HW2
Harrel Blatt
September 14, 2016
Section 2.1
2 Number, percentage
3 1
5
Washing your hands, 61%
Drinking orange juice, 2%
25 %
13
- Never 125 0,026172529 Rarely 324 0,067839196 someti. 552 0,115577889 Most 1257 0,263190955 Always 2518 0,52721943 Total 4776 1
datt <- c(125, 324, 552, 1257, 2518)
rel.freqq <- datt/sum(datt)
categoriess <- c("Never", "Rarely", "Sometimes", "Most of time", "Always")
answerr <- data.frame(categoriess,rel.freqq)
answerr## categoriess rel.freqq
## 1 Never 0.02617253
## 2 Rarely 0.06783920
## 3 Sometimes 0.11557789
## 4 Most of time 0.26319095
## 5 Always 0.5272194352,7 %
9,4 %
Never 125
Rarely 324
someti. 552
Most 1257
Always 2518
barplot(datt,main="Seat Belt Usage",names=categoriess, col =c("red","blue","green","yellow","orange"))
- Never 0,026172529 Rarely 0,067839196 someti. 0,115577889 Most 0,263190955 Always 0,52721943
barplot(rel.freqq,main="Seat Belt Usage",names=categoriess, col =c("red","blue","green","yellow","orange"))
- Never 9,422110553 Rarely 24,42211055 sometimes 41,6080402 Most of 94,74874372 Always 189,798995 Total 360
pie(datt,main="Seat Belt Usage",labels=categoriess, col =c("red","blue","green","yellow","orange"))
- A descriptive statement describing the statistic and not a parameter.
15
- More than 1 377 0,367804878 Up to 1 192 0,187317073 Weekly 132 0,128780488 Monthly 81 0,07902439 Never 243 0,237073171 Total 1025 1
dat <- c(377,192,132,81,243)
rel.freq <- dat/sum(dat)
categories <- c("More 1", "Up to 1", "Few a week", "Few a month", "Never")
answer <- data.frame(categories,rel.freq)
answer## categories rel.freq
## 1 More 1 0.36780488
## 2 Up to 1 0.18731707
## 3 Few a week 0.12878049
## 4 Few a month 0.07902439
## 5 Never 0.23707317About 0,237 never use the internet
More than 1 377 Up to 1 192
Weekly 132
Monthly 81
Never 243
barplot(dat,main="Internet Usage",names=categories, col =c("red","blue","green","yellow","orange"))
- More than 1 0,367804878 Up to 1 0,187317073 Weekly 0,128780488 Monthly 0,07902439 Never 0,237073171
barplot(rel.freq,main="Internet Usage(Relative Freq)",names=categories, col =c("red","blue","green","yellow","orange"))
- 132,4097561 67,43414634 46,36097561 28,44878049 85,34634146
pie(dat,main="Internet Usage",labels=categories, col =c("red","blue","green","yellow","orange"))
Section 2.2
7 False, it is left skewed
8 True, it is bell shaped
9 type answer here
8
2
15
Around 4 (it is hard to tell when all numbers aren’t shown up the vertical axis. I’ll say there’s around 7 4’s and 11 5’s so 4 more 4’s than 5’s)
15 % of the time
Bell shaped
10
4
9 weeks
17,3 % of time were two cars sold
It is more or less symmetric and bell shaped. Theoretically there is a higher chance of it being right skewed than left skewed, as it is imposible to sell less than 0 cars but posible to sell more than 10. This frequency histogram is however bell shaped.
13
Right skewed. There is a natural barrier to the left as people cannot make less than 0, so the tail should be longer to the right.
Bell shaped as there is a good chance that the median and the mean is the same value. There are more average students than students scoring either really good or really bad.
Righ skewed because there is a natural barrier of 0 but a possibility that data might be skewed to the right. The mean will probably be around 4 but the median might be further to the right as some households are bigger
Left skewed, as people probably are older and the observations will mainly centered aroung high ages. Thus the mean will be far out the right of the horisontal axis but the median might be to the left as some people are diagnozed young.
14
Right skewed because there is a natural barrier of 0 but a possibility that data might be skewed to the right because some might drink way more than other so the median will be to the right of the meann
Bell shaped as most students are the same age due to rules on when to start in school, but some might start younger and some might be older.
Left skewed as people are most likely older when they need at hearing aid but a few are young.
Bell shaped as very short or very tall people are outliers but do exist. However the mean and the median are probably close to be the same.
15
- 0,32 0,36 0,24 0,06 0,02
dattt <- c(16, 18, 12, 3, 1)
rel.freqqq <- dattt/sum(dattt)
categoriesss <- c("Zero", "One", "Two", "Three", "Four")
answerrr <- data.frame(categoriesss,rel.freqqq)
answerrr## categoriesss rel.freqqq
## 1 Zero 0.32
## 2 One 0.36
## 3 Two 0.24
## 4 Three 0.06
## 5 Four 0.0224 %
60 %
16
- 0,32 0,22 0,18 0,14 0,04 0,06 0 0,02 0 0,02
free_throws <- c(16, 11, 9, 7, 2,3,0,1,0,1)
rel.freqqq <- free_throws/sum(free_throws)
categoriesss <- c("1", "2", "3", "4","5","6","7","8","9","10")
answerrr <- data.frame(categoriesss,rel.freqqq)
answerrr## categoriesss rel.freqqq
## 1 1 0.32
## 2 2 0.22
## 3 3 0.18
## 4 4 0.14
## 5 5 0.04
## 6 6 0.06
## 7 7 0.00
## 8 8 0.02
## 9 9 0.00
## 10 10 0.0214 %
2 %
14 % (five throws and everything above)
25
The data are discrete, as you can not own 1,5 TVs. You have to count the number of TVs in the discrete numbers without digits.
1 14 14 8 2 1
tv <- c(1, 1, 1, 2, 1,
1, 2, 2, 3, 2,
4, 2, 2, 2, 2,
2, 4, 1, 2, 2,
3, 1, 3, 1, 2,
3, 1, 1, 2, 1,
5, 0 ,1, 3, 3,
1, 3, 3, 2, 1)
#table(tv)
tv <- c(1,14,14,8,2,1)
tv.freq <- tv/sum(tv)
tv.cat <- c("0", "1", "2", "3","4","5")
freq.tab <- data.frame(tv.cat,tv)
rfreq.tab <- data.frame(tv.cat,tv.freq)
freq.tab## tv.cat tv
## 1 0 1
## 2 1 14
## 3 2 14
## 4 3 8
## 5 4 2
## 6 5 1- 0,025 0,35 0,35 0,2 0,05 0,025
rfreq.tab## tv.cat tv.freq
## 1 0 0.025
## 2 1 0.350
## 3 2 0.350
## 4 3 0.200
## 5 4 0.050
## 6 5 0.02520 %
7,5 %