What percent of a standard normal distribution \(N(\mu = 0, \sigma = 1)\) is found in each region? Be sure to draw a graph.

1. Z < -1.35
2. Z > 1.48
3. -0.4 < Z < 1.5
4. |Z| > 2

First, we define in R code the Z score and then use the pnorm function to determine the percentage on the left tail.

zLt <- -1.35
pLt <- pnorm(zLt)
pLt

## [1] 0.08850799

The percent of the standard normal distribution found in the region Z < -1.35 is 8.85%.

Again, we define in R code the Z score and then use the pnorm function to determine the percentage on the left tail. This time we subtract the value returned from pnorm from 1 to convert to the right tail value.

zGt <- 1.48
pGt <- 1 - pnorm(zGt)
pGt

## [1] 0.06943662

The percent of the standard normal distribution found in the region Z > 1.48 is 6.94%.

Again, we define in R code the Z scores and use pnorm. This time we have a middle region with a portion below zero and a portion above zero, therefore we substract using 0.5 depending on which side we are on.

z1 <- -0.4
p1 <- 0.5 - pnorm(z1) # Left of mean 0

z2 <- 1.5
p2 <- pnorm(z2) - 0.5 # Right of mean 0

pT <- p1 + p2
pT

## [1] 0.5886145

The percent of the standard normal distribution found in the region -0.4 < Z < 1.5 is 58.86%.

Once again, we define in R code the Z score and use pnorm. We have a two separate tail regions with a portion below 0 and a portion above zero (not contiguous). In this case, the regions are identical due to the absolute value, therefore we can simply multiply by 2.

z <- 2
p1 <- 1 - pnorm(z) # Right tail

pT <- 2 * p1
pT

## [1] 0.04550026

The percent of the standard normal distribution found in the region ( |Z| > 2) is 4.55%.