Normal Distribution
Scott Yanco
March 26, 2018
Learning Objectives
- Be able to describe the charateristics of the normal distribution
- Be able to use the mean and standard deviation to describe any normal distribution
- Be able to convert a normally distributed variable to the “standard normal distribution” and calculate a Z score.
- Get p values from a standard anormal table (or equivalent procedure in R).
- Relate the Central Limit Theorem (CLT) to sampling distributions
- Relate the normal and binomial distributions
The Normal Distribution
The Normal Distribution
The Normal Distribution
Normal Distribution Characteristics
- Continuous variables
- Positive or negative numbers \(-\infty\) to \(\infty\)
- Symetrical
- Described using \(\mu\) and \(\sigma\)
- The normal probability density function: \[ \frac{1}{\sqrt{2 \pi \sigma^{2}}} e^{- \frac{(x- \mu )^{2}}{2 \sigma ^{2}}} \]
You don’t need to memorize this, but note how this is functionally equivalent to the binomial function
An example (with code!)
Let’s work an example with owl masses (from my research). 
Owl Masses
library(mosaic)
owls <- read.file("owl_morph.csv")
head(owls)
## owl_mass owl_sex
## 1 56.5 M
## 2 58.0 M
## 3 51.5 M
## 4 50.5 M
## 5 51.0 M
## 6 48.0 M
Owl Masses
mean.mass <- mean(owls$owl_mass)
print(mean.mass)
## [1] 54.22727
sd.mass <- sd(owls$owl_mass)
print(sd.mass)
## [1] 6.555181
Owl Masses
hist(owls$owl_mass, freq = F)
x.mass <- seq(35, 70)
y.mass <- dnorm(x.mass, mean = mean.mass, sd = sd.mass)
lines(x.mass, y.mass)
One small problem…
The following code will give us the probability mass of the normal distribution for values less than 0:
pnorm(0, mean = mean.mass, sd = sd.mass)
## [1] 6.562843e-17
Discuss with your neighbors: what’s the problem with this?
One small problem…
Non-negative probability of having owls with negative mass!
The Standard Normal (it’s like a medium regular)
Sometimes it is useful to convert a data set to the “standard normal” distribution. This distribution is a specific parameterization of the normal where: \(\mu = 0\) and \(\sigma = 1\)
There are some very convenient properties of this distribution. What are they?
Z-Score
We can convert data sets to the standard normal with a simple function:
\(\vec{Z} = \frac{\vec{Y} - \mu}{\sigma}\)
Let’s do that with the owl data…
Z Score Example
owls$z <- (owls$owl_mass - mean.mass)/sd.mass #note all the cool stuff we did with that code (new data frame column and used scalars on vectors...)
head(owls)
## owl_mass owl_sex z
## 1 56.5 M 0.3467070
## 2 58.0 M 0.5755336
## 3 51.5 M -0.4160484
## 4 50.5 M -0.5685995
## 5 51.0 M -0.4923239
## 6 48.0 M -0.9499772
Standard Normal Table
Using the table on page 279 of Whitlock and Schluter, what’s the probability of a z-score more extreme than \(|1.50|\)?
What about if the z-score were derived from a wing length data set?
But just use R though…
2*pnorm(1.5, lower.tail = F)
## [1] 0.1336144
It even works without having to standardize the distribution…
pnorm(64, mean = mean.mass, sd = sd.mass, lower.tail = F) #this is the one-tailed version
## [1] 0.06800174
Easier, right?
Sampling From the Sample
Getting back to the owl data set, let’s imagine we took a bunch of random samples from this population and look at our sampling distribution (someone remind us what a sampling distribution is…)
In groups consider:
- What do we know about the sample distribution? (i.e. is it normal?)
- What sort of distribution should we expect from the sampling distribution (think back to recitation?
