The normal distribution

The preceding pages introduced continuous distributions through their probability density functions. Suppose a phonetic model represents an F1 measurement as varying symmetrically around 550 Hz. The density is highest at 550 and decreases as distance from 550 increases in either direction.

The normal distribution is a continuous location and scale family with this symmetric shape. We write

X\sim\mathcal{N}(\mu,\sigma^2).

The first parameter is the mean:

\mathbb{E}[X]=\mu.

The second parameter is the variance:

\operatorname{Var}(X)=\sigma^2.

Thus the standard deviation is \sigma, not \sigma^2.

Reading the density from the center outward

The density is

f_X(x) \equiv\frac{1}{\sqrt{2\pi\sigma^2}} \exp\!\left[-\frac{(x-\mu)^2}{2\sigma^2}\right].

At x=\mu, the centered distance x-\mu is zero, so the density reaches its maximum. Moving the same distance above or below \mu produces the same squared distance and thus the same density.

This equality gives the family’s symmetry:

f_X(\mu-d)=f_X(\mu+d).

The density decreases smoothly as |d| grows.

Changing location while holding scale fixed

Increasing \mu shifts the full density to larger values. Decreasing \mu shifts it to smaller values. The spread and symmetric shape remain unchanged when \sigma is held fixed.

For instance, moving from \mathcal{N}(550,40^2) to \mathcal{N}(600,40^2) adds 50 Hz to the center without changing the standard deviation.

Changing scale while holding location fixed

Increasing \sigma spreads probability across a wider range. The density peak becomes lower because the total area must remain one. Decreasing \sigma concentrates probability nearer \mu.

Compare \mathcal{N}(550,40^2) with \mathcal{N}(550,80^2). Both have mean 550 Hz, but the second standard deviation is twice as large.

Code

f1 <- seq(300, 800, length.out = 400)

plot(f1, dnorm(f1, mean = 550, sd = 40), type = "l",
     xlab = "F1 in Hz", ylab = "density")
lines(f1, dnorm(f1, mean = 600, sd = 40), lty = 2)
lines(f1, dnorm(f1, mean = 550, sd = 80), lty = 3)

The first comparison isolates location; the second isolates scale.

Standardizing one value

Standardization subtracts the mean and divides by the standard deviation. The resulting z-score

Z\equiv\frac{X-\mu}{\sigma}

records signed distance from the mean in standard deviation units. Under the normal model,

Z\sim\mathcal{N}(0,1).

For X\sim\mathcal{N}(550,40^2), the values 510 and 590 are one standard deviation below and above the mean:

\frac{510-550}{40}=-1

and

\frac{590-550}{40}=1.

The probability between them is

\mathbb{P}(510\leq X\leq590) =\mathbb{P}(-1\leq Z\leq1) \approx.683.

In R,

Code

pnorm(590, mean = 550, sd = 40) -
  pnorm(510, mean = 550, sd = 40)

This interval calculation is a property of the normal family. It is not a universal rule for every distribution with mean 550 and standard deviation 40.

Checking support against the measurement

A normal variable has support across the full real line. It assigns some density to negative values even when the measurement, such as F1, cannot be negative.

The approximation may still be useful when the mean is many standard deviations from the boundary and the modeled density below zero is negligible. It may be poor for a strongly right skewed response or one concentrated near a hard boundary.

A histogram is not enough

A roughly bell shaped histogram does not by itself justify a normal model. A finite histogram may hide tail asymmetry, boundary problems, or repeated structure across speakers and items.

Check the support, the symmetry claim, and the deviations relevant to the linguistic analysis. Visual familiarity alone does not establish those properties.

Check your understanding

In \mathcal{N}(100,225), what are the mean, variance, and standard deviation?
Standardize X=130 under the model in the first question.
Which parameter changes when a density shifts right without changing shape?
Give one linguistic response for which support on the full real line may be a poor direct choice.

The normal family supplies the standard normal variables used to construct the chi-squared distribution on the next page.

