Question

according to an almanac, 80% 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?
(a) \\( \mu _ { x } = 320 \\)
\\( \sigma _ { x } = 8.0 \\) (round to the nearest tenth as needed.)
(b) what is the correct interpretation of the mean?
a. it is expected that in a random sample of 400 adult smokers, 320 will have started smoking before turning 18
b. it is expected that in a random sample of 400 adult smokers, 320 will have started smoking after turning 18
c. it is expected that in 50% of random samples of 400 adult smokers, 320 will have started smoking before
turning 18.

Explanation:

Step1: Recall the formula for the mean of a binomial distribution

The mean of a binomial distribution is given by \(\mu = np\), where \(n = 400\) (number of trials) and \(p=0.8\) (probability of success).
\(\mu = 400\times0.8=320\)

Step2: Recall the formula for the standard deviation of a binomial distribution

The standard deviation of a binomial distribution is \(\sigma=\sqrt{np(1 - p)}\).
Substitute \(n = 400\) and \(p = 0.8\) into the formula:
\(\sigma=\sqrt{400\times0.8\times(1 - 0.8)}=\sqrt{400\times0.8\times0.2}=\sqrt{64}=8.0\)

Step3: Interpret the mean

The mean of a binomial distribution represents the expected value. In the context of this problem, \(n = 400\) (number of adult smokers in the sample) and \(p = 0.8\) (probability that a smoker started before 18). So, the mean \(\mu=320\) means that in a random sample of \(n = 400\) adult smokers, we expect \(320\) to have started smoking before turning 18.

Answer:

(a) \(\mu_X = 320\), \(\sigma_X=8.0\)
(b) A. It is expected that in a random sample of 400 adult smokers, 320 will have started smoking before turning 18.