2.1

7

  1. China

  2. 50 Million users

  3. 420 mil-52 mil = 368 million users

  4. The graph disregards relative population. For example, China has the most internet users, but it also is the most populous nation. So it disregards relative percentages to a country’s respective population.

9

  1. 69%

  2. 23% believe it is morally wrong. So 240,000,000 * 0.23 = 55,200,000 people.

  3. Inferential. This is because “depends on situation” doesn’t necessarily mean its acceptable in a situation. It can either be acceptable or morally wrong depending on the situation. So the 8% represents both of these opinions, not just being “acceptable depending on the situation.”

11

  1. 18-34 : 42% 35-44 : 61%

  2. 55+

  3. 18-34

  4. The general trend is that the older the consumers are, the more likely they are to have bought “made in America” products 13

Never: 0.0262

Rarely: 0.0678

Sometimes: 0.1156

Most of the time: 0.2632

Always: 0.5272

  1. 52.72%

  2. 9.40%

d e f

my_data <- c(125, 324, 552, 1257, 2518)

groups <- c("Never", "Rarely", "Sometimes", "Most", "Always")

barplot(my_data, main = "Wearing Seatbelts", names.arg = groups)

barplot(my_data, main = "Wearing Seatbelts", names.arg = groups, col = c("red","blue","green","yellow", "black"))

rel_freq <- my_data / sum(my_data)

barplot(rel_freq, main = "Wearing Seatbelts", names.arg = groups, col = c("red","blue","green","yellow","black"))

pie(my_data, labels = groups, main = "Wearing Seatbelts")

  1. This is a descriptive statement, as there is no room for interpretation for the phrase “Always wears a seatbelt.” According to the data, 52.7% of college students always wear seatbelts.

15

More then 1 hour: 0.3678

Up to 1 hour: 0.1873

A few time a week: 0.1288

A few times a month:0.0790

Never: 0.2371

  1. 23.7%

c d e

my_data <- c(377, 192, 132, 81, 243)

groups <- c("More 1", "Up to 1", "Few times week", "Few times month", "Never")

barplot(my_data, main = "Use the internet", names.arg = groups)

barplot(my_data, main = "Use the internet", names.arg = groups, col = c("red","blue","green","yellow", "black"))

rel_freq <- my_data / sum(my_data)

barplot(rel_freq, main = "Use the internet", names.arg = groups, col = c("red","blue","green","yellow","black"))

pie(my_data, labels = groups, main = "Use the internet")

  1. The statements “A few times a week” and “A few times a month or less” don’t specify the duration of each time they use the internet. These statements only reflect frequency. So although a person may not use it frequently, they may use it for more than an hour for every time they use it.

2.2

9

  1. 8

  2. 2

  3. 15 times

  4. 11-7 = 4 more times

  5. 20%

  6. It is a fairly symmetric bar graph distribution with a slightly elongated tail on the left side of the graph.

10

  1. 4

  2. 9 weeks

  3. 52 weeks in one year. 9/52*100= 17.31% of the time.

  4. Bar chart like histogram is not very symmetrical. The peak is at 4. Frequences from 0-4 fluctuate greatly. While values from 4-8 follow a gradual decreasing frequency trend. No value is recorded at 9. An outlier frequency of 1 occurs at 10 cars sold.

11

  1. 200 students

  2. Class ranged from IQ 60 to IQ 160. The range is 100 IQ score units.

  3. 60-70 IQ: 0.01 70-80 IQ: 0.015 80-90 IQ: 0.065 90-100 IQ: 0.21 100-110 IQ: 0.29 110-120 IQ: 0.20 120-130 IQ: 0.155 130-140 IQ: 0.04 140-150 IQ: 0.01 150-160 IQ: 0.005

  4. The highest frequency scored between 100 and 110 on their IQ test.

  5. The lowest frequency scored between 150 and 160 on their IQ test.

  6. 5.5% of the students scored at least 130 on the IQ test.

  7. No students scored above 160.

12

  1. The class width is between 0 and 1600 fatalities. The range of 1600.

  2. Skip this problem

  3. 0-200 fatalities had the highest frequency

  4. The bar type histogram has a general downward trend going from 0-1600. However the frequency was 0 between 600 and 1000 fatalities, and 0 again between 1200 and 1400 fatalities.

  5. Texas is a more populous state than Vermont. This means that with more people, the higher the probability there is that an accident will occur. Also, the accidents were most likely caused by alcohol, rather than the roads not being safe. 13

  6. I would expect a skewed left histogram, as there are more people with low/middle class incomes than high income households in America.

  7. Bell shaped curve, as most students would have an average score in the middle while few will do very poorly and few will do extremely well.

  8. Bell shaped curve as the average household has 4 people in United states. Fewer houses will have fewer/more people.

  9. I would expect a skewed right histogram, as alzheimers is predominantly diagnosed in elderly people.

14

  1. Bell Shaped curve, as there would be a symmetric distribution, with few people who don’t drink, many people who drink a moderate amount, and few people who drink far too much.

  2. I would expect a uniform graph, as in the frequencies would be relatively similar. Public schools grades usually contain around the same amount of students.

  3. I would expect a skewed righ graph, as hearing aids are predominantly given to elderly due to age related hearing loss.

  4. I would expect a bell shaped curve, as most people would be an average height, with fewer people who are shorter/taller than that height.