Intro to R Part 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

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.

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

[1] 20

sd(Robert_HRs)

[1] 16.02082

max(Robert_HRs)

[1] 44

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 

**Question 1* 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?

Juan_Soto_Walks = c(79, 108, 41, 145, 135)
Juan_Soto_Walks

[1]  79 108  41 145 135

# 100 = (s1 + s2 + s3 + s4 + s5 + x) / 6
# 100 * 6 - (s1 + s2 + s3 + s4 + s5) = x
100* 6 - sum(Juan_Soto_Walks)

[1] 92

The answer is 92

Case-scenario 2 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.

Solution We can easily find the joined mean by adding both mean and dividing by the total number of people.

Let n1=10 denote the number of baseball players, and y1=72000 their mean salary. Let n2=4 the number of soccer players and y2=84000 their mean salary. Then the mean salary of all 16 individuals is: n1x1+n2x2n1+n2

We can compute this in R as follows

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

Question 2 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.

n_basketball_players = 7
n_NFL_players = 9
av_salary_basketball = 102000
av_salary_NFL = 91000
av_salary_all = (n_basketball_players *av_salary_basketball + n_NFL_players * av_salary_NFL) / (n_basketball_players + n_NFL_players)
av_salary_all

[1] 95812.5

Case-scenario 3 The frequency distribution below lists the number of active players in the Barclays Premier League and the time left in their contract.

Years Number of players 6 28 5 72 4 201 3 109 2 56 1 34

Find the mean,the median and the standard deviation.

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

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

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

Draw a histogram to illustrate the data.

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

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

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

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

# 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_hit.csv and triples_hit.csv

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

#mean and median
mean(doublesnumber)

[1] 23.55

median(doublesnumber)

[1] 23.5

#percentage of data within one standard deviation of the mean
doubles_n = length(doublesnumber)
doubles_sd = sd(doublesnumber)
doubles_w1sd = sum((doublesnumber - mean(doublesnumber))/doubles_sd < 1)/ doubles_n
doubles_w1sd

[1] 0.79

#difference from empirical
doubles_w1sd - 0.68

[1] 0.11

#percentage of data within two standard deviation of the mean
doubles_w2sd = sum((doublesnumber - mean(doublesnumber))/doubles_sd < 2)/ doubles_n
doubles_w2sd

[1] 1

## Difference from empirical 
doubles_w2sd - 0.95

[1] 0.05

#percentage of data within three standard deviation of the mean
doubles_w3sd = sum((doublesnumber - mean(doublesnumber))/doubles_sd < 3)/ doubles_n
doubles_w3sd

[1] 1

## Difference from empirical 
contracts_w3sd - 0.9973

[1] 0.0027

# Create histogram
hist(doublesnumber, xlab = "Number of Doubles", ylab ="Frequency",col = "green",border = "red", xlim = c(0,55), ylim = c(0,8),
   breaks = 35)

This histogram shows how most of players frequent between 5 and 40 doubles historically. Additionally, it looks like no player in this dataset was able to get beyond 50 doubles.