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Math problem set Grade 6 Math

Dissolved Oxygen at Silver Pines Pond: Mean, Median, and Range

A grade 6 math problem set where students find the mean, median, and range of fictional dissolved-oxygen readings and compare them to a real fish-safety threshold.

Objectives

  1. Students will find the mean, median, and range of a set of decimal dissolved-oxygen readings.
  2. Students will compare a data set's readings against a real dissolved-oxygen safety threshold for fish.
  3. Students will explain how an added data point affects the mean differently than the median.

Vocabulary

Dissolved oxygen
Dissolved oxygen is the oxygen gas mixed into water that fish and other underwater animals breathe. Cold, moving water holds more oxygen than warm, still water.

Also: mean, median, range

Teacher overview

A wildlife club at Silver Pines Pond (a fictional pond used for this problem set) measures dissolved oxygen (DO) every morning for a week. Most fish need dissolved oxygen above 5 mg/L to thrive, so the club compares its readings against that real threshold while also practicing mean, median, and range on the fictional data set.

A common misconception is assuming the mean and median must shift by the same amount whenever a new data point is added. Problem 8 asks students to explain why one measure can shift much more than the other when a single unusual reading joins the set.

Dissolved Oxygen at Silver Pines Pond: Mean, Median, and Range — Grade 6 Math
Name: ____________________________ Date: ______________

Problems

Silver Pines Pond's (fictional) wildlife club measures dissolved oxygen six mornings in a row: 7.2, 6.8, 7.5, 5.9, 6.4, and 7.0 mg/L.

  1. Find the sum of the six readings.

  2. Find the mean (average) of the six readings by dividing your sum from problem 1 by 6.

  3. List the six readings in order from least to greatest, then find the median. (With six values, the median is the average of the two middle values.)

  4. Find the range of the six readings (the greatest value minus the least value).

  5. Most fish need dissolved oxygen above 5 mg/L to thrive. How many of the six mornings, if any, had a reading at or below that threshold?

  6. A seventh reading is added: 4.6 mg/L, taken after a warm afternoon rainstorm. Recompute the mean using all seven readings. Round to the nearest hundredth.

  7. With the seventh reading included, which single reading, if any, now falls at or below the 5 mg/L fish-safety threshold?

  8. Explain why the mean changed more than the median would if you recalculated the median with the new seventh reading included.

  9. Challenge. Find the new median using all seven readings (sorted from least to greatest, the median of seven values is the fourth value). Compare it to your answer from problem 3 and explain what you notice.

From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.

Answer Key

  1. Answer: 40.8 mg/L.
    7.2 + 6.8 + 7.5 + 5.9 + 6.4 + 7.0 = 40.8.

  2. Answer: 6.8 mg/L.
    40.8 ÷ 6 = 6.8.

  3. Answer: 6.9 mg/L.
    Sorted readings: 5.9, 6.4, 6.8, 7.0, 7.2, 7.5. The two middle values are 6.8 and 7.0, so the median is (6.8 + 7.0) ÷ 2 = 6.9.

  4. Answer: 1.6 mg/L.
    7.5 − 5.9 = 1.6.

  5. Answer: 0 mornings.
    The lowest of the six readings, 5.9 mg/L, is still above the 5 mg/L threshold, so no morning fell at or below it.

  6. Answer: 6.49 mg/L.
    New sum = 40.8 + 4.6 = 45.4. Mean = 45.4 ÷ 7 = 6.4857..., which rounds to 6.49.

  7. Answer: The seventh reading (4.6 mg/L).
    4.6 mg/L is the only one of the seven readings at or below the 5 mg/L threshold.

  8. Answer: The mean uses every value in its calculation, so one unusual reading shifts it directly; the median depends only on the position of the middle value or values.
    Adding 4.6 mg/L pulls the mean down because that low value is included in the sum that gets divided by 7. The median, however, only looks at which value sits in the middle position once the data is sorted, so a single low outlier at one end of the sorted list may not change the median much, or at all.

  9. Answer: 6.8 mg/L, down 0.1 from the 6.9 mg/L median in problem 3.
    Sorted with the new reading: 4.6, 5.9, 6.4, 6.8, 7.0, 7.2, 7.5. The fourth value is 6.8 mg/L, only 0.1 below the median found in problem 3, because the new low reading joined the low end of the list and moved the middle position by just half a place.

Differentiation

Support (below grade level)

Provide the six original readings already sorted from least to greatest, and give a labeled formula card: "mean = sum ÷ count" and "median = middle value(s)." Allow a calculator for all division.

Extension (above grade level)

Ask students to find how low an eighth reading would need to be to pull the mean below the 5 mg/L fish-safety threshold, keeping the seven readings from problem 6 the same. They should find that it would have to be a negative number, then explain why no real reading could do this, since dissolved oxygen cannot fall below 0 mg/L.

ELL supports

Pre-teach "mean," "median," and "range" with a simple number-line diagram showing the six sorted readings, marking the middle two values and the highest and lowest. Allow the explanation in problem 8 to be supported with the diagram and short phrases instead of full sentences.