Intro to R P2

# 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

According to the calculations above, Robert must hit 44 home-runs or better on this season to get an average number of home-runs per season of at least 20.

#Roberts actual performance was 
Robert_HR<- c(11, 13, 12, 38)
mean(Robert_HR)

[1] 18.5

sd(Robert_HR)

[1] 13.02562

Most he hit in a single season, one of the best and highest potential players in the MLB

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

[1] 38

min(Robert_HR)

[1] 11

This was in 2023, but after that the players performance did not change dramatically

summary(Robert_HR)

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  11.00   11.75   12.50   18.50   19.25   38.00 

Q1: Now, you must complete the problem below which represents a similar case scenario. You may use the steps that we executed in Case-scenario 1 as a template for your solution.

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, how many does he need on this season for his overall number of walks per season to be at least 100?

# Walks during the first five seasons
BB_before <- c(79, 108, 41, 145, 135)

# Average number of walks per season wanted
wanted_BB <- 100

# Number of seasons
n_seasons <- 6

# Needed walks in season 6
x_6 <- n_seasons * wanted_BB - sum(BB_before)

# Minimum number of walks needed by Juan Soto
x_6

[1] 92

The average salary of 10 baseball players is 72,000 dollars a week and the average salary of 4 soccer players is 84,000. Find the mean salary of all 14 professional players.

# Number of players
baseball_players <- 10
soccer_players <- 4

# Average salaries
baseball_avg <- 72000
soccer_avg <- 84000

# Mean salary of all players
mean_salary_overall <- (baseball_players * baseball_avg +
                 soccer_players * soccer_avg) /
                (baseball_players + soccer_players)

mean_salary_overall

[1] 75428.57

OR simpler abbreviated

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

Q2: The average salary of 7 basketball players is 102,000 dollars a week and the average salary of 9 NFL players is 91,000. Find the mean salary of all 16 professional players.

basketball_players <- 7
nfl_players <- 9

# Average salaries
basketball_avg <- 102000
nfl_avg <- 91000

# Mean salary of all players
overall_mean <- (basketball_players * basketball_avg +
                 nfl_players * nfl_avg) /
                (basketball_players + nfl_players)

overall_mean

[1] 95812.5

getwd()

[1] "/cloud/project"

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

1. To find the mean and the standard deviation

# Mean 
contracts_mean  <- mean(contract_years)
contracts_mean

[1] 3.458918

# Median
contracts_median <- median(contract_years)
contracts_median

[1] 3

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

2. 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

3. 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

4. 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

5. Draw a histogram

hist(contract_years,xlab = "Years Left in Contract",col = "green",border = "red", xlim = c(0,6), ylim = c(0,250),
   breaks = 3)

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_hit.csv

SOLUTION:

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

1. To find the mean and the standard deviation

doublesnumber_mean <- mean(doublesnumber)
doublesnumber_mean

[1] 23.55

1.1 Median

doublesnumber_median <- median(doublesnumber)
doublesnumber_median

[1] 23.5

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

players_n <- length(doublesnumber)
players_sd <- sd(doublesnumber)

doublesnumber_w1sd <- sum(abs((doublesnumber - doublesnumber_mean) / players_sd) < 1) / players_n

doublesnumber_w1sd

[1] 0.58

doublesnumber_w1sd - 0.68

[1] 0.32

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

doublesnumber_w2sd <- sum(abs((doublesnumber - doublesnumber_mean) / players_sd) < 2) / players_n

doublesnumber_w2sd

[1] 1

doublesnumber_w2sd - 0.95

[1] 0.05

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

doublesnumber_w3sd <- sum(abs((doublesnumber - doublesnumber_mean) / players_sd) < 3) / players_n

doublesnumber_w3sd

[1] 1

doublesnumber_w3sd - 0.9973

[1] 0.0027

5. Histogram

hist(doublesnumber,
     main = "Histogram of Doubles Hit",
     xlab = "Number of Doubles",
     col = "green",
     border = "red")