Worksheet Grade 11 Math
Lammermoor Brook Bacteria Counts: The Geometric Mean on a Log Scale
A grade 11 worksheet comparing the arithmetic and geometric means of five skewed bacteria counts, using base-ten logarithms and one added low sample.
Objectives
- Students will compute the arithmetic mean, the median, and the geometric mean of a skewed data set.
- Students will use base-ten logarithms to find a geometric mean.
- Students will explain why a water rule specifies one measure of center rather than another.
Vocabulary
Key words: geometric mean, logarithm, skewed distribution, colony-forming unit, swimming advisory
Teacher overview
Bacteria counts in a creek spread across orders of magnitude within a single month, so one storm sample can sit a hundred times above a dry-weather sample taken a week earlier. Averaging such counts the ordinary way lets that one storm sample decide the answer, which is why health rules written for recreational water specify a geometric mean instead. This worksheet uses five fictional weekly samples from the fictional Lammermoor Brook at Dunthorne, walks students through the logarithm route to a geometric mean, and then adds a sixth low sample to test how each measure of center responds. The misconception it addresses is that the geometric mean is just a gentler average, when in fact it is the ordinary mean of the data after a change of scale.
Student Worksheet
Five weekly samples were taken from the fictional Lammermoor Brook at Dunthorne. Counts are colony-forming units per 100 milliliters.
| Week | Count |
|---|---|
| 1 | 40 |
| 2 | 120 |
| 3 | 300 |
| 4 | 1,500 |
| 5 | 9,000 |
The fictional Dunthorne county health office posts a swimming advisory when the geometric mean of five samples goes above 200.
- Find the arithmetic mean of the five counts.
- Find the median of the five counts.
- In one or two sentences, explain why the arithmetic mean sits so far above the median.
- Find the base-ten logarithm of the week 1 count, to three decimal places.
- Find the base-ten logarithm of the week 5 count, to three decimal places.
- Take the base-ten logarithm of all five counts. Find the mean of those five logarithms, to three decimal places.
- Raise 10 to the power of your answer to item 6. Round to the nearest whole count. This is the geometric mean.
- State whether the geometric mean is above or below the advisory level of 200.
- Find the arithmetic mean divided by the geometric mean, to two decimal places.
- A sixth sample is taken and reads 30. Find the geometric mean of all six counts, to the nearest whole count.
- Find the arithmetic mean of all six counts, to one decimal place.
- In one or two sentences, explain why a bacteria rule would name the geometric mean rather than the arithmetic mean.
From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.
Answer Key
Answer: 2,192 CFU per 100 mL.
The five counts total 40 + 120 + 300 + 1,500 + 9,000 = 10,960, and 10,960 divided by 5 = 2,192 colony-forming units per 100 milliliters.Answer: 300 CFU per 100 mL.
Sorted, the counts are 40, 120, 300, 1,500, and 9,000, so the third value is the median.Answer: One count is far larger than the rest.
The week 5 sample alone supplies more than four-fifths of the total, so it drags the mean upward, while the median only records which value sits in the middle and ignores how large the largest one is.Answer: 1.602.
The base-ten logarithm of 40 is 1.60206, which rounds to 1.602.Answer: 3.954.
The base-ten logarithm of 9,000 is 3.95424, which rounds to 3.954.Answer: 2.658.
The five logarithms are 1.602, 2.079, 2.477, 3.176, and 3.954. They total 13.289, and 13.289 divided by 5 = 2.658.Answer: 455 CFU per 100 mL.
Raising 10 to the power 2.658 gives 455.0, which rounds to 455 colony-forming units per 100 milliliters. Using the unrounded mean of the logarithms, 2.6577, gives 454.7, the value item 9 uses.Answer: Above.
The geometric mean of 455 sits above the advisory level of 200, so the fictional county would post the advisory.Answer: 4.82.
Dividing gives 2,192 divided by 454.7 = 4.82, so the ordinary mean is nearly five times the geometric mean for this set.Answer: 289 CFU per 100 mL.
The logarithm of 30 is 1.477. The six logarithms total 14.766, and 14.766 divided by 6 = 2.461. Raising 10 to that power gives 289.Answer: 1,831.7 CFU per 100 mL.
The six counts total 10,990, and 10,990 divided by 6 = 1,831.7 colony-forming units per 100 milliliters.Answer: The geometric mean is not controlled by a single high sample.
Counts that range across orders of magnitude are close to symmetric once logarithms are taken, so the mean of the logarithms describes the typical week while the ordinary mean mostly reports the worst one.
Differentiation
Support
Give these students the logarithm column already filled in for the five fictional Lammermoor counts, so items 6 and 7 are an average and a power rather than a calculator sequence. Provide a number line marked in powers of ten with the five counts plotted, which makes the skew in item 3 visible before it has to be explained.
Extension
Ask these students to replace the fictional week 5 count of 9,000 with 90,000 and recompute both measures of center, then to state how much each one moved. Have them explain, using the logarithm step, why the geometric mean moves by a predictable factor while the arithmetic mean does not.
ELL supports
Pre-teach "count," "advisory," and "geometric mean" with a picture of a plate of bacterial colonies beside a posted beach sign. Provide sentence frames for items 3 and 12, such as "The mean is larger than the median because ___," and "A rule uses the geometric mean because one high sample ___."
Extensions
Students can look up whether their own state posts beach or river advisories and describe, in words alone, how many samples such a rule requires before it acts. A second task is to explain why the fictional Dunthorne office would want samples spread across several weeks rather than five taken on one afternoon.