Mathematics, 21.07.2020 20:01, matt416760

Data collected by the Substance Abuse and Mental Health Services Administration (SAMSHA) suggests that 69.7% of 18-20-year-olds consumed alcoholic beverages in 2008. (a) Suppose a random sample of the ten 18-20-year-olds is taken. Is the use of the binomial distribution appropriate for calculating the probability that exactly six consumed alcoholic beverages?
i. No, this follows the bimodal distribution.
ii. Yes, there are 10 independent trials, each with exactly two possible outcomes, and a constant probability associated with each possible outcome.
iii. No, the trials are not independent.
iv. No, the normal distribution should be used.
(b) Calculate the probability that exactly 6 out of 10 randomly sampled 18- 20-year-olds consumed an alcoholic drink.
(c) What is the probability that exactly four out of the ten 18-20-year-olds have not consumed an alcoholic beverage?
(d) What is the probability that at most 2 out of 5 randomly sampled 18-20-year-olds have consumed alcoholic beverages?

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(a) Yes, there are 10 independent trials, each with exactly two possible outcomes, and a constant probability associated with each possible outcome.

(b) The probability that exactly 6 out of 10 randomly sampled 18- 20-year-olds consumed an alcoholic drink is 0.203.

(c) The probability that exactly 4 out of 10 randomly sampled 18- 20-year-olds have not consumed an alcoholic drink is 0.203.

(d) The probability that at most 2 out of 5 randomly sampled 18-20-year-olds have consumed alcoholic beverages is 0.167.

Step-by-step explanation:

We are given that data collected by the Substance Abuse and Mental Health Services Administration (SAMSHA) suggests that 69.7% of 18-20-year-olds consumed alcoholic beverages in 2008.

(a) The conditions required for any variable to be considered as a random variable is given by;

The experiment consists of identical trials.Each trial must have only two possibilities: success or failure.The trials must be independent of each other.

So, in our question; all these conditions are satisfied which means the use of the binomial distribution is appropriate for calculating the probability that exactly six consumed alcoholic beverages.

Yes, there are 10 independent trials, each with exactly two possible outcomes, and a constant probability associated with each possible outcome.

(b) Let X = Number of 18- 20-year-olds people who consumed an alcoholic drink

The above situation can be represented through binomial distribution;

$P(X = r) = \binom{n}{r}\times p^{r} \times (1-p)^{n-r}; x = 0,1,2,......$

where, n = number of trials (samples) taken = 10 people

r = number of success = exactly 6

p = probability of success which in our question is % 18-20

year-olds consumed alcoholic beverages in 2008, i.e; 69.7%.

So, X ~ Binom(n = 10, p = 0.697)

Now, the probability that exactly 6 out of 10 randomly sampled 18- 20-year-olds consumed an alcoholic drink is given by = P(X = 6)

P(X = 3) = $\binom{10}{6}\times 0.697^{6} \times (1-0.697)^{10-6}$

= $210\times 0.697^{6} \times 0.303^{4}$

= 0.203

(c) The probability that exactly 4 out of 10 randomly sampled 18- 20-year-olds have not consumed an alcoholic drink is given by = P(X = 4)

Here p = 1 - 0.697 = 0.303 because here our success is that people who have not consumed an alcoholic drink.

P(X = 4) = $\binom{10}{4}\times 0.303^{4} \times (1-0.303)^{10-4}$

= $210\times 0.303^{4} \times 0.697^{6}$

= 0.203

(d) Let X = Number of 18- 20-year-olds people who consumed an alcoholic drink

The above situation can be represented through binomial distribution;

$P(X = r) = \binom{n}{r}\times p^{r} \times (1-p)^{n-r}; x = 0,1,2,......$

where, n = number of trials (samples) taken = 5 people

r = number of success = at most 2

p = probability of success which in our question is % 18-20

year-olds consumed alcoholic beverages in 2008, i.e; 69.7%.

So, X ~ Binom(n = 5, p = 0.697)

Now, the probability that at most 2 out of 5 randomly sampled 18-20-year-olds have consumed alcoholic beverages is given by = P(X $\leq$ 2)

P(X $\leq$ 2) = P(X = 0) + P(X = 1) + P(X = 3)

= $\binom{5}{0}\times 0.697^{0} \times (1-0.697)^{5-0}+\binom{5}{1}\times 0.697^{1} \times (1-0.697)^{5-1}+\binom{5}{2}\times 0.697^{2} \times (1-0.697)^{5-2}$