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“ (RCook 2019)

Chapter 8. Probability in R

”Probability theory is the foundation of statistics, and R has plenty of machinery for working with probability, R probability distributions, and R random variables. The R recipes in this chapter show you how to R calculate probabilities from quantiles, R calculate quantiles from probabilities, R generate random variables drawn from distributions, R plot distributions, and so forth.

R has an abbreviated name for every R probability distribution. This name is used to identify the R functions associated with the distribution. For example, the name of the normal distribution is “norm,” which is the root of the function names listed in Table 8-1.

Table 8-1. Normal R distribution functions

Function Purpose dnorm

Normal density

pnorm

Normal distribution function

qnorm

Normal quantile function

rnorm

Normal random variates

Table 8-2 describes some common discrete distributions, and Table 8-3 describes several common continuous distributions.

Table 8-2. Common discrete distributions Discrete distribution R name Parameters Binomial

binom

n = number of trials; p = probability of success for one trial

Geometric

geom

p = probability of success for one trial

Hypergeometric

hyper

m = number of white balls in urn; n = number of black balls in urn; k = number of balls drawn from urn

Negative binomial (NegBinomial)

nbinom

size = number of successful trials; either prob = probability of successful trial or mu = mean

Poisson

pois

lambda = mean

Table 8-3. Common continuous distributions Continuous distribution R name Parameters Beta

beta

shape1; shape2

Cauchy

cauchy

location; scale

Chi-squared (Chisquare)

chisq

df = degrees of freedom

Exponential

exp

rate

F

f

df1 and df2 = degrees of freedom

Gamma

gamma

rate or scale

Log-normal (Lognormal)

lnorm

meanlog = mean on logarithmic scale; sdlog = standard deviation on logarithmic scale

Logistic

logis

location; scale

Normal

norm

mean; sd = standard deviation

Student’s t (TDist)

t

df = degrees of freedom

Uniform

unif

min = lower limit; max = upper limit

Weibull

weibull

shape; scale

Wilcoxon

wilcox

m = number of observations in first sample; n = number of observations in second sample

All distribution-related functions require distributional parameters, such as size and prob for the binomial or prob for the geometric. The big “gotcha” is that the distributional parameters may not be what you expect. For example, we would expect the parameter of an exponential distribution to be β, the mean. The R convention, however, is for the exponential distribution to be defined by the rate = 1/β, so we often supply the wrong value. The moral is, study the help page before you use a function related to a distribution. Be sure you’ve got the parameters right.

To see the R functions related to a particular R probability distribution, use the R help command and the full name of the distribution. For example, this will show the functions related to the normal distribution:

?Normal Some distributions have names that don’t work well with the help command, such as “Student’s t.” They have special help names, as noted in Table 8-2 and Table 8-3: NegBinomial, Chisquare, Lognormal, and TDist. Thus, to get help on the Student’s t distribution, use this:

?TDist See Also There are many other distributions implemented in downloadable packages; see the CRAN task view devoted to probability distributions. The SuppDists package is part of the R base, and it includes 10 supplemental distributions. The MASS package, which is also part of the base, provides additional support for distributions, such as maximum-likelihood fitting for some common distributions as well as sampling from a multivariate normal distribution.

8.1 Counting the Number of Combinations

Problem

You want to calculate the number of combinations of n items taken k at a time.

Solution

Use the choose function:

choose(n, k)

Discussion

A common problem in computing probabilities of discrete variables is counting combinations: the number of distinct subsets of size k that can be created from n items. The number is given by n!/r!(n – r)!, but it’s much more convenient to use the choose function—especially as n and k grow larger:

choose(5, 3) # How many ways can we select 3 items from 5 items?

1. > [1] 10

choose(50, 3) # How many ways can we select 3 items from 50 items?

1. > [1] 19600

choose(50, 30) # How many ways can we select 30 items from 50 items?

1. > [1] 4.71e+13

These numbers are also known as binomial coefficients.

See Also

This recipe merely counts the combinations; see Recipe 8.2 to actually generate them.

Problem

You want to generate all combinations of n items taken k at a time.

