We learned in Exercise 4.17 that about 70% of 18-20 year olds consumed alcoholic beverages in any given year. We now consider a random sample of fifty 18-20 year olds.

Part A

How many people would you expect to have consumed alcoholic beverages? And with what standard deviation?

\(\mu = np\)

n <- 50
p <- 0.697

# Average
n*p
## [1] 34.85

\(\sigma = \sqrt[]{np(1-p)}\)

# Standard deviation
sqrt(n*p*(1-p))
## [1] 3.249546

Answer: Average: 34.85. Standard deviation: 3.25.

Part B

Would you be surprised if there were 45 or more people who have consumed alcoholic beverages?

sd <- sqrt(n*p*(1-p))
avg <- n*p

# Z-score
(45-avg)/sd
## [1] 3.123513

Answer: Yes, I would be surprised as 45 is greater than 3 standard deviations away from the mean.

Part C

What is the probability that 45 or more people in this sample have consumed alcoholic beverages? How does this probability relate to your answer to part (b)?

# Normal approximation with 0.5 correction
normalPlot(mean = avg, sd = sd, bounds = c(44.5, Inf), tails = FALSE)

Answer: 0.00149, with a 0.5 correction.