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

knitr::opts_chunk$set(echo = TRUE)

Answer 1

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

# 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

Answer 1 (a)

Distribution of LungCap:

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hist(readD$LungCap)

The distribution is a normal distribution. ## Answer 1 (b)

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boxplot(readD$LungCap ~ readD$Gender)

The mean of males appear higher than females.

Answer 1 (c)

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readD%>%
  group_by(Smoke) %>% 
  summarize(Mean=mean(LungCap))

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

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readD%>%
  group_by(Smoke) %>% 
  summarize(stdev=sd(LungCap))

# A tibble: 2 × 2
  Smoke stdev
  <chr> <dbl>
1 no     2.73
2 yes    1.88

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ggplot(readD, aes(x=LungCap, y=Smoke))+geom_boxplot()

The mean of smokers is higher than the mean of non smokers and therefore it is not sensible.

Answer 1 (d)

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class(readD$Age)

[1] "numeric"

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readD <- mutate(readD, AgeGroup = case_when(Age <= 13 ~ "13 and below", 
                                            Age == 14 | Age == 15 ~ "14 to 15", 
                                            Age == 16 | Age == 17 ~ "16 to 17", 
                                            Age >= 18 ~ "18 and above"))
ggplot(readD, aes(x = LungCap)) +
  geom_histogram() +
  facet_grid(AgeGroup~Smoke)

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

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readD %>%
  ggplot(aes(x = Age, y = LungCap, color = Smoke)) +
  geom_line() +
  facet_wrap(vars(Smoke)) +
  labs(y = "Lung Capacity", x = "Age")

From the above results we can say that people from age group 10 and above smoke.

Answer 1 (f)

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cor(readD$LungCap,readD$Age)

[1] 0.8196749

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cov(readD$LungCap,readD$Age)

[1] 8.738289

From the data we can see that the covariance is positive and it shows that there is a direct relationship between age and lung capacity. And the correlation is also positive, so they move in same direction. Therefore as the age increases, the lung capacity also increases that is they are directly proportional to each other.

Answer 2

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X<-c(0, 1, 2, 3, 4)
Frequency<-c(128, 434, 160, 64, 24)
C<- data.frame(X, Frequency)
C

  X Frequency
1 0       128
2 1       434
3 2       160
4 3        64
5 4        24

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C<-rename(C, PriorConvictions=X)
C

  PriorConvictions Frequency
1                0       128
2                1       434
3                2       160
4                3        64
5                4        24

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#visualizing df using bar chart
ggplot(C, aes(x=PriorConvictions, y=Frequency))+geom_bar(stat="identity")+geom_text(aes(label = Frequency), vjust = -.3)

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#There are 810 obs in df
sum(Frequency)

[1] 810

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PO<-Frequency/810
PO

[1] 0.15802469 0.53580247 0.19753086 0.07901235 0.02962963

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#A
# P(x=2)=160/810
160/810

[1] 0.1975309

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#B
#P(x<2)=P(0)+P(1)
(128+434)/810

[1] 0.6938272

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#C
#P(x<=2)=P(0)+P(1)+P(2)
(128+434+160)/810

[1] 0.891358

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#D
#1-P(above)
1-((128+434+160)/810)

[1] 0.108642

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#E
#Expected value=sum of probabilities*each value (0, 1, 2, 3 or 4)
weighted.mean(X, PO)

[1] 1.28642

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#F
#Calculating the Variance using the formula for variance
(sum(Frequency*((X-1.28642)^2)))/(sum(Frequency)-1)

[1] 0.8572937

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#Calculating the sample standard deviation from the variance
sqrt(0.8572937)

[1] 0.9259016

Answer

a: 19.75% b :9.38% c :89.14% d :10.86% e :1.28642 f: variance: 0.8572937 standard deviation: 0.9259016