3.1 Area under the curve.

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

(a) Z < -1.35

\[Z=\frac{x-\mu}{\sigma}\]

mu <- 0
sd <- 1
Z <- -1.35
# finding value for "x"
x <- Z * sd + mu
x
## [1] -1.35

Since we have that \(P(Z < -1.35)\) then, we can find \(P(x < -1.35)\)

# Finding probability for the left tail
pnorm(x, mean = 0, sd = 1)
## [1] 0.08850799
# Probability curve plot
normalPlot(mean = 0, sd = 1, bounds = c(-Inf, x), tails = FALSE)

Answer: The percentage represented in the region is: 8.85%

(b) Z > 1.48

\[Z=\frac{x-\mu}{\sigma}\]

mu <- 0
sd <- 1
Z <- 1.48
# finding value for "x"
x <- Z * sd + mu
x
## [1] 1.48

Since we have that \(P(Z > 1.48)\) then, we can find \(P(x > 1.48)\)

# Finding probability for the right tail (I am taking it as the complement)
1 - pnorm(x, mean = 0, sd = 1)
## [1] 0.06943662
# Probability curve plot
normalPlot(mean = 0, sd = 1, bounds = c(x, Inf), tails = FALSE)

Answer: The percentage represented in the region is: 6.94%

(c) -0.4 < Z < 1.5

\[Z=\frac{x-\mu}{\sigma}\]

mu <- 0
sd <- 1
Z1 <- -0.4
Z2 <- 1.5
# finding value for "x"
x1 <- Z1 * sd + mu
x2 <- Z2 * sd + mu
x1
## [1] -0.4
x2
## [1] 1.5

Since we have that \(P(-0.4 < Z < 1.5)\) then, we can find \(P(-0.4 < x < 1.5)\)

# Finding probabilities in order tofind our desired probability.
p1 <- pnorm(x1, mean = 0, sd = 1)
p2 <- pnorm(x2, mean = 0, sd = 1)
p2 - p1
## [1] 0.5886145
# Probability curve plot
normalPlot(mean = 0, sd = 1, bounds = c(x1, x2), tails = FALSE)

Answer: The percentage represented in the region is: 58.86%

(d) |Z| > 2

\[Z=\frac{x-\mu}{\sigma}\]

mu <- 0
sd <- 1
Z <- 2
# finding value for "x"
x <- Z * sd + mu
# Since it's an absolute value, we have a negative value x1 and a positve value x2
x1 <- -x
x2 <- x

Since we have that \(P(|Z| > 2)\) then, we can find \(P(|x| > 2) = P(x < -2 \: or \: x > 2\)

Do not try to write it as \(P(-2 > x > 2)\) <– WRONG!

# Finding probability fo |x| > 2 which means P(x < -2) or P(x > 2)
p1 <- pnorm(x1, mean = 0, sd = 1)
p2 <- 1- pnorm(x2, mean = 0, sd = 1)
p1 +  p2
## [1] 0.04550026
# Probability curve plot
normalPlot(mean = 0, sd = 1, bounds = c(x1, x2), tails = TRUE)

Answer: The percentage represented on the region is: 4.55%

Thank you!

Questions?