Pokemon <- read.csv("Pokemon.csv")
download.file("http://www.openintro.org/stat/data/bdims.RData", destfile = "bdims.RData")
load("bdims.RData")
head(bdims)
## bia.di bii.di bit.di che.de che.di elb.di wri.di kne.di ank.di sho.gi che.gi
## 1 42.9 26.0 31.5 17.7 28.0 13.1 10.4 18.8 14.1 106.2 89.5
## 2 43.7 28.5 33.5 16.9 30.8 14.0 11.8 20.6 15.1 110.5 97.0
## 3 40.1 28.2 33.3 20.9 31.7 13.9 10.9 19.7 14.1 115.1 97.5
## 4 44.3 29.9 34.0 18.4 28.2 13.9 11.2 20.9 15.0 104.5 97.0
## 5 42.5 29.9 34.0 21.5 29.4 15.2 11.6 20.7 14.9 107.5 97.5
## 6 43.3 27.0 31.5 19.6 31.3 14.0 11.5 18.8 13.9 119.8 99.9
## wai.gi nav.gi hip.gi thi.gi bic.gi for.gi kne.gi cal.gi ank.gi wri.gi age
## 1 71.5 74.5 93.5 51.5 32.5 26.0 34.5 36.5 23.5 16.5 21
## 2 79.0 86.5 94.8 51.5 34.4 28.0 36.5 37.5 24.5 17.0 23
## 3 83.2 82.9 95.0 57.3 33.4 28.8 37.0 37.3 21.9 16.9 28
## 4 77.8 78.8 94.0 53.0 31.0 26.2 37.0 34.8 23.0 16.6 23
## 5 80.0 82.5 98.5 55.4 32.0 28.4 37.7 38.6 24.4 18.0 22
## 6 82.5 80.1 95.3 57.5 33.0 28.0 36.6 36.1 23.5 16.9 21
## wgt hgt sex
## 1 65.6 174.0 1
## 2 71.8 175.3 1
## 3 80.7 193.5 1
## 4 72.6 186.5 1
## 5 78.8 187.2 1
## 6 74.8 181.5 1
mdims <- subset(bdims, sex == 1)
fdims <- subset(bdims, sex == 0)
fhgtmean <- mean(fdims$hgt)
fhgtsd <- sd(fdims$hgt)
hist(fdims$hgt, probability = TRUE)
x <- 140:190
y <- dnorm(x = x, mean = fhgtmean, sd = fhgtsd)
lines(x = x, y = y, col = "blue")
qqnorm(fdims$hgt)
qqline(fdims$hgt)
sim_norm <- rnorm(n = length(fdims$hgt), mean = fhgtmean, sd = fhgtsd)
qqnormsim(fdims$hgt)
1 - pnorm(q = 182, mean = fhgtmean, sd = fhgtsd)
## [1] 0.004434387
sum(fdims$hgt > 182) / length(fdims$hgt)
## [1] 0.003846154
#1.a) The histogram for female biiliac (pelvic) diameter (bii.di) belongs to normal probability plot letter B
qqnorm(fdims$bii.di)
qqline(fdims$bii.di)
#1.b) The histogram for female elbow diameter (elb.di) belongs to normal probability plot letter C
qqnorm(fdims$elb.di)
qqline(fdims$elb.di)
#1.c) The histogram for general age (age) belongs to normal probability plot letter D
qqnorm(bdims$age)
qqline(bdims$age)
#1.d) The histogram for female chest depth (che.de) belongs to normal probability plot letter A
qqnorm(fdims$che.de)
qqline(fdims$che.de)
#2.) Note that normal probability plots C and D have a slight stepwise pattern.Why do you think this is the case?
#Because the amount of theoritical quantiles has repeated for some amount until it steps up to next samples quantile
#3.) As you can see, normal probability plots can be used both to assess normality and visualize skewness. Make a normal probability plot for female knee diameter (kne.di). Based on this normal probability plot, is this variable left skewed, symmetric, or right skewed? Use a histogram to confirm your findings.
qqnorm(fdims$kne.di)
qqline(fdims$kne.di)
hist(fdims$kne.di, breaks = 40)
## MILESTONE DATA
#1. Restate your probability question from Lab 3, On Your Own - Milestone Data, Question 1.
# What is the probaility of a Pokemon has generation over 5?
#2.Identify the numeric variable you used to answer the probability question.
Pokemon$Generation
## [1] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
## [38] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
## [75] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
## [112] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
## [149] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
## [186] 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
## [223] 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
## [260] 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3
## [297] 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3
## [334] 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3
## [371] 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3
## [408] 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 4
## [445] 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4
## [482] 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4
## [519] 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 5 5
## [556] 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5
## [593] 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5
## [630] 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5
## [667] 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5
## [704] 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6
## [741] 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6
## [778] 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6
#3. Produce a histogram and a Normal Q-Q plot of the numeric variable.
five_gen_up <- subset(Pokemon, Generation>5 )
qqnorm(five_gen_up$Attack)
five_gen_up$Attack
## [1] 61 78 107 45 59 69 56 63 95 36 56 50 73 81 35 22 52 50 68
## [20] 38 45 65 65 100 82 124 80 48 48 48 80 110 150 50 52 72 48 80
## [39] 54 92 52 105 60 75 53 73 38 55 89 121 59 77 65 92 58 50 50
## [58] 75 100 80 70 110 66 66 66 66 90 85 95 100 69 117 30 70 131 131
## [77] 100 100 160 110 160 110
hist(five_gen_up$Attack, breaks=10, main="Generation for Pokemon", xlab="Attack Points")
#4. Assess the normality of the numeric variable. Are you surprised by what you see? Why or why not?
#Not at all the emperical data is pretty much what I expect. It is accurate because we have our data ahead of time so basically we know what to expect from data and probablity.
#5. Use the theoretical normal distribution to answer your probability question.
POKE_mean=mean(five_gen_up$Attack)
POKE_sd=sd(five_gen_up$Attack)
1-pnorm(60,POKE_mean,POKE_sd)
## [1] 0.7059679
#6.Compare the answer computed empirically (i.e., the answer from Lab 3, On Your Own - Milestone Data, Question 1) to the answer using the theoretical normal distribution. Which answer do you trust more? Why?
# the probablity from emperical method is 68.95 compared to the normal distribution method of 72.40%. This is a difference of about 3.45%. I trust the empirical data more because the normal distribution shows more error.
## Documentaton: I work with all my teammate.