Solution

Use the combn function:

items ← 2:5 k ← 2 combn(items, k)

1. > [,1] [,2] [,3] [,4] [,5] [,6]
2. > [1,] 2 2 2 3 3 4
3. > [2,] 3 4 5 4 5 5

Discussion We can use combn(1:5,3) to generate all combinations of the numbers 1 through 5 taken three at a time:

combn(1:5, 3)

1. > [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
2. > [1,] 1 1 1 1 1 1 2 2 2 3
3. > [2,] 2 2 2 3 3 4 3 3 4 4
4. > [3,] 3 4 5 4 5 5 4 5 5 5

The function is not restricted to numbers. We can generate combinations of strings, too. Here are all combinations of five treatments taken three at a time:

combn(c(“T1”, “T2”, “T3”, “T4”, “T5”), 3)

1. > [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
2. > [1,] “T1” “T1” “T1” “T1” “T1” “T1” “T2” “T2” “T2” “T3”
3. > [2,] “T2” “T2” “T2” “T3” “T3” “T4” “T3” “T3” “T4” “T4”
4. > [3,] “T3” “T4” “T5” “T4” “T5” “T5” “T4” “T5” “T5” “T5”

WARNING As the number of items, n, increases, the number of combinations can explode—especially if k is not near to 1 or n.

See Also See Recipe 8.1 to count the number of possible combinations before you generate a huge set.

8.3 Generating Random Numbers Problem You want to generate random numbers.

Solution The simple case of generating a uniform random number between 0 and 1 is handled by the runif function. This example generates one uniform random number:

runif(1)

1. > [1] 0.915

NOTE If you are saying runif out loud (or even in your head), you should pronounce it “are unif” instead of “run if.” The term runif is a portmanteau of “random uniform” so should not sound as if it’s a flow control function.

R can generate random variates from other distributions as well. For a given distribution, the name of the random number generator is “r” prefixed to the distribution’s abbreviated name (e.g., rnorm for the normal distribution’s random number generator). This example generates one random value from the standard normal distribution:

rnorm(1)

1. > [1] 1.53

Discussion Most programming languages have a wimpy random number generator that generates one random number, uniformly distributed between 0.0 and 1.0, and that’s all. Not R.

R can generate random numbers from many probability distributions other than the uniform distribution. The simple case of generating uniform random numbers between 0 and 1 is handled by the runif function:

runif(1)

1. > [1] 0.83

The argument of runif is the number of random values to be generated. Generating a vector of 10 such values is as easy as generating one:

runif(10)

1. > [1] 0.642 0.519 0.737 0.135 0.657 0.705 0.458 0.719 0.935 0.255

There are random number generators for all built-in distributions. Simply prefix the distribution name with “r” and you have the name of the corresponding random number generator. Here are some common ones:

runif(1, min = -3, max = 3) # One uniform variate between -3 and +3

1. > [1] 2.49

rnorm(1) # One standard Normal variate

1. > [1] 1.53

rnorm(1, mean = 100, sd = 15) # One Normal variate, mean 100 and SD 15

1. > [1] 114

rbinom(1, size = 10, prob = 0.5) # One binomial variate

1. > [1] 5

rpois(1, lambda = 10) # One Poisson variate

1. > [1] 12

rexp(1, rate = 0.1) # One exponential variate

1. > [1] 3.14

rgamma(1, shape = 2, rate = 0.1) # One gamma variate

1. > [1] 22.3

As with runif, the first argument is the number of random values to be generated. Subsequent arguments are the parameters of the distribution, such as mean and sd for the normal distribution or size and prob for the binomial. See the function’s R help page for details.

The examples given so far use simple scalars for distributional parameters. Yet the parameters can also be vectors, in which case R will cycle through the vector while generating random values. The following example generates three normal random values drawn from distributions with means of –10, 0, and +10, respectively (all distributions have a standard deviation of 1.0):

rnorm(3, mean = c(-10, 0, +10), sd = 1)

1. > [1] -9.420 -0.658 11.555

That is a powerful capability in cases such as hierarchical models, where the parameters are themselves random. The next example calculates 30 draws of a normal variate whose mean is itself randomly distributed and with hyperparameters of μ = 0 and σ = 0.2:

means ← rnorm(30, mean = 0, sd = 0.2) rnorm(30, mean = means, sd = 1)

1. > [1] -0.5549 -2.9232 -1.2203 0.6962 0.1673 -1.0779 -0.3138 -3.3165
2. > [9] 1.5952 0.8184 -0.1251 0.3601 -0.8142 0.1050 2.1264 0.6943
3. > [17] -2.7771 0.9026 0.0389 0.2280 -0.5599 0.9572 0.1972 0.2602
4. > [25] -0.4423 1.9707 0.4553 0.0467 1.5229 0.3176

If you are generating many random values and the vector of parameters is too short, R will apply the Recycling Rule to the parameter vector.

