HW1

Code

library(tidyverse)
library(readxl)
library(ggplot2)
library(stats)



knitr::opts_chunk$set(echo = TRUE)

Question 1

Read the data from xls file

Code

RE <- read_excel("_data/LungCapData.xls")
RE

# A tibble: 725 × 6
   LungCap   Age Height Smoke Gender Caesarean
     <dbl> <dbl>  <dbl> <chr> <chr>  <chr>    
 1    6.48     6   62.1 no    male   no       
 2   10.1     18   74.7 yes   female no       
 3    9.55    16   69.7 no    female yes      
 4   11.1     14   71   no    male   no       
 5    4.8      5   56.9 no    male   no       
 6    6.22    11   58.7 no    female no       
 7    4.95     8   63.3 no    male   yes      
 8    7.32    11   70.4 no    male   no       
 9    8.88    15   70.5 no    male   no       
10    6.8     11   59.2 no    male   no       
# … with 715 more rows

##A

Code

RE %>% 
  ggplot(aes(LungCap))+
  geom_histogram(bins=20)

The histogram looks close to normal distributed.

B

Code

RE %>%
  ggplot(aes (LungCap, color=Gender)) +
  geom_boxplot() +
  theme_classic() 

The probability density of the female is higher than the males.

C

Code

Mean_Smoker <- RE %>%
  group_by(Smoke) %>%
  summarise(mean = mean(LungCap))
Mean_Smoker

# A tibble: 2 × 2
  Smoke  mean
  <chr> <dbl>
1 no     7.77
2 yes    8.65

Code

ggplot(RE, aes(LungCap,Smoke))+
  geom_boxplot()

From this sample, it appears that smokers have a higher mean lung capacity than non-smokers.

D

Code

RE<-RE %>% 
  mutate(Category = as.factor(case_when(Age <= 13 ~ "13 and under", 
                           Age == 14 |Age ==15 ~ "14-15", 
                           Age == 16 | Age==17 ~ "16-17",
                           Age >= 18 ~ "18 or over"
                           )))
RE

# A tibble: 725 × 7
   LungCap   Age Height Smoke Gender Caesarean Category    
     <dbl> <dbl>  <dbl> <chr> <chr>  <chr>     <fct>       
 1    6.48     6   62.1 no    male   no        13 and under
 2   10.1     18   74.7 yes   female no        18 or over  
 3    9.55    16   69.7 no    female yes       16-17       
 4   11.1     14   71   no    male   no        14-15       
 5    4.8      5   56.9 no    male   no        13 and under
 6    6.22    11   58.7 no    female no        13 and under
 7    4.95     8   63.3 no    male   yes       13 and under
 8    7.32    11   70.4 no    male   no        13 and under
 9    8.88    15   70.5 no    male   no        14-15       
10    6.8     11   59.2 no    male   no        13 and under
# … with 715 more rows

Code

RE %>%
  ggplot(aes( LungCap, color = Smoke)) +
  geom_histogram()+
  facet_grid(Smoke ~ Category)

`stat_bin()` using `bins = 30`. Pick better value with `binwidth`.

The people who smoke are few in age group of “less than or equal to 13”. From the result we can say age is inversely proportional to the lung capacity.

E

Form the above data we can say the output are pretty similar that smokers have a lower lung capacity compared to non-smokers

F

correlation and covariance between lung capacity and age

Code

cov(RE$LungCap,RE$Age)

[1] 8.738289

Code

cor(RE$LungCap,RE$Age)

[1] 0.8196749

Covariance is positive and indicates that age and lung capacity are directly related. Correlation is also positive,from these results we can conclude that the lung capacity increases with age.

Question 2

Code

x <- c(0:4)
freq <- c(128, 434, 160, 64, 24)
convictions <- data_frame(x, freq)

Warning: `data_frame()` was deprecated in tibble 1.1.0.
Please use `tibble()` instead.

Code

convictions

# A tibble: 5 × 2
      x  freq
  <int> <dbl>
1     0   128
2     1   434
3     2   160
4     3    64
5     4    24

Code

convictions <- convictions %>% mutate(probability = freq/sum(freq))
convictions

# A tibble: 5 × 3
      x  freq probability
  <int> <dbl>       <dbl>
1     0   128      0.158 
2     1   434      0.536 
3     2   160      0.198 
4     3    64      0.0790
5     4    24      0.0296

A

Probability of exactly 2 is 19.75%

B

Code

a <-head(convictions,2)
sum(a$probability)

[1] 0.6938272

Probability that a randomly selected inmate has fewer than 2 prior convictions : 69.38%

C

Code

a <-head(convictions,3)
sum(a$probability)

[1] 0.891358

The probability that a randomly selected inmate has 2 or fewer prior convictions : 89.13%

D

Code

a <-tail(convictions,2)
sum(a$probability)

[1] 0.108642

The probability that a randomly selected inmate has more than 2 prior convictions? : 10.86%

E

Code

WE <- weighted.mean(convictions$x,convictions$probability)
WE

[1] 1.28642

The expected value for the number of prior convictions : 1.28

F

The variance is 0.857 and the standard deviation is 0.925

Code

AB <- (sum(freq*((x-WE)^2)))/(sum(freq)-1)
AB

[1] 0.8572937

Code

sqrt(AB)

[1] 0.9259016