Grade 6

Mid-Unit Assessment

Mid-Unit Assessment
1.

Select all the statistical questions.

A.

How many nickels does it take to make a dollar?

B.

In what year was the first penny made in the United States?

C.

Among all the pennies at a bank, what is the most frequent year the pennies were made?

D.

Which coin (penny, nickel, dime, or quarter) is used most frequently in transactions at a bank?

E.

On average, how many pennies do people receive in change when they make a purchase at a store?

Answer: C, D, E

Teaching Notes

Students who select Choice A did not recognize that this requires only a calculation. Students who select Choice B did not understand that a question requiring research is not necessarily a statistical question, because there is no variability in the answer. Students who don’t select Choice C may not have understood that different pennies have different years. Students who don’t select Choice D may think that because the answer is the name of a coin, it is not statistical. Students who don’t select Choice E most likely have a deep misunderstanding of the nature of statistical questions.

2.

This dot plot shows the number of hours 20 sixth grade students slept on a Saturday night.

Dot plot from 4 to 9 by 1’s. Sleep in hours.
Dot plot from 4 to 9 by 1’s. Sleep in hours. Beginning at 4, number of dots above each increment is 1, 0, 2, 4 at 6 point 5, 4, 3 at 7 point 5, 5, 1 at 8 point 5, 0.

Select all the true statements about the data used to build the dot plot.

A.

Six students slept for at least 8 hours.

B.

The mean amount of sleep was 6.25 hours.

C.

More than half of the students slept 7 hours or less.

D.

Only 1% of the students slept more than 8 hours.

E.

The difference between the most hours of sleep and the least for these students was 2.5 hours.

Answer: A, C

Teaching Notes

Students analyze a dot plot, including simple questions about spread and distribution. Students are asked to calculate percentages: There are 20 data points to make calculations simpler.

Students who don’t select Choice A may have mistaken the group of 5 as one, may have miscounted, or may misunderstand the difference between “at least” and “more than.” Students who select Choice B have confused “mean” with “midrange” (the average of the largest and smallest values). Students who don’t select Choice C may not be accounting for the frequencies when determining information in a dot plot. Students who select Choice D misunderstand the difference in meaning between “1% of students” and the data point indicating one student. Students who select Choice E likely excluded 4 hours as an outlier.

3.

Students responded to a survey. The survey asked students to report the amount of time they spent doing homework during a week, to the nearest hour. This histogram displays the data. Which of these statements must be true?

Histogram from 0 to 20 by 5's. Homework time in hours. Vertical axis from 0 to10 by 1's labeled students.
Histogram from 0 to 20 by 5's. Homework time in hours. Vertical axis from 0 to10 by 1's labeled students. Beginning at 0 up to but not including 5, height of bar at each interval is 8, 5, 3, 1.

A.

A total of 4 students participated in the survey.

B.

Every student spent at least 1 hour doing homework.

C.

More students spend less than 5 hours on homework than those who spend between 5 and 10 hours on homework.

D.

More students spent 7 hours doing homework than spent 12 hours.

Answer:

More students spend less than 5 hours on homework than those who spend between 5 and 10 hours on homework.

Teaching Notes

Students who select Choice A are misreading the number of intervals as the number of observations. Students who select Choice B may misunderstand the height of the bars as the number of hours spent doing homework. Students who select Choice D are looking at the size of the interval and drawing a conclusion that a histogram is not capable of representing.

4.

Here is the height of 20 flowers in the school garden, in centimeters.

  • 5
  • 5
  • 10
  • 10
  • 15
  • 25
  • 25
  • 25
  • 25
  • 30
  • 30
  • 35
  • 45
  • 45
  • 50
  • 55
  • 65
  • 70
  • 105
  • 110

  1. Draw a histogram to display the data.
  2. Based on the histogram, what is a typical length for these 20 flowers?

Blank grid. Height in centimeters. Horizontal axis 0 to 140 by 20's. Vertical axis 0 to 10 by 2's.

Answer:

  1.  

    <p>Histogram. Height in centimeters. </p>

  2. Sample response: About 20–40 centimeters. Most of the flowers are in the 20–40 centimeter interval with a few more flowers greater than 40 than less than 20, so maybe closer to 40.

Tier 1 response:

  • Accurate, correct work.
  • All histogram bar heights are correct, and the typical length given is in the 20–40 cm range.

Tier 2 response:

  • Work shows general conceptual understanding and mastery, with some errors.
  • Sample errors: One or two mistakes in histogram bar heights; correct histogram but a typical length above 40 cm given.

Tier 3 response:

  • Significant errors in work demonstrate lack of conceptual understanding or mastery.
  • Sample errors: Major mistakes in histogram bar heights; attempt to draw a different type of plot; mistakes in histogram and an incorrect typical length; empty or nonsensical answer to typical length question.