See Also

See the introduction to this chapter.

Problem

You want to generate a sequence of random numbers, but you want to reproduce the same sequence every time your program runs.

Solution

Before running your R code, call the set.seed function to initialize the random number generator to a known state:

set.seed(42) # Or use any other positive integer…

Discussion

After generating random numbers, you may often want to reproduce the same sequence of “random” numbers every time your program executes. That way, you get the same results from run to run. One of the authors once supported a complicated Monte Carlo analysis of a huge portfolio of securities. The users complained about getting slightly different results each time the program ran. No kidding! The analysis was driven entirely by random numbers, so of course there was randomness in the output. The solution was to set the random number generator to a known state at the beginning of the program. That way, it would generate the same (quasi-)random numbers each time and thus yield consistent, reproducible results.

In R, the set.seed function sets the random number generator to a known state. The function takes one argument, an integer. Any positive integer will work, but you must use the same one in order to get the same initial state.

The function returns nothing. It works behind the scenes, initializing (or reinitializing) the random number generator. The key here is that using the same seed restarts the random number generator back at the same place:

set.seed(165) # Initialize generator to known state runif(10) # Generate ten random numbers

1. > [1] 0.116 0.450 0.996 0.611 0.616 0.426 0.666 0.168 0.788 0.442

set.seed(165) # Reinitialize to the same known state runif(10) # Generate the same ten “random” numbers

1. > [1] 0.116 0.450 0.996 0.611 0.616 0.426 0.666 0.168 0.788 0.442

WARNING When you set the seed value and freeze your sequence of random numbers, you are eliminating a source of randomness that may be critical to algorithms such as Monte Carlo simulations. Before you call set.seed in your application, ask yourself: am I undercutting the value of my program or perhaps even damaging its logic?

See Also See Recipe 8.3 for more about generating random numbers.

8.5 Generating a Random Sample Problem You want to sample a dataset randomly.

Solution The sample function will randomly select n items from a set:

sample(set, n) Discussion Suppose your World Series data contains a vector of years when the Series was played. You can select 10 years at random using sample:

world_series ← read_csv(“./data/world_series.csv”) sample(world_series$year, 10)

1. > [1] 2010 1961 1906 1992 1982 1948 1910 1973 1967 1931

The items are randomly selected, so running sample again (usually) produces a different result:

sample(world_series$year, 10)

1. > [1] 1941 1973 1921 1958 1979 1946 1932 1919 1971 1974

The sample function normally samples without replacement, meaning it will not select the same item twice. Some statistical procedures (especially the bootstrap) require sampling with replacement, which means that one item can appear multiple times in the sample. Specify replace=TRUE to sample with replacement.

It’s easy to implement a simple bootstrap using sampling with replacement. Suppose we have a vector, x, of 1,000 random numbers, drawn from a normal distribution with mean 4 and standard deviation 10:

set.seed(42) x ← rnorm(1000, 4, 10) This code fragment samples 1,000 times from x and calculates the median of each sample:

medians ← numeric(1000) # empty vector of 1000 numbers for (i in 1:1000) {

 medians[i] <- median(sample(x, replace = TRUE))

} From the bootstrap estimates, we can estimate the confidence interval for the median:

ci ← quantile(medians, c(0.025, 0.975)) cat(“95% confidence interval is (”, ci, “)\n”)

1. > 95% confidence interval is ( 3.16 4.49 )

We know that x was created from a normal distribution with a mean of 4, and hence the sample median should be 4 also. (In a symmetrical distribution like this one, the mean and the median are the same.) Our confidence interval easily contains the value.

See Also See Recipe 8.7 for randomly permuting a vector and Recipe 13.8 for more about bootstrapping. Recipe 8.4 discusses setting seeds for quasi-random numbers.

8.6 Generating Random Sequences Problem You want to generate a random sequence, such as a series of simulated coin tosses or a simulated sequence of Bernoulli trials.

Solution Use the sample function. Sample n draws from the set of possible values, and set replace=TRUE:

sample(set, n, replace = TRUE) Discussion The sample function randomly selects items from a set. It normally samples without replacement, which means that it will not select the same item twice and will return an error if you try to sample more items than exist in the set. With replace=TRUE, however, sample can select items over and over; this allows you to generate long, random sequences of items.

