Unit 8 Data Sets And Distributions — Unit Plan

1. Categorical
2. Categorical
3. Numerical
4. Categorical
5. Categorical
6. Numerical

1. • Question A is a statistical question. Sample reasoning: The temperature in Miami in December changes from day to day and from year to year.
   • Question B is not a statistical question. Water freezes at sea level at 32 degrees Fahrenheit. This is a known fact.
2. To answer Question A (about the warmest temperature), find the temperature records for the past ten years and look for the highest value in degrees Fahrenheit.

1. No. He could use a bar graph because the home states of swimmers are categorical data.
2. 16 students are in the class.
3. 1. Agree. Sample reasoning: The number of dots stacked over a number is never greater than 3 for this dot plot.
   2. Agree. On the number line, eight of the 16 data points, or half of the class, are placed to the right of 24 months and to the left of 36 months.

1. There are 20 dots and each corresponds to one student in the group.
2. 75%. 15 out of 20 students answered that there are 2 or more children in the family.
3.  Sample response: A typical number of children for this group of families is around 2 because the center is around 2.5 or so, but some families had many more children than others. The distribution is not very spread out with most families having 1–3 children and only a few of them having more.

1. In general, Variety A tomatoes are about 5.49 ounces and Variety B tomatoes are about 5.5 ounces.

2. She should choose Variety B. Sample reasoning: The two varieties of tomatoes have about the same center of their distributions, but there is much less variability in Variety B tomato weights. The weights are much more consistent than the weights for Variety A, so the tomatoes are more likely to be the same size and closer to 5.5 ounces in weight.

1. rainfall (inches)	frequency
   0–2	1
   2–4	5
   4–6	1
   6–8	2
   8–10	3

   

2. Sample response: The center of the distribution appears to be between 4 and 6 inches of rain.

Sample response:

1. In both seasons, the player typically scored around 15 to 20 points in a game. In each histogram, there seems to be a similar frequency of values on each side of this interval.
2. The spread of the distribution for 2008 seems less than the spread for the 2016 distribution. There are only 5 games in which the player did not score between 10 and 25 points per game in 2008, but in 2016 the data are more spread out. This means that, from game to game, the player was more consistent in 2008 than in 2016.

Sample response: The distribution is not symmetrical because there is a peak on the left. The histogram shows a gap between 35 and 40, so there is no game in which the player scored 35, 36, 37, 38 or 39 points. There was one game that was unusually high scoring, between 40 and 44 points. The peak is between 15 and 20 points. The center is around 20 points, and the data is spread out pretty far above this center.

1. 6 degrees Celsius. The sum of the temperatures divided by the total number of recorded temperatures is $(5+ 8+ 6+ 5+ 10+ 7+ 1)\div 7 = 6$.
2. 9. Sample reasoning: The 4 numbers must be distributed evenly around 7. Because 2 of the numbers are 7, and the third number is two less than 7, the fourth number must be 2 more than 7.

1. Jada's mean is 5, since $\frac{4+4+4+6+6+6}{6}=\frac{30}{6}=5$. Diego's mean is 6, since $\frac{4+5+5+6+8+8}{6}=\frac{36}{6}=6$. Lin's mean is 4, since $\frac{1+1+2+2+9+9}{6}=\frac{24}{6}=4$. On average, Diego sent the most text messages to his parents per day, and Lin sent the fewest text messages per day to her parents.
2. Sample response: Lin's data has the highest variability. The sum of the distances to each side of the mean is the greatest.

1. Lin's data set has the largest MAD, because the data have the most variability. Jada's data set has the smallest MAD, because the data have the least variability.
2. Jada's MAD is $\frac{1+1+1+1+1+1}{6}=1$. Diego's MAD is $\frac{2+1+1+0+2+2}{6}=1.33$. Lin's MAD is $\frac{3+3+2+2+5+5}{6}=3.33$.

1. South Africa, because it has the largest MAD.
2. Sample response: The mean travel times are similar, so students in these countries typically take about the same amount of time to get to school. The MAD for New Zealand students is less than the MAD for Canadian students. This means that the travel time for students in New Zealand is more consistent than the travel times for students in Canada.
3. Sample response: On average, the students in Australia have a longer commute to school than students in Canada, but the travel times of students in both countries have the same variability. The data points are, on average, about 8 minutes from the mean.

1. Jada's median: 20 minutes. Diego's median: 22.5 minutes.
2. Sample response: Half of Jada's practices are 20 minutes or shorter and the other half of her practices are 20 minutes or longer. Half of Diego's practices are 22.5 minutes or shorter, and the other half are 22.5 minutes or longer.

Sample responses:

1. Player heights
   1. The mean would be approximately equal to the median, because the data are roughly symmetric.
   2. Since I think the values would be pretty close, either the mean or the median would describe a typical height pretty well.
2. Backpack weights
   1. The mean would be higher than the median. The value of 16 kilograms would bring the mean up and move it away from the center of the data.
   2. The median would better describe a typical backpack weight, since that value would lie in the center of the large cluster of data points.
3. People's ages
   1. The mean would be lower than the median, because even though a large fraction of the people at the dinner party are 40 or older, the ages of the people that span from 5 to 40 would bring the average age down.
   2. The median would better describe the center of the distribution of around 40–45 years old.

1. The median is 51.5 feet. $(50+53)\div 2=51.5$. The IQR is 10, because Q1 is 47, Q3 is 57, and $57-47 = 10$.
2. 1. Worse. Sample reasoning: The median of the second data set is 49 feet, which is 2.5 feet lower than in the first round. 
   2. Less variable. Sample reasoning: The IQR of the second data set is smaller, so the values are less spread out.

1. 1. Disagree
   2. Agree
   3. Agree.
   4. Disagree

2. Box plot A goes with Data set 2. Box plot B goes with Data set 1. Sample reasonings:

• The maximum values tell which box plot goes with which dot plot.
• The middle half of the points in data set 1 are more spread out compared to those in data set 2, so box plot B, which has a longer box, goes with data set 1.
• Three quarters of the points in data set 2 are between 0 and 5, which matches the box and left whisker in Box plot A.

1. The longest whale is about 55 feet long and is a female.
2. A typical male humpback whale is about 44.5 feet long.
3. 1. Disagree. Sample explanation: The upper quartile of the data for the male humpbacks is 46 feet, which means a quarter of the whales are longer than 46 feet.
   2. Disagree. Sample explanation: The entire distribution for the lengths of female humpbacks is greater than that for male humpbacks, so female humpbacks tend to be longer than their male counterparts.
   3. Agree. Sample explanation: The IQR of the data for male humpbacks is slightly greater than that for female humpbacks, and the range of the data for the males is larger than that for females, so the lengths of male humpbacks tend to vary more.

Samples responses:

1. Disagree. The dot plot makes it possible to find the median, the minimum, and the maximum fairly easily because it shows each data value individually. The histogram makes it possible to estimate these values, but it is impossible to tell the exact values because the data points are grouped together.
2. Lin should use the median and IQR because the data is not approximately symmetrical and has values far from the center. There are a few larger values that are not similar to most of the other values