Question

Underage drinking, Part I: 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? No, this follows the bimodal distribution No, the normal distribution should be used Yes, there are 10 independent trials, each with exactly two possible outcomes, and a constant probability associated with each possible outcome No, the trials are not independent (please round to four (b) Calculate the probability that exactly 6 out of 10 randomly sampled 18-20 year olds consumed an alcoholic drink. (please round to four decimal places) (c) What is the probability that exactly four out of the ten 18-20 year olds have not consumed an alcoholic beverage? decimal places) (d) What is the probability that at most 2 out of 5 randomly sampled 18-20 year olds have consumed alcoholic beverages? (please round to four decimal places) (e) What is the probability that at least 1 out of 5 randomly sampled 18-20 year olds have consumed alcoholic beverages? (please round to four decimal places)

Answer 1

Answer:

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 is 0.2029 or 20.29%

c. The probability is 0.2029 or 20.29%

d. The probability is 0.1671 or 16.71%

e. The probability is 0.9975 or 99.75%

Step-by-step explanation:

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

b. Let's use the binomial distribution table, this way:

Binomial distribution (n=10, p=0.697)

 f(x) F(x) 1 - F(x)

x Pr[X = x] Pr[X ≤ x]

0 0.0000 0.0000

1 0.0002 0.0002

2 0.0016 0.0017

3 0.0095 0.0112

4 0.0384 0.0496

5 0.1059 0.1555

6 0.2029 0.3584

7 0.2668 0.6252

8 0.2301 0.8553

9 0.1176 0.9729

10 0.0271 1.0000

The probability is 0.2029 or 20.29%

c. If 69.7% of 18-20 years old consumed alcoholic beverages in 2008, therefore, 30.3% did not and the binomial distribution table is:

Binomial distribution (n=10, p=0.303)

 f(x) F(x) 1 - F(x)

x Pr[X = x] Pr[X ≤ x]

0 0.0271 0.0271

1 0.1176 0.1447

2 0.2301 0.3748

3 0.2668 0.6416

4 0.2029 0.8445

5 0.1059 0.9504

6 0.0384 0.9888

7 0.0095 0.9983

8 0.0016 0.9998

9 0.0002 1.0000

10 0.0000 1.0000

The probability is 0.2029 or 20.29%

d. Let's use the binomial distribution table, this way:

Binomial distribution (n=5, p=0.697)

 f(x) F(x) 1 - F(x)

x Pr[X = x] Pr[X ≤ x]

0 0.0026 0.0026

1 0.0294 0.0319

2 0.1351 0.1671

3 0.3109 0.4779

4 0.3576 0.8355

5 0.1645 1.0000

P(0) + P(1) + P (2) = 0.0026 + 0.0294 + 0.1351

The probability is 0.1671 or 16.71%

e. Using the same binomial distribution table we used in d. we have:

P(1) + P (2) + P(3) + P(4) + P (5) = 0.0294 + 0.1351 + 0.3109 + 0.3576 + 0.1645

The probability is 0.9975 or 99.75%