--- title: "The normal distribution" --- The preceding pages introduced continuous distributions through their [probability density functions](probability-density-functions.qmd). Suppose a phonetic model represents an F1 measurement as varying symmetrically around 550 Hz. The density is highest at 550 and decreases as distance from 550 increases in either direction. The [**normal distribution**](https://bruno.nicenboim.me/bayescogsci/ch-intro.html) is a continuous location and scale family with this symmetric shape. We write $$ X\sim\mathcal{N}(\mu,\sigma^2). $$ The first parameter is the mean: $$ \mathbb{E}[X]=\mu. $$ The second parameter is the variance: $$ \operatorname{Var}(X)=\sigma^2. $$ Thus the standard deviation is $\sigma$, not $\sigma^2$. ## Reading the density from the center outward The density is $$ f_X(x) \equiv\frac{1}{\sqrt{2\pi\sigma^2}} \exp\!\left[-\frac{(x-\mu)^2}{2\sigma^2}\right]. $$ At $x=\mu$, the centered distance $x-\mu$ is zero, so the density reaches its maximum. Moving the same distance above or below $\mu$ produces the same squared distance and thus the same density. This equality gives the family's symmetry: $$ f_X(\mu-d)=f_X(\mu+d). $$ The density decreases smoothly as $|d|$ grows. ## Changing location while holding scale fixed Increasing $\mu$ shifts the full density to larger values. Decreasing $\mu$ shifts it to smaller values. The spread and symmetric shape remain unchanged when $\sigma$ is held fixed. For instance, moving from $\mathcal{N}(550,40^2)$ to $\mathcal{N}(600,40^2)$ adds 50 Hz to the center without changing the standard deviation. ## Changing scale while holding location fixed Increasing $\sigma$ spreads probability across a wider range. The density peak becomes lower because the total area must remain one. Decreasing $\sigma$ concentrates probability nearer $\mu$. Compare $\mathcal{N}(550,40^2)$ with $\mathcal{N}(550,80^2)$. Both have mean 550 Hz, but the second standard deviation is twice as large. ```{r} #| eval: false f1 <- seq(300, 800, length.out = 400) plot(f1, dnorm(f1, mean = 550, sd = 40), type = "l", xlab = "F1 in Hz", ylab = "density") lines(f1, dnorm(f1, mean = 600, sd = 40), lty = 2) lines(f1, dnorm(f1, mean = 550, sd = 80), lty = 3) ``` The first comparison isolates location; the second isolates scale. ## Standardizing one value [**Standardization**](https://appliedstatisticsforlinguists.org/bwinter_stats_proofs.pdf#page=102) subtracts the mean and divides by the standard deviation. The resulting [**z-score**](https://openstax.org/books/introductory-statistics-2e/pages/6-1-the-standard-normal-distribution) $$ Z\equiv\frac{X-\mu}{\sigma} $$ records signed distance from the mean in standard deviation units. Under the normal model, $$ Z\sim\mathcal{N}(0,1). $$ For $X\sim\mathcal{N}(550,40^2)$, the values 510 and 590 are one standard deviation below and above the mean: $$ \frac{510-550}{40}=-1 $$ and $$ \frac{590-550}{40}=1. $$ The probability between them is $$ \mathbb{P}(510\leq X\leq590) =\mathbb{P}(-1\leq Z\leq1) \approx.683. $$ In R, ```{r} #| eval: false pnorm(590, mean = 550, sd = 40) - pnorm(510, mean = 550, sd = 40) ``` This interval calculation is a property of the normal family. It is not a universal rule for every distribution with mean 550 and standard deviation 40. ## Checking support against the measurement A normal variable has support across the full real line. It assigns some density to negative values even when the measurement, such as F1, cannot be negative. The approximation may still be useful when the mean is many standard deviations from the boundary and the modeled density below zero is negligible. It may be poor for a strongly right skewed response or one concentrated near a hard boundary. ## A histogram is not enough A roughly bell shaped histogram does not by itself justify a normal model. A finite histogram may hide tail asymmetry, boundary problems, or repeated structure across speakers and items. Check the support, the symmetry claim, and the deviations relevant to the linguistic analysis. Visual familiarity alone does not establish those properties. ## Check your understanding 1. In $\mathcal{N}(100,225)$, what are the mean, variance, and standard deviation? 2. Standardize $X=130$ under the model in the first question. 3. Which parameter changes when a density shifts right without changing shape? 4. Give one linguistic response for which support on the full real line may be a poor direct choice. The normal family supplies the standard normal variables used to construct the [chi-squared distribution on the next page](chi-squared-distribution.qmd).