A B C D

Solutions for Session 10, Part B

See solutions for Problems: B1 | B2 | B3 | B4 | B5 | B6

Problem B1

 a. Statistical ideas included using the mode, median, and mean as measures of center, and the range as an indicator of variation. b. Answers will vary. Examples of questions would be, "You stated that the range is 3. What does this tell us about the data set?" or "You said that the median is 4. If this is the only information I had asked you to figure out, what wouldn't I know about the data?"

Problem B2

Here are some possible statements that children might make:

 • There is a bump at 4. • The data are really bunched together. • There is a cluster at 3 and 4. • There's a big gap from 6 to 11. • The data are not very spread out.

Problem B3

 • The bump at 4 is the size of families that occurred most often for our class. • Because the data are all bunched together, we know that families in our class are very similar in size. • The cluster at 3 and 4 indicates that most families in our class have three or four people. • The gap from 6 to 11 tells us that no families in our class have six, seven, eight, nine, 10, or 11 children. • The lack of "spread" in our data tells us that our class's families are similar in size.

Problem B4

 • A response based on the mode might be to make the prices of all nine bags exactly \$1.38. Another response based on the mode is to price four bags at \$1.38 and the others at \$1.30, \$1.32, \$1.36, \$1.37, and \$1.50. The reasoning is to place more bags at \$1.38 than at any other price. • A response that is based on the median is to make three bags cost \$1.38 and the others cost \$1.30, \$1.30, \$1.35, \$1.40, \$1.47, and \$1.49. The reasoning is to put some bags at \$1.38 and then to place an equal number of bags at prices lower and higher than \$1.38. Here, three bags cost more than \$1.38 and three bags cost less than \$1.38. • A response that is based on the mean is to make the bags cost \$1.38, \$1.37, \$1.39, \$1.36, \$1.40, \$1.35, \$1.41, \$1.34, and \$1.42. Since there's an odd number of bags, the reasoning is to place one bag at \$1.38 and then add and subtract the same amount to create new prices. Here, 1 cent was subtracted from \$1.38 to get \$1.37, then 1 cent was added to \$1.38 to get \$1.39, and so on.

Problem B5

 a. Median b. Mode c. Mean d. Mode or median e. Mean

 Problem B6 Answers will vary. You may want to use the suggestions for action research to assess your own students' understanding of average. How would they respond to the potato-chip task?