2.1

7

  1. China

  2. 50 million

  3. 350-370 million more

  4. It could be misleading because the bars are tough to get exact numbers off of, or because the y-axis is listed in 10’s and you have to notice that it is millions in the label. Honestly not sure what it’s looking for here, it’s pretty straightforward…

9

  1. Just under 70%; call it 68%.

  2. Let’s call it 23%, and 23 percent of 240 million is 55.2 million.

  3. It’s an inferential statement, because they are making a claim about a population based on sample data.

11

  1. 18-34: about 45% ; 35-44: about 60%

  2. 55+

  3. 18-34

  4. In general, it appears as though younger people are less likely to buy American made, while older are more likely to buy American made.

13

Never: 2.6%

Rarely: 6.8%

Sometimes: 11.6%

Most of the time: 26.3%

Always: 52.7%

  1. 52.7%

  2. 9.4%

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. While it says it was a “national survey,” it has to be unlikely that it was a census, so it’s still an inferential statement (sample data > population statement)

15

More then 1 hour: 36.8%

Up to 1 hour: 18.7%

A few time a week: 12.9%

A few times a month: 7.9%

Never: 23.7%

  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. I guess what’s wrong is that the statement is inferential so it is not certain that 37% of American adults actually use the internet more than 1 hour per day. That said, the question is kindy weird, because there was nothing “wrong” with the inferential statements made earlier. So I guess the newscaster should have said, “37% of adults surveyed said blablabla.”

2.2

9

  1. 8

  2. 2

  3. 15

  4. call it 11 and 6, so 5 more

  5. 15%

  6. The shape of the distribution is approximately normal (which I think the textbook still calls “bell-shaped” so far).

10

  1. 4

  2. 9

  3. 17.3%

  4. The distribution is slightly right skewed.

11

  1. 200

  2. 10

  3. 60-70: 2/1%; 70-80: 3/1.5%; 80-90: 13/6.5%; 90-100: 42/21%; 100-110: 58/29%; 110-120: 40/20%; 120-130: 31/15.5%; 130-140: 8/4%; 140-150: 2/1%; 150-160: 1/0.5%

  4. 100-110

  5. 150-160

  6. 5.5%

  7. nope

12

  1. 200

  2. Skip this problem (meh, 0-200, 200-400, 400-600, 600-800, 800-1000, 1000-1200, 1200-1400, 1400-1600)

  3. 0-200

  4. The distribution is very right skewed.

  5. I imagine the problem being highlighted is that the data provided does not tell us anything about the safety of roads in general. It merely demonstrates that drunk driving is a larger problem in Texas than it is in Vermont. General road safety includes lots of other factors; maybe Vermont has more car accidents in winter than Texas has all year, the data in this chart just doesn’t allow for such a sweeping conclusion.

13

  1. It would be great if they were normally distrubuted, or even uniform, but I’m fairly sure they would be heavily left skewed.

  2. Definitely normally distributed, almost by definition; I’m pretty sure the SAT is scored relative to the other people who took it at the same time as you.

  3. Probably slightly right skewed. I would think there are more households with less people than there are with a lot.

  4. Definitely left skewed. Most people who are diagnosed with Alzheimer’s are older.

14

  1. Hopefully/probably right skewed, but I wouldn’t be shocked to find out if it was normally distrubuted.

  2. Could be mostly uniform, but a slight right skew would make sense to account for people dropping out in higher grade levels.

  3. Could be normally distributed (hearing disabled people on the low end, older people losing their hearing on the high) but I imagine it would be left skewed, eventually those hearing disabled people grow up to join the elderly groups.

  4. Depends how you define full-grown men, but I imagine it would be normally distributed around whatever “average height” is.