This is a continuous Symmetric, unimodal bell-shaped Distribution

- it has two parameters: mean $\mu$ and std $\sigma$, denoted as $N(\mu, \sigma)$
- Standard Normal Distribution is $N(\mu = 0, \sigma = 1)$

x = seq(from=-3, to=3, length=15) normalDensity = dnorm(x, mean=0, sd=1) r = round(normalDensity, 2) bp = barplot(r) xspline(x=bp, y=r, lwd=2, shape=1, border="blue") text(x=bp, y=r+0.03, labels=as.character(r), xpd=TRUE, cex=0.7)

Also referred as the "rule of 3 sigmas"

- most of the data lay within 3 $\sigma$s from $\mu$

$Z$-score of an observation is the number of standard deviations for the mean

- 1 sdt above - $z = +1$
- 1.5 std below - $z = -1.5$
- $z = \cfrac{x - \mu}{\sigma}$

we can use $z$-scores to identify unusual observations

- $x_1$ is more unusual than $x_2$ if $| z_1 | > | z_2 |$

- so $Z$-scores are used to standardize the observations
- in effect, it normalizes any normal distribution $N(\mu, \sigma$) to $N(0, 1)$
- see Normalization

Example:

- Scores of SAT takers are distributed normally
- parameters: $\mu = 1500, \sigma = 300$
- Ann earned 1800 on SAT,
- so Ann's $z = 1$

Ann's *percentile* - percent of people who earned lower SAT score

- shaded - individuals who scored below Ann
- so knowing the $z$-score we can calculate the percentile
- Ann is the 84th percentile of SAT takers

- and vise-versa: we can also find $z$-score for given percentile

Example 2

- Shannon is a randomly selected SAT-taker.
- What's the probability that she'll score 1630 or more?
- Can find the $z$-score for that - it's $z = \cfrac{x - \mu}{\sigma} = 0.43$
- so we calculate the percentiles
- probability of getting below $z=0.43$ is 2/3
- so probability of getting above $z=0.43$ is 1 - 2/3 = 1/3

Always draw the bell shape first and then shade the area of interest

it may be useful for

- calculating Confidence Intervals, e.g. Confidence Intervals for Means or Binomial Proportion Confidence Intervals
- doing Hypothesis Testing, e.g.

Many processes can be approximated well by normal distribution

- e.g. SAT, height of USA males, etc

But need to check if it's reasonable to use the normal approximation

2 visual methods for checking the assumption of normality

- simple histogram + best fit of normal shape
- Q-Q Plot (or Normal Probability Plot)

Code to produce the first figure:

load(url("http://www.openintro.org/stat/data/bdims.RData")) fdims = subset(bdims, bdims$sex == 0) hist(fdims$hgt, probability=TRUE, ylim=c(0, 0.07)) x = 140:190 y = dnorm(x=x, mean=mean(fdims$hgt), sd=sd(fdims$hgt)) lines(x=x, y=y, col="blue")

Code to produce Q-Q Plots

qqnorm(fdims$hgt, col="orange", pch=19) qqline(fdims$hgt, lwd=2)