# 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) 
y metric probability
0 binomial 0
1 binomial 0
2 binomial 0
3 binomial 0
4 binomial 0
5 binomial 0

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()
percentile binom norm
5% 147 147.2819
25% 159 158.7177
50% 167 166.6667
75% 175 174.6156
95% 186 186.0515