4.17; Underage drinking pt. 1: Data collected by the Substance Abuse and Mental HEalth Service Administration (SAMSHA) suggests that 69.7% of 18-20 year olds consumed alcoholic beverages in any given year.

  1. Suppose a random sample of of ten 18-20 year olds is taken. Is the use of the binomial distribution appropriate for calculating the probability that exactly 6 consumed alcohol? Explain

The binomial distribution is appropriate for this calculation because the sample of 18-20 year olds is fixed (10), and the probability that they’ve consumed alcohol is a success or failure event (consumed (P = 0.697) or not consumed (Q = 0.303))

  1. Calculate the probabiltiy that exactly 6/10 randomly sampled 18-20 year olds have consumed alcohol.

Binomial formula: (N choose K) * P^K * Q^(N-K) N = 10 P = 0.067 K = 6 Q = 0.303

choose(10,6) * (0.697)^6 * (0.303)^4
## [1] 0.2029488
  1. What is the probability that exactly 4/10 18-20 year olds have not consumed alcohol?
choose(10,4) * (0.303)^4 * (0.697)^6 
## [1] 0.2029488
  1. What is the probabiltiy that at most 2/5 randomly sampled 18-20 year olds have consumed alcohol?

Idea: P(k = 0, or 1, or 2,…., or n) = P(k=0) + P(k=1) + … + P(k=n) = P(k = “at most”) Where P is calculated with the binomial formula:

(N choose K) * P^K * Q^(N-K) N = 5 P = 0.697 K = 0,1,2 Q = 0.303

p_zero <- choose(5,0) * (0.697)^0 * (0.303)^5
p_one <- choose(5,1) * (0.697)^1 * (0.303)^4
P_two <- choose(5,2) * (0.697)^2 * (0.303)^3

at_most_two <- p_zero + p_one + P_two
at_most_two
## [1] 0.1670716
  1. What is the probability that at least 1/5 randomly sampled 18-20 year olds have consumed alcohol? > This part can be done in two ways: The first follows the above method of adding all probabilities following the binomial distribution. The other method entails finding the compliment of the probability that nobody has consumed alcohol–the compliment is illustrated as the Q variable in the described binomial formula above, which follows Q = 1 - P > Both methods will be exemplified to demonstrate effectiveness

Sum of all probabilities: P(k=1) + P(k=2) + P(k=3) + P(k=4) + P(k=5)

p_one <- choose(5,1) * (0.697)^1 * (0.303)^4
P_two <- choose(5,2) * (0.697)^2 * (0.303)^3
P_three <- choose(5,3) * (0.697)^3 * (0.303)^2
p_four <- choose(5,4) * (0.697)^4 * (0.303)^1
p_five <- choose(5,5) * (0.697)^5 * (0.303)^0

at_least_one = p_one + P_two + P_three + p_four + p_five
at_least_one
## [1] 0.997446

Via the compliment of nobody drinking

p_at_least_one_comp <- (1 - choose(5,0) * (0.697)^0 * (0.303)^5)
p_at_least_one_comp
## [1] 0.997446

As we compare both answers and methods above, we see it is much easier to find the compliment of nobody drinking, which is the same as asking/solving for the probability of at least one drinking alcohol.