MATH SOLVE

6 months ago

Q:
# According to an almanac, 60% of adult smokers started smoking before turning 18 years old. (a) compute the mean and standard deviation of the random variable x, the number of smokers who started before 18 in 400 trials of the probability experiment. (b) interpret the mean. (c) would it be unusual to observe 340 smokers who started smoking before turning 18 years old in a random sample of 400 adult smokers? why?

Accepted Solution

A:

The given problem describes a binomial distribution with p = 60% = 0.6. Given that there are 400 trials, i.e. n = 400.

a.) The mean is given by:

[tex]\mu=np=400\times0.6=240[/tex]

The standard deviation is given by:

[tex]\sigma=\sqrt{npq} \\ \\ =\sqrt{np(1-p)} \\ \\ =\sqrt{400(0.6)(0.4)} \\ \\ =\sqrt{96}=9.80[/tex]

b.) The mean means that in an experiment of 400 adult smokers, we expect on the average to get about 240 smokers who started smoking before turning 18 years.

c.) It would be unusual to observe 340 smokers who started smoking before turning 18 years old in a random sample of 400 adult smokers because 340 is far greater than the mean of the distribution.

340 is greater than 3 standard deviations from the mean of the distribution.

a.) The mean is given by:

[tex]\mu=np=400\times0.6=240[/tex]

The standard deviation is given by:

[tex]\sigma=\sqrt{npq} \\ \\ =\sqrt{np(1-p)} \\ \\ =\sqrt{400(0.6)(0.4)} \\ \\ =\sqrt{96}=9.80[/tex]

b.) The mean means that in an experiment of 400 adult smokers, we expect on the average to get about 240 smokers who started smoking before turning 18 years.

c.) It would be unusual to observe 340 smokers who started smoking before turning 18 years old in a random sample of 400 adult smokers because 340 is far greater than the mean of the distribution.

340 is greater than 3 standard deviations from the mean of the distribution.