The following example generates a random sequence of 10 simulated flips of a coin:

sample(c(“H”, “T”), 10, replace = TRUE)

1. > [1] “H” “T” “H” “T” “T” “T” “H” “T” “T” “H”

The next example generates a sequence of 20 Bernoulli trials—random successes or failures. We use TRUE to signify a success:

sample(c(FALSE, TRUE), 20, replace = TRUE)

1. > [1] TRUE FALSE TRUE TRUE FALSE TRUE FALSE FALSE TRUE TRUE FALSE
2. > [12] TRUE TRUE FALSE TRUE TRUE FALSE FALSE FALSE FALSE

By default sample will choose equally among the set elements, so the probability of selecting either TRUE or FALSE is 0.5. With a Bernoulli trial, the probability p of success is not necessarily 0.5. You can bias the sample by using the prob argument of sample; this argument is a vector of probabilities, one for each set element. Suppose we want to generate 20 Bernoulli trials with a probability of success p = 0.8. We set the probability of FALSE to be 0.2 and the probability of TRUE to 0.8:

sample(c(FALSE, TRUE), 20, replace = TRUE, prob = c(0.2, 0.8))

1. > [1] TRUE TRUE FALSE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
2. > [12] TRUE TRUE TRUE TRUE TRUE FALSE FALSE TRUE TRUE

The resulting sequence is clearly biased toward TRUE. We chose this example because it’s a simple demonstration of a general technique. For the special case of a binary-valued sequence you can use rbinom, the random generator for binomial variates:

rbinom(10, 1, 0.8)

1. > [1] 1 0 1 1 1 1 1 0 1 1

8.7 Randomly Permuting a Vector Problem You want to generate a random permutation of a vector.

Solution If v is your vector, then sample(v) returns a random permutation.

Discussion We typically think of the sample function for sampling from large datasets. However, the default parameters enable you to create a random rearrangement of the dataset. The function call sample(v) is equivalent to:

sample(v, size = length(v), replace = FALSE) which means “select all the elements of v in random order while using each element exactly once.” That is a random permutation. Here is a random permutation of 1, …, 10:

sample(1:10)

1. > [1] 7 3 6 1 5 2 4 8 10 9

See Also See Recipe 8.5 for more about sample.

8.8 Calculating Probabilities for Discrete Distributions Problem You want to calculate either the simple or the cumulative probability associated with a discrete random variable.

Solution For a simple probability, P(X = x), use the density function. All built-in probability distributions have a density function whose name is “d” prefixed to the distribution name; for example, dbinom for the binomial distribution.

For a cumulative probability, P(X ≤ x), use the distribution function. All built-in probability distributions have a distribution function whose name is “p” prefixed to the distribution name; thus, pbinom is the distribution function for the binomial distribution.

Discussion Suppose we have a binomial random variable X over 10 trials, where each trial has a success probability of 1/2. Then we can calculate the probability of observing x = 7 by calling dbinom:

dbinom(7, size = 10, prob = 0.5)

1. > [1] 0.117

That calculates a probability of about 0.117. R calls dbinom the density function. Some textbooks call it the probability mass function or the probability function. Calling it a density function keeps the terminology consistent between discrete and continuous distributions (see Recipe 8.9).

The cumulative probability, P(X ≤ x), is given by the distribution function, which is sometimes called the cumulative probability function. The distribution function for the binomial distribution is pbinom. Here is the cumulative probability for x = 7 (i.e., P(X ≤ 7)):

pbinom(7, size = 10, prob = 0.5)

1. > [1] 0.945

It appears the probability of observing X ≤ 7 is about 0.945.

The density functions and distribution functions for some common discrete distributions are shown in Table 8-4.

Table 8-4. Discrete distributions Distribution Density function: P(X = x) Distribution function: P(X ≤ x) Binomial

dbinom(x, size, prob)

pbinom(x, size, prob)

Geometric

dgeom(x, prob)

pgeom(x, prob)

Poisson

dpois(x, lambda)

ppois(x, lambda)

The complement of the cumulative probability is the survival function, P(X > x). All of the distribution functions let you find this right-tail probability simply by specifying lower.tail=FALSE:

pbinom(7, size = 10, prob = 0.5, lower.tail = FALSE)

1. > [1] 0.0547

Thus we see that the probability of observing X > 7 is about 0.055.

