an​ almanac, 60​% of adult smokers started smoking before turning 18 years old. ​(a) If 100 adult smokers are randomly​ selected, how many would we expect to have started smoking before turning 18 years​ old? ​(b) Would it be unusual to observe 80 smokers who started smoking before turning 18 years old in a random sample of 100 adult​ smokers? Why?

A- We would expect about ____adult smokers to have started smoking before turning 18 years old.

B-Would it be unusual to observe 80 smokers who started smoking before turning 18 years old in a random sample of 100 adult​ smokers?

A. No​, because 80 is greater than mu plus 2 sigma .

B. ​Yes, because 80 is between mu minus 2 sigma and mu plus 2 sigma.

C. No​, because 80 is between mu minus 2 sigma and mu plus 2 sigma .

D. Yes​, because 80 is greater than mu plus 2 sigma .

E. ​No, because 80 is less than mu minus 2 sigma.

Answer 1

Let p , Probability of success that adult smokers start smoking before 18 be 0.6 (60% given).

Now n = 100.

A) Expected Value = np

= 100(0.6)

= 60 adult smokers.

We would expect about 60 adult smokers to have started smoking before turning 18 years old.

B ) Mean = 60

Standard deviation =

=

=

standard deviation = 4.9 (approx)

Now , = 60 +/- 2(4.9)

= 60 +/- 9.8

= (50.2 , 69.8)

Thus, 80 is unusual.

Ans. D. Yes , because 80 is greater than mu plus 2 sigma.

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