Setup

# Load Libriaries
library(tidyverse)
library(knitr)

# Read data
super <- read.csv("2013Coupes.csv")
dat <- filter(super,Price < 100000)

Identify Variables

#### Calculate and assign mean
m <- mean(dat$Price)

#### Assign sample size 
n <- 4

#### Calculate new standard deviation and assign 
o <- sd(dat$Price)/sqrt(n)

#### Return results
df <- data.frame(m,n,o)
kable(df)
m n o
21145.8 4 3096.83

Problem 1: For the next 4 cars that are sampled, what is the probability that the price will be less than $500 dollars below the mean?

# Calculate mean-500 and assign
x <- m-500

# Calculate z
z <- (x-m)/o

# Calculate probability where P(X <= J) = P(X < J)
p <- pnorm(z)

## Interpret results
cat("For the next 4 cars that are sampled, there is a ",round(p*100,digits = 2),"% probability that the price will be less than $500 dollars below the mean." )
## For the next 4 cars that are sampled, there is a  43.59 % probability that the price will be less than $500 dollars below the mean.

Problem 2: For the next 4 cars that are sampled, what is the probability that the price will be higher than $1000 dollars above the mean?

# Calculate mean+1000 and assign
x <- m+1000

# Calculate z
z <- (x-m)/o

# Calculate probability where P(X => J) = 1-P(X < J)
p <- 1-pnorm(z)

## Interpret results
cat("For the next 4 cars that are sampled, there is a ",round(p*100,digits = 2),"% probability that the price will be higher than $1000 dollars above the mean.")
## For the next 4 cars that are sampled, there is a  37.34 % probability that the price will be higher than $1000 dollars above the mean.

Problem 3: For the next 4 cars that are sampled, what is the probability that the price will be equal to the mean?

# Calculate z2, using the methodology as described in Example 7.6 problem 5
z2 <- ((m+.5)-m)/o

# Calculate z1, using the methodology as described in Example 7.6 problem 5
z1 <- ((m-.5)-m)/o

# Calculate probability where P(J < X < K) = P(X < K) - P(X < J)
p <- pnorm(z2)-pnorm(z1)

## Interpret results
cat("For the next 4 cars that are sampled, there is a ",round(p*100,digits = 2),"% probability that the price will be equal to the mean.")
## For the next 4 cars that are sampled, there is a  0.01 % probability that the price will be equal to the mean.

Problem 4: For the next 4 cars that are sampled, what is the probability that the price will be $1500 within the mean?

# Calculate z2
z2 <- ((m+1500)-m)/o

# Calculate z1
z1 <- ((m-1500)-m)/o

# Calculate probability where P(J < X < K) = P(X < K) - P(X < J)
p <- pnorm(z2)-pnorm(z1)

## Interpret results
cat("For the next 4 cars that are sampled, there is a ",round(p*100,digits = 2),"% probability that the price will be $1500 within the mean.")
## For the next 4 cars that are sampled, there is a  37.19 % probability that the price will be $1500 within the mean.

Conclusion

The results are consistent with expectations. In the first problem, p= 43.59 % < 50 %, which makes sense since we are looking for the probability of a price slightly less than the middle In the second problem, p = 37.34%, which is double the distance from the middle as the first problem. In the third problem, p is effectively 0 which makes sense since the chance the price will exactly equal the mean to the cent is very small. The methodology used here does return a 0.01% chance, but when I ran it using dnorm(), a function intended to return this figure I get 0%. In problem 4, p=37.19 %, which represents the probability for a price within a range of values in the middle of the distribution.

Reference

kable(dat)
Vehicle.Type Year Make Model Price MPG..city. MPG..highway. Horsepower Cylinders
Coupe 2013 Jaguar XK 21807 16 24 385 8
Coupe 2013 Chevrolet Camero 27795 15 24 426 8
Coupe 2013 Ford Mustang 29145 15 26 420 8
Coupe 2013 Mercedes E550 14403 17 27 402 8
Coupe 2013 Audi S5 17209 18 28 333 6
Coupe 2013 BMW M3 25732 14 20 414 8
Coupe 2013 Mini Coupe 2D 13674 26 35 208 4
Coupe 2013 Dodge Challenger 13774 16 25 375 8
Coupe 2013 Cadillac CTS-V 27742 12 18 556 8
Coupe 2013 Nissan 370Z 20177 19 26 332 6