BDA3 Chapter 2 Exercise 3

Here’s my solution to exercise 3, chapter 2, of Gelman’s Bayesian Data Analysis (BDA), 3rd edition. There are solutions to some of the exercises on the book’s webpage.

For 1000 rolls of a fair die, The mean number of sixs is 1000/6 = 166.667, the variance is 138.889, and the standard deviation is 11.7851. Let’s compare the binomial distribution to the normal approximation.

N <- 1000
p <- 1 / 6
mu <- N * p
sigma <- sqrt(N * p * (1 - p))

ex3 <- tibble(
    y = seq(0, N),
    binomial = dbinom(y, N, p),
    normal_approx = dnorm(y, mu, sigma)
  ) %>% 
  gather(metric, probability, -y) 

The two curves are visually indistinguishable. The percentiles are listed in the table below.

percentiles <- c(0.05, 0.25, 0.5, 0.75, 0.95)

tibble(
    percentile = scales::percent(percentiles),
    binom = qbinom(percentiles, N, p),
    norm = qnorm(percentiles, mu, sigma)
) %>% kable() %>% kable_styling()