The interval probability, P(x1 < X ≤ x2), is the probability of observing X between the limits x1 and x2. It is calculated as the difference between two cumulative probabilities: P(X ≤ x2) – P(X ≤ x1). Here is P(3 < X ≤ 7) for our binomial variable:

pbinom(7, size = 10, prob = 0.5) - pbinom(3, size = 10, prob = 0.5)

1. > [1] 0.773

R lets you specify multiple values of x for these functions and will return a vector of the corresponding probabilities. Here we calculate two cumulative probabilities, P(X ≤ 3) and P(X ≤ 7), in one call to pbinom:

pbinom(c(3, 7), size = 10, prob = 0.5)

1. > [1] 0.172 0.945

This leads to a one-liner for calculating interval probabilities. The diff function calculates the difference between successive elements of a vector. We apply it to the output of pbinom to obtain the difference in cumulative probabilities—in other words, the interval probability:

diff(pbinom(c(3, 7), size = 10, prob = 0.5))

1. > [1] 0.773

See Also See this chapter’s introduction for more about the built-in probability distributions.

8.9 Calculating Probabilities for Continuous Distributions Problem You want to calculate the distribution function (DF) or cumulative distribution function (CDF) for a continuous random variable.

Solution Use the distribution function, which calculates P(X ≤ x). All built-in probability distributions have a distribution function whose name is “p” prefixed to the distribution’s abbreviated name—for instance, pnorm for the normal distribution.

For example, we can calculate the probability of a draw being from a random standard normal distribution being below 0.8 as follows:

pnorm(q = .8, mean = 0, sd = 1)

1. > [1] 0.788

Discussion The R functions for probability distributions follow a consistent pattern, so the solution to this recipe is essentially identical to the solution for discrete random variables (see Recipe 8.8). The significant difference is that continuous variables have no “probability” at a single point, P(X = x). Instead, they have a “density” at a point.

Given that consistency, the discussion of distribution functions in Recipe 8.8 is applicable here, too. Table 8-5 gives the distribution functions for several continuous distributions.

Table 8-5. Continuous distributions Distribution Distribution function: P(X ≤ x) Normal

pnorm(x, mean, sd)

Student’s t

pt(x, df)

Exponential

pexp(x, rate)

Gamma

pgamma(x, shape, rate)

Chi-squared (χ2)

pchisq(x, df)

We can use pnorm to calculate the probability that a man is shorter than 66 inches, assuming that men’s heights are normally distributed with a mean of 70 inches and a standard deviation of 3 inches. Mathematically speaking, we want P(X ≤ 66) given that X ~ N(70, 3):

pnorm(66, mean = 70, sd = 3)

1. > [1] 0.0912

Likewise, we can use pexp to calculate the probability that an exponential variable with a mean of 40 could be less than 20:

pexp(20, rate = 1 / 40)

1. > [1] 0.393

Just as for discrete probabilities, the functions for continuous probabilities use lower.tail=FALSE to specify the survival function, P(X > x). This call to pexp gives the probability that the same exponential variable could be greater than 50:

pexp(50, rate = 1 / 40, lower.tail = FALSE)

1. > [1] 0.287

Also like discrete probabilities, the interval probability for a continuous variable, P(x1 < X < x2), is computed as the difference between two cumulative probabilities, P(X < x2) – P(X < x1). For the same exponential variable, here is P(20 < X < 50), the probability that it could fall between 20 and 50:

pexp(50, rate = 1 / 40) - pexp(20, rate = 1 / 40)

1. > [1] 0.32

See Also See this chapter’s introduction for more about the built-in probability distributions.

8.10 Converting Probabilities to Quantiles Problem Given a probability p and a distribution, you want to determine the corresponding quantile for p: the value x such that P(X ≤ x) = p.

Solution Every built-in distribution includes a quantile function that converts probabilities to quantiles. The function’s name is “q” prefixed to the distribution name; thus, for instance, qnorm is the quantile function for the normal distribution.

The first argument of the quantile function is the probability. The remaining arguments are the distribution’s parameters, such as mean, shape, or rate:

qnorm(0.05, mean = 100, sd = 15)

1. > [1] 75.3

Discussion A common example of computing quantiles is when we compute the limits of a confidence interval. If we want to know the 95% confidence interval (α = 0.05) of a standard Normal variable, then we need the quantiles with probabilities of α/2 = 0.025 and (1 – α)/2 = 0.975:

qnorm(0.025)

1. > [1] -1.96

qnorm(0.975)

1. > [1] 1.96

In the true spirit of R, the first argument of the quantile functions can be a vector of probabilities, in which case we get a vector of quantiles. We can simplify this example into a one-liner:

qnorm(c(0.025, 0.975))

1. > [1] -1.96 1.96

All the built-in probability distributions provide a quantile function. Table 8-6 shows the quantile functions for some common discrete distributions.

