Lesson plan Grade 10 Math

Vasserly Creek Sampling: Comparing Two Sites by Center and Spread

A 50-minute grade 10 lesson where students compare eight weeks of turbidity readings from two fictional creek sites using mean, median, IQR, and an outlier test.

Objectives

  1. Students will calculate the mean, median, and interquartile range for two sets of turbidity readings.
  2. Students will apply the 1.5 times IQR rule to identify an outlier in a monitoring data set.
  3. Students will explain which measure of center better represents a site with one extreme reading.

Vocabulary

Turbidity
Turbidity is a measure of how cloudy or murky water looks because of tiny particles floating in it. Clear water has low turbidity, and muddy water has high turbidity.

Also: mean, median, interquartile range, outlier

Teacher overview

Vasserly Creek and its two sampling sites are fictional, and the eight weekly readings used here are set for this lesson. Turbidity, measured in nephelometric turbidity units, is one of the first things a technician records at a stream site, and a single storm week can produce a reading many times higher than the rest. The misconception this lesson addresses is treating the mean as the typical value regardless of the data's shape; one extreme reading pulls the mean of the downstream site up by nearly a full unit while leaving the median almost unchanged. Students finish by deciding which statistic they would report to a watershed board and defending the choice.

Procedure

Hook (5 min): Write two numbers on the board, a mean of 3.60 and a median of 2.85, and tell students both describe the same eight readings from one fictional creek site. Ask what could make those two numbers disagree by that much, and collect three or four ideas without confirming any of them.

Direct instruction (12 min): Review the three statistics the lesson uses. Model each one on a short teacher example that is not on the handout: the five fictional readings 1.5, 1.6, 1.8, 2.0, and 6.0. The mean is 12.9 divided by 5, which equals 2.58, while the median is 1.8. Show the quartile positions on that same list, then state the outlier rule: a value above the third quartile plus 1.5 times the interquartile range is treated as an outlier. Do not work any handout number during this segment.

Guided practice (12 min): Distribute the handout. Work items 1 through 4 together, pausing after item 3 so pairs can confirm they divided by 8 and not by 7. Ask students to write one sentence predicting whether the two sites will have similar spreads before they compute anything in item 7.

Independent practice (13 min): Students complete items 5 through 12 on their own or in pairs. Circulate and check that quartiles are being read from the sorted list rather than the original order, which is the most common error at this stage.

Closure and formative check (8 min): Collect the handouts and read item 12 aloud as an exit question: which measure of center better represents a typical week at the downstream site? Ask for a show of hands for mean and for median. If more than one-third of the class chooses the mean, reteach the outlier segment before the next class using the item 10 recomputation as the anchor example.

Answer Key

  1. Answer: 2.25 NTU. The eight readings total 18.0, and 18.0 ÷ 8 = 2.25 NTU.

  2. Answer: 2.25 NTU. Sorted, the readings are 1.9, 2.0, 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, so the median is (2.2 + 2.3) ÷ 2 = 2.25 NTU.

  3. Answer: 3.60 NTU. The eight readings total 28.8, and 28.8 ÷ 8 = 3.60 NTU.

  4. Answer: 2.85 NTU. Sorted, the readings are 2.3, 2.6, 2.7, 2.8, 2.9, 3.0, 3.1, 9.4, so the median is (2.8 + 2.9) ÷ 2 = 2.85 NTU.

  5. Answer: 2.65 NTU. The lower half is 2.3, 2.6, 2.7, 2.8, and its median is (2.6 + 2.7) ÷ 2 = 2.65 NTU.

  6. Answer: 3.05 NTU. The upper half is 2.9, 3.0, 3.1, 9.4, and its median is (3.0 + 3.1) ÷ 2 = 3.05 NTU.

  7. Answer: 0.40 NTU. The interquartile range is 3.05 − 2.65 = 0.40 NTU, the same spread Site A shows.

  8. Answer: 3.65 NTU. The boundary is 3.05 + 1.5 × 0.40 = 3.05 + 0.60 = 3.65 NTU.

  9. Answer: Yes. The 9.4 NTU reading is well above the 3.65 NTU boundary, so it counts as an outlier.

  10. Answer: 2.77 NTU. The remaining seven readings total 28.8 − 9.4 = 19.4, and 19.4 ÷ 7 = 2.7714, which rounds to 2.77 NTU.

  11. Answer: 0.52 NTU. The difference is 2.77 − 2.25 = 0.52 NTU, far smaller than the 1.35 NTU gap between the two full-sample means.

  12. Answer: The median describes a typical week better. Seven of the eight Site B readings sit between 2.3 and 3.1, and the median of 2.85 falls inside that range, while the mean of 3.60 is higher than all but one reading. The single storm week pulls the mean upward without changing what most weeks look like.

Differentiation

Support

Give students both data sets already sorted from least to greatest, with the quartile positions marked by short vertical lines, so items 5 through 7 become a matter of reading the marked positions. Provide the outlier rule as a filled-in formula with the item 6 and item 7 values already substituted.

Extension

Ask students to find the largest value that could replace the 9.4 reading and still fall below the outlier boundary, then explain what that tells them about how sensitive the rule is. A second challenge is to compute the mean absolute deviation for both fictional sites and compare it with the interquartile range as a measure of spread.

ELL supports

Provide a vocabulary card pairing mean with "add and divide," median with "middle of the sorted list," and interquartile range with "middle half," each beside a small labeled number line. Allow students to answer item 12 by marking the mean and the median on a dot plot of the Site B readings and writing one short sentence.

Extensions

Have students collect eight weekly readings from a public monitoring site their teacher selects, then repeat items 1 through 11 with that record in place of the fictional Vasserly Creek data. A short written comparison of the two sites' interquartile ranges makes a good bridge to a later lesson on control charts and sampling frequency.