Sampling From the Sample
Let’s imagine we took a bunch of random samples from this population and look at our sampling distribution (someone remind us what a sampling distribution is…)
sampling.mass <- 0
for (i in 1:1000) {
samp <- sample(owls$owl_mass, length(owls$owl_mass), replace = T)
sampling.mass[i] <- mean(samp)
}
Sampling From the Sample
hist(sampling.mass, xlab = "mean owl mass", main = "Sampling Distribution of Mean Owl Masses")
Is it normal?
x.mass <- seq(20, 80, by = .01)
y.mass <- dnorm(x.mass, mean = mean(sampling.mass), sd = sd(sampling.mass))
hist(sampling.mass, xlab = "mean owl mass", main = "Sampling Distribution of Mean Owl Masses", freq = F)
lines(x.mass, y.mass)
The Central Limit Theorem
“…the sum or mean of a large number of measurements randomly sampled from a non-normal population is approximately normal.” (Whitlock and Schluter pg 286)
What does this mean?
The Central Limit Theorem
Let’s simulate a distinctly non-normal data set and see what happens when we invoke the CLT.
Let’s consider the average height of human’s in a daycare - what might we expect that histogram to look like?
The Central Limit Theorem - The Daycare
Let’s simulate a distinctly non-normal data set and see what happens when we invoke the CLT.
Let’s consider the average height of human’s in a daycare - what might we expect that histogram to look like?
#code simulates the bimodal distribution
ad.ht <- rnorm(30, 67, 3)
kid.ht <- rnorm(100, 30, 5)
daycare.heights <- c(ad.ht, kid.ht)
The Central Limit Theorem - The Daycare
#let's look at it
hist(daycare.heights, main = "Freq. distribution of heights at a daycare", xlab = "Heights (in.)")
The Central Limit Theorem - The Daycare
Let’s take a look at the mean of that population:
pop.mean.ht <- mean(daycare.heights)
print(pop.mean.ht)
## [1] 39.12092
The Central Limit Theorem - The Daycare
Let’s take a look at the mean of that population:
hist(daycare.heights, main = "Freq. distribution of heights at a daycare", xlab = "Heights (in.)")
abline(v = pop.mean.ht, col = "red")
Daycare Sample
OK, now we’ve looked at our population, which we typically cannot - let’s do this with a sample, which is more realistic.
We’ll take a random sample of 20, plot it and look at the mean:
dc.samp <- sample(daycare.heights, 20)
samp.mean <- mean(dc.samp)
print(samp.mean)
## [1] 37.4384
Daycare Sample
hist(daycare.heights, main = "Freq. distribution of heights at a daycare", xlab = "Heights (in.)")
abline(v = samp.mean, col = "red")
Daycare Sample Again
OK, now we’ve looked at our population, which we typically cannot - let’s do this with a sample, which is more realistic.
We’ll take a random sample of 20, plot it and look at the mean:
## [1] 40.59754
and again
## [1] 35.6178
and again
## [1] 36.15207
and again
## [1] 38.11664
But Why Though?
Why are the results different each time? Discuss with your neighbors.
SamplING Distribution
Let’s do that sample trick a bunch of times and look at that distribution (sampling distribution).
reps <- 1000
sampling.mean.dc <- 0
for (i in 1:reps) {
dc.samp <- sample(daycare.heights, 20, replace = F)
samp.mean <- mean(dc.samp)
sampling.mean.dc[i] <- samp.mean
}
SamplING Distribution
SamplING Distribution
SamplING Distribution
SamplING Distribution
Binomial to Normal
This concept is really just invoking the CLT again.
Let’s use the owl sex data - we’ll first simulate taking a bunch of random samples to get the \(\mu\) and \(\sigma\).
The we’ll use the shortcut the book showed us.
Binomial to Normal
p.sex <- props(owls$owl_sex)
print(p.sex)
## prop_F prop_M
## 0.25 0.75
Binomial to Normal
successes <- 0
for (i in 1:reps) {
samp <- rbinom(1, size = 20, prob = p.sex[1])
successes[i] <- samp
}
Binomial to Normal
## [1] 5.01
## [1] 1.975805
or use the shortcut \(mean = np\) and \(SD = \sqrt{np(1-p)}\)
## [1] 5
## [1] 1.936492
In fact…
These shortcuts work precisely BECAUSE of the CLT! Cool right?
Remember that \(SE = SD\) of the Sampling distribution? It’s because of the CLT.
Remember \(95 \% CI \approx 2 \times SE\) ? It’s because fo the CLT.
A large proportion of inferential statistics rely upon the CLT to work… as you will see in the remainder of the semester.
