Section 2.1

2 Number, Proportions

3 The relative frequencies should add up to 1 though rounding may cause some answers to vary.

5

  1. The most common way to beat the flu is by “Washing your hands” as 61% of respondents chose that option.

  2. The least common approach is drinking orange juice in which 2% of respondents chose this option.

  3. 25% of the population thinks flu shots are the best way to beat the flu.

13

  1. Total number of students surveyed = 125 + 324 + 552 + 1257 + 2518 = 4776. Therefore “Never” 125/4776=0.0262, “Rarely” 324/4776=0.0678, “Sometimes” 552/4776=0.1156, “Most of the time” 1257/4776=0.2632, “Always” 2518/4776=0.5272
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 0.52721943

  1. 52.72% answered always.

  2. 0.0262 + 0.0678 = 0.0940. Which equals 9.40%

barplot(datt,main="Seat Belt Usage",names=categoriess, col =c("red","blue","green","yellow","orange"))

barplot(rel.freqq,main="Seat Belt Usage",names=categoriess, col =c("red","blue","green","yellow","orange"))

pie(datt,main="Seat Belt Usage",labels=categoriess, col =c("red","blue","green","yellow","orange"))

  1. The data is inferential because the statement is making an inference about the entire student population based on the results from the survey.

15

  1. Total adults surveyed = 377 + 192 + 132 + 81 + 243 = 1025, Then we do 377/1025 = 0.3678 and so on with the rest.
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 0.23707317

  1. 0.2371 or 23.70% of the population.

barplot(dat,main="Internet Usage",names=categories, col =c("red","blue","green","yellow","orange"))

barplot(rel.freq,main="Internet Usage(Relative Freq)",names=categories, col =c("red","blue","green","yellow","orange"))

pie(dat,main="Internet Usage",labels=categories, col =c("red","blue","green","yellow","orange"))

Section 2.2

7 False, the distribution is skewed to the right.

8 False, the distribution is more of a bell curve.

9

  1. The value of 8 has the highest frequency.

  2. The value with the lowest frequency is 2

  3. The value of 7 was observed 15 times.

  4. 5 was observed 11 times and 4 was observed 7 times. Therefore, the difference between times observed is 11-7 which is 4.

  5. 15/100 = 0.15. 15% of the time a 7 was observed.

  6. The distribution is close to being bell-shaped.

10

  1. 4 cars.

  2. 9 weeks

  3. 52 total weeks in a year. 9/52 = 0.1731. For 17.31% of the time only 2 cars were sold.

  4. The shape is approximately skewed right.

13

  1. It is skewed right because most household incomes will be on the left with fewer high incomes on the right.

  2. It is bell shaped because most scores will be around the middle area, with some scores going off equally in both direcitons.

  3. It is skewed right. Most households will commonly have 1-4 people with fewer households having more people in them.

  4. It is skewed left. Most Alzheimer’s patients will be in older aged categories whilst there will be fewer younger Alzheimer’s patients.

14

  1. It is skewed right. People are less likely to have a lot of drinks per week whilst most people are commonly going to have a few.

  2. It is skewed right. As the age increases, their are less students in the public school system as they may have moved on to college and may be too old for the public school system etc.

  3. It is skewed left. As people get older, they are more likely to have to use hearing aids. It is significantly less common in younger people.

  4. It is bell shaped. Men probably reach a similar height when they are fully grown with some men being shorter or taller than the average based on other circumstances like race or environment.

15

  1. Total number of housholds = 16 + 18 + 12 + 3 + 1 = 50. Thus the relative frequency of 0 children is 16/50 = 0.32 etc. and so on.
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       0.02

  1. 12/50 = 0.24 so 24% of housholds have two children under the age of 5.

  2. (18+12)/50 = 0.6 so 60% of households have one or two children under the age of 5.

16

  1. Total number of free throws is 50. Therefore we do 16/50 = 0.32 to find the frequency in which a 1 free throw misses. 11/50 = 0.22 which is the frequency that there is a miss after 2 free throws and so on.
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       0.02

  1. 7/50 = 0.14, so 14% of the time she missed on her 4th free throw.

  2. 1/50 = 0.02, so 2% of the time she missed on her 10th free throw.

  3. 2 + 3 + 1 + 1 = 7 so 7/50 = 0.14. She made at least 5 in a row 14% of the time.

25

  1. The data is discrete because the possible values for the number of colour tvs in a household is countable.

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
  1. The relative frequency for 0 colour televisions is 1/40 = 0.025 etc and so on.
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   0.025

  1. The relative frequency is 0.2 therefore 20% of the households had 3 colour telvisions.

  2. 0.050 + 0.025 = 0.075 so 7.5 % of the households in the survey had 4 or more colour tvs.