Table 8-6. Discrete quantile distributions Distribution Quantile function Binomial

qbinom(p, size, prob)

Geometric

qgeom(p, prob)

Poisson

qpois(p, lambda)

Table 8-7 shows the quantile functions for common continuous distributions.

Table 8-7. Continuous quantile distributions Distribution Quantile function Normal

qnorm(p, mean, sd)

Student’s t

qt(p, df)

Exponential

qexp(p, rate)

Gamma

qgamma(p, shape, rate) or qgamma(p, shape, scale)

Chi-squared (χ2)

qchisq(p, df)

See Also Determining the quantiles of a dataset is different from determining the quantiles of a distribution—see Recipe 9.5.

8.11 Plotting a Density Function Problem You want to plot the density function of a probability distribution.

Solution Define a vector x over the domain. Apply the distribution’s density function to x and then plot the result. If x is a vector of points over the domain you care about plotting, you then calculate the density using one of the d_____ density functions, like dlnorm for lognormal or dnorm for normal:

dens ← data.frame(x = x,

                  y = d_____(x))

ggplot(dens, aes(x, y)) + geom_line() Here is a specific example that plots the standard normal distribution for the interval –3 to +3:

library(ggplot2)

x ← seq(-3, +3, 0.1) dens ← data.frame(x = x, y = dnorm(x))

ggplot(dens, aes(x, y)) + geom_line() Figure 8-1 shows the smooth density function.

rcbk 0801 Figure 8-1. Smooth density function Discussion All the built-in probability distributions include a density function. For a particular density, the function name is “d” prepended to the distribution name. The density function for the normal distribution is dnorm, the density for the gamma distribution is dgamma, and so forth.

If the first argument of the density function is a vector, then the function calculates the density at each point and returns the vector of densities.

The following code creates a 2 × 2 plot of four densities (Figure 8-2):

x ← seq(from = 0, to = 6, length.out = 100) # Define the density domains ylim ← c(0, 0.6)

1. Make a data.frame with densities of several distributions

df ← rbind(

 data.frame(x = x, dist_name = "Uniform"=, y = dunif(x, min   = 2, max = 4)),
 data.frame(x = x, dist_name = "Normal"=, y = dnorm(x, mean  = 3, sd = 1)),
 data.frame(x = x, dist_name = "Exponential", y = dexp(x, rate  = 1 / 2)),
 data.frame(x = x, dist_name = "Gamma"=, y = dgamma(x, shape = 2, rate = 1)) )

1. Make a line plot like before, but use facet_wrap to create the grid

ggplot(data = df, aes(x = x, y = y)) +

 geom_line() +
 facet_wrap(~dist_name)   # facet and wrap by the variable dist_name

rcbk 0802 Figure 8-2. Multiple density plots Figure 8-2 shows four density plots. However, a raw density plot is rarely useful or interesting by itself, and we often shade a region of interest.

Figure 8-3 is a normal distribution with shading from the 75th percentile to the 95th percentile.

We create the plot by plotting the density and then creating a shaded region with the geom_ribbon function from ggplot2.

First, we create some data and draw a density curve like the one shown in Figure 8-4:

x ← seq(from = -3, to = 3, length.out = 100) df ← data.frame(x = x, y = dnorm(x, mean = 0, sd = 1))

p ← ggplot(df, aes(x, y)) +

 geom_line() +
 labs(
   title = "Standard Normal Distribution",
   y = "Density",
   x = "Quantile"
 )

p rcbk 0803 Figure 8-3. Standard normal with shading rcbk 0804 Figure 8-4. Density plot Next, we define the region of interest by calculating the x values for the quantiles we’re interested in. Finally, we use geom_ribbon to add a subset of our original data as a colored region:

q75 ← quantile(df$x, .75) q95 ← quantile(df$x, .95)

p +

 geom_ribbon(
   data = subset(df, x > q75 & x < q95),
   aes(ymax = y),
   ymin = 0,
   fill = "blue",
   color = NA,
   alpha = 0.5
 )

The resulting plot is shown in Figure 8-5.

rcbk 0805 Figure 8-5. Normal density with shading

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