Teaching Notes

For the second question, accept any answer from 20 to 40 centimeters. The mean of the data is 39.25 centimeters, but students should not need this calculation to answer the question. Watch for students making minor errors in building the histogram. The problem does not specify the intervals that students must use, so accept alternate correct histograms. Nearly all students will use the intervals provided.

5.

The mean of four numbers is 25. Three of the four numbers are 20, 26, and 26. What is the fourth number?

Answer:

28. (One number is 5 less than the mean, and two numbers are each 1 more than the mean. Because the four numbers must be evenly distributed around 25, the last number must be 3 more than the mean.)

Teaching Notes

There is more than one way to do this problem. Thinking of the mean as a fair share is the intended method. Students may also recognize that if the mean of four numbers is 25, the sum of the four numbers is 100.

6.

  1. Draw two dot plots, each with 7 or fewer data points, so that:

    • Both dot plots display data with approximately the same mean.

    • The data displayed in Dot plot A has a much larger MAD (mean absolute deviation) than the data displayed in Dot plot B.

      Dot plot A
      A blank dot plots labeled “dot plot A”. It has the numbers 0 through 20, in increments of 2, indicated. There are tick marks midway between.

      Dot plot B
      A blank dot plots labeled "dot plot B.” It has the numbers 0 through 20, in increments of 2, indicated. There are tick marks midway between.

       

  2. How can you tell, visually, that one dot plot displays data with a larger MAD than another?

Answer:

  1. The dot plots should show roughly the same center, with Dot plot B showing a much tighter clustering than Dot plot A.
  2. Sample response: Because MAD is a measure of spread, the data in a dot plot with a larger MAD will have the wider spread.

Tier 1 response:

  • Accurate, correct work.
  • Dot plots show roughly the same center; Dot plot A has visibly larger MAD; appropriate description of spread given.
  • Acceptable errors: Dot plots include more than 7 points.

Tier 2 response:

  • Work shows general conceptual understanding and mastery, with some errors.
  • Sample errors: Dot plots show a significantly different center but correct MAD; dot plots show correct center but similar MAD; dot plots are correct but backward; no reasonable description given.

Tier 3 response:

  • Significant errors in work demonstrate lack of conceptual understanding or mastery.
  • Sample errors: Two or more error types from Tier 2 response; failure to draw two dot plots.

Teaching Notes

Some students may have trouble here constructing the dot plots because the problem is quite open-ended. The guideline of 7 or fewer data points is intended only to limit the amount of time students spend on this problem. Also, note that students do not have to make the means equal, just approximately the same.

7.

Here are two dot plots, the ages of the five children in each of two families.

family 1
A dot plot, family 1, from 0 to 26 by 2's. Age in years.
A dot plot, family 1, from 0 to 26 by 2's. Age in years. Beginning at 5, number of dots above each increment by 1's is 1, 0, 0, 0, 0, 1, 0, 0, ,0 ,0 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0.

family 2
A dot plot, family 2, from 0 to 26 by 2's. Age in years.
A dot plot, family 2, from 0 to 26 by 2's. Age in years. Beginning at 5, number of dots above each increment by 1's is 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 1.

  1. For each family, identify anyone whose age is unusual compared to the rest of the family.
  2. Calculate the MAD (mean absolute deviation) of each data set. Which family has the wider spread of ages?

Answer:

  1. Family 1: No unusual values, or 5 and 25 (both acceptable). Family 2: One unusual value, 5.
  2. Family 1 has the larger spread. Family 1: 6 years, Family 2: 3.2 years.

Tier 1 response:

  • Accurate, correct work.
  • Correct list of unusual data, correct calculations of MAD, correct selection of Family 1 as having larger spread.

Tier 2 response:

  • Work shows good conceptual understanding and mastery, with minor errors.
  • Sample errors: A calculation error causes one mean or MAD to be incorrect; incorrect list of unusual data; failure to correctly compare MADs with otherwise accurate work. Acceptable errors: An error in calculation causes an incorrect conclusion about which family has the larger spread.

Tier 3 response:

  • Work shows a developing but incomplete conceptual understanding, with significant errors.
  • Sample errors: Two or more error types from Tier 2 response; multiple calculation errors; an incorrect MAD with no work shown; minor errors in chosen method of determining mean or MAD, including failure to use absolute value of deviations.

Tier 4 response:

  • Work includes major errors or omissions that demonstrate a lack of conceptual understanding and mastery.
  • Sample errors: Badly incorrect algorithm for calculating mean or MAD; no calculation of MAD.

Teaching Notes

Students analyze the MAD of two data sets. One data set has an unusually low value, while the other is evenly spread. The first set has a larger MAD because “on average” it is more spread out.

Some students may decide that 5 and 25 are unusual in Family 1. Because formal definitions for outliers have not yet been learned, this is an acceptable response.