R practice pt 2

# Home-runs so far
HR_before <- c(11, 13, 12)
# Average Number of Home-runs per season wanted
wanted_HR <- 20
# Number of seasons
n_seasons <- 4
# Needed Home-runs on season 4
x_4 <- n_seasons*wanted_HR - sum(HR_before)
# Minimum number of Home-runs needed by Robert
x_4

[1] 44

# Robert's performance
Robert_HRs <- c(11, 13, 12,44)
# Find mean
mean(Robert_HRs)

[1] 20

# Find standard deviation
sd(Robert_HRs)

[1] 16.02082

# Find the maximum number of home-runs during the four seasons period
max(Robert_HRs)

[1] 44

# Find the minimum number of home-runs during the four seasons period
min(Robert_HRs)

[1] 11

summary(Robert_HRs)

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  11.00   11.75   12.50   20.00   20.75   44.00 

# This is the sixth season of outfielder Juan Soto in the majors. If during the first five seasons he received 79, 108,41,145, and 135 walks
HR_Juan <- c(79, 108, 41, 145, 135)

w_hr <- 100

n_seasons <- 6

x_6 <- n_seasons*w_hr - sum(HR_Juan)

x_6

[1] 92

n_1 <- 10
n_2 <- 4
y_1 <- 72000
y_2 <- 84000
# Mean salary overall
salary_ave <-  (n_1*y_1 + n_2*y_2)/(n_1+n_2)
salary_ave

[1] 75428.57

d_1 <- 7
d_2 <- 9
s_1 <- 102000
s_2 <- 91000
salary_avg <- (d_1*s_1 + d_2*s_2)/(d_1+d_2)
salary_avg

[1] 95812.5

getwd()

[1] "/cloud/project"

contract_length <- read.table("allcontracts.csv", header = TRUE, sep = ",")
contract_years <- contract_length$years

# Mean 
contracts_mean  <- mean(contract_years)
contracts_mean

[1] 3.458918

contracts_median <- median(contract_years)
contracts_median

# Find number of observations
contracts_n <- length(contract_years)
# Find standard deviation
contracts_sd <- sd(contract_years)

What percentage of the data lies within one standard deviation of the mean?

contracts_w1sd <- sum((contract_years - contracts_mean)/contracts_sd < 1)/ contracts_n
# Percentage of observation within one standard deviation of the mean
contracts_w1sd

[1] 0.8416834

## Difference from empirical 
contracts_w1sd - 0.68

[1] 0.1616834

What percentage of the data lies within two standard deviations of the mean?

## Within 2 sd
contracts_w2sd <- sum((contract_years - contracts_mean)/ contracts_sd < 2)/contracts_n
contracts_w2sd

[1] 1

## Difference from empirical 
contracts_w2sd - 0.95

[1] 0.05

What percent of the data lies within three standard deviations of the mean?

## Within 3 sd 
contracts_w3sd <- sum((contract_years - contracts_mean)/ contracts_sd < 3)/contracts_n
contracts_w3sd

[1] 1

## Difference from empirical 
contracts_w3sd - 0.9973

[1] 0.0027

# Create histogram
hist(contract_years,xlab = "Years Left in Contract",col = "green",border = "red", xlim = c(0,8), ylim = c(0,225),
   breaks = 5)

Question 3

Use the skills learned in case scenario number 3 on one the following data sets. You may choose only one dataset. They are both available in Canvas.

doubles <- read.table("doubles_hit.csv", header = TRUE, sep = ",")
doublesnumber <- doubles$doubles_hit

doublesnumber_mean <- mean(doublesnumber) 
doublesnumber_mean

[1] 23.55

doublesnumber_median <- median(doublesnumber)
doublesnumber_median

[1] 23.5

doublesnumber_n <- length(doublesnumber)
doublesnumber_sd <- sd(doublesnumber)

doublesnumber_wld <- sum((doublesnumber - doublesnumber_mean)/doublesnumber_sd < 1)/ doublesnumber_n

doublesnumber_wld

[1] 0.79

doublesnumber_wld - 0.68

[1] 0.11

doublesnumber_w2ld <- sum((doublesnumber - doublesnumber_mean)/doublesnumber_sd < 2)/ doublesnumber_n

doublesnumber_w2ld

[1] 1

doublesnumber_w2ld - 0.95

[1] 0.05

doublesnumber_w3ld <- sum((doublesnumber - doublesnumber_mean)/doublesnumber_sd < 3)/ doublesnumber_n
doublesnumber_w3ld

[1] 1

doublesnumber_w3ld - 0.9973

[1] 0.0027

# Create histogram
hist(doublesnumber,xlab = "Years Left in Contract",col = "green",border = "red", xlim = c(0,8), ylim = c(0,225),
   breaks = 5)