2.1

7

  1. China

  2. 50 Million

  3. 350 Million

  4. Should have used relative frequency

9

  1. 69%

  2. 22% = 55.2 million

  3. inferential. The statement is inferential because the survey was done with random sample of adult Americans

11

  1. 0.44 = 951 adults; 0.61 = 1,319 Adults

  2. 55+

  3. 18-34

  4. As you get older it’s more likely you’ll buy products “made in America”

13

Sometimes: 552/4776 = .1155

Most of the time: 1257/4776 = .2631

  1. Always: 2518/4776 = .5272

  2. Never: 125/4776 = .0261

Rarely: 324/4776 = .0678

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. Inferential

15

  1. more then 1 hour:377/1025 = .3678

Up to 1 hour: 192/1025 = .1873

A few time a week: 132/1025 = .1287

A few times a month: 81/1025 = .0790

Never: 243/1025 = .2370

  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. He should have said “almost” 37% of adults Americans use the internet more than 1 hour a day, according to the chart it’s 36.7%

2.2

9

  1. the sum of 8 was the most frequent outcome of the experiment, 20 times

  2. the sum of 2 was the least frequent outcome, probably twice

  3. 15 times

  4. 4 times

  5. 15%

  6. bell shaped

10

  1. 4

  2. 9 weeks

  3. 17.3%

  4. skewed right

11

  1. 200 students

  2. class width = 160-60 = 100/10 classes = 10 width

  3. 60-70 class has a frequency of 2; 70-80 class has a frequency of 3; 80-90 class has a frequency of 13; 90-100 class has a frequency of 42; 100-110 class has a frequency of 58; 110-120 class has a frequency of 40; 120-130 class has a frequency of 31; 130-140 class has a frequency of 8; 140-150 class has a frequency of 2; 150-160 class has a frequency of 1

  4. 100-110 IQs class

  5. 150-160 IQs class

  6. 4%

  7. no

12

  1. class width = 1600-0 = 1600/8 classes = 200 width

  2. Skip this problem

  3. 0-200 class

  4. skewed right

  5. Vermont is under the 0-200 class with a frequency of 26 and Texas in under on 1400-1600 class with a frequency os 1. 51 is the total of frequencies. 26/51 = .509 and 1/51 = .019;the proportion of fatal accidents is higher under 0-200 class where Vermont is under. The statement is misleading because TX is a big State compared to Vermont; the population in Texas is higher than Vermont which means Texas has more drivers.

13

  1. Skewed right; from low - high; there are more poor people than rich

  2. bell shaped

  3. uniform

  4. skewed left - from youngest to oldest; young is less like to have Alzheimer

14

  1. skewed right- from young to older people, yougest people drink more

  2. skewed right- from young to older, more young people in public school

  3. skewed-left; old people m=are more likely to have hearing aid

  4. uniform?