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Lesson plan Grade 12 Math

Aldous and Brenmoor Plants: Comparing Residual Consistency

A grade 12 statistics lesson comparing the mean and spread of daily chlorine residuals from two fictional plants, and testing how one outlier shifts each measure.

Objectives

  1. Students will compute the mean and standard deviation of two sets of residual readings.
  2. Students will compare two plants by center and spread to judge operational consistency.
  3. Students will explain how a single outlier shifts the mean more than the median.

Vocabulary

Key words: mean, standard deviation, median, spread, outlier, residual

Teacher overview

Two treatment plants can hold the same average chlorine residual and still differ sharply in how steady that residual is from day to day, and a regulator cares about the steadiness. This lesson uses a week of daily residual readings from the fictional Aldous and Brenmoor plants, which share an identical mean, to show that the standard deviation, not the average, reveals which plant runs under tighter control. Students then add one bad analyzer reading and watch it drag the mean while barely moving the median, a common misconception being that the mean and median respond to an outlier equally.

Procedure

Hook (5 min): Project the two lists of daily chlorine residual readings from the fictional Aldous and Brenmoor plants without any statistics attached. Ask students to eyeball which plant they would trust to hold a steady residual, and to defend their guess in one sentence. Record the split of opinion on the board.

Direct instruction (12 min): Define center (mean) and spread (standard deviation), and stress that comparing two data sets fairly means comparing both. Work the Aldous readings on the board: 1.0, 1.1, 0.9, 1.0, 1.2, 0.8, 1.0 mg/L. Show the mean is 7.0 ÷ 7 = 1.0 mg/L, then build the squared deviations (0, 0.01, 0.01, 0, 0.04, 0.04, 0), sum to 0.10, divide by n − 1 = 6 to get 0.0167, and take the square root for a sample standard deviation of about 0.13 mg/L.

Guided practice (15 min): In pairs, students compute the mean and sample standard deviation for the Brenmoor readings: 1.2, 0.7, 1.4, 0.6, 1.3, 0.5, 1.3 mg/L. They should find the same mean of 1.0 mg/L and a much larger standard deviation of about 0.38 mg/L. Circulate and check that pairs are dividing by 6, not 7, for the sample standard deviation.

Formative check (thumbs poll): Ask, "Which plant holds the steadier residual, and which statistic told you?" Require a thumbs-up from at least 80 percent of students naming Aldous and the standard deviation before moving on; reteach the spread idea to any pair still pointing to the mean.

Outlier mini-investigation (10 min): Tell students Brenmoor's analyzer glitched one day and logged an eighth reading of 2.4 mg/L. Have pairs recompute Brenmoor's mean with all eight values and find the new median, then compare how far each moved from the original.

Independent practice (8 min): Each student writes two to three sentences answering: if you were the regulator, which plant would you flag for a closer look, and would you rely on the mean, the standard deviation, or the median to make that call?

Closure / exit ticket (5 min): Students answer one question on a card: "Two plants both average 1.0 mg/L. Plant X has a standard deviation of 0.10 mg/L and Plant Y has 0.35 mg/L. Which plant's residual is more predictable, and why?" Collect the cards on the way out.

Aldous and Brenmoor Plants: Comparing Residual Consistency — Grade 12 Math
Name: ____________________________ Date: ______________

Student Handout

Aldous and Brenmoor are two fictional treatment plants. Each list below holds one week of daily chlorine residual readings, in milligrams per liter (mg/L). Use the sample standard deviation throughout, dividing the sum of the squared deviations by n − 1.

Aldous readings: 1.0, 1.1, 0.9, 1.0, 1.2, 0.8, 1.0 mg/L

Brenmoor readings: 1.2, 0.7, 1.4, 0.6, 1.3, 0.5, 1.3 mg/L

  1. What is the mean of the seven Aldous readings, in mg/L?
  2. What is the sample standard deviation of the seven Aldous readings, in mg/L, rounded to two decimal places?
  3. What is the mean of the seven Brenmoor readings, in mg/L?
  4. What is the sample standard deviation of the seven Brenmoor readings, in mg/L, rounded to two decimal places?
  5. Which plant holds the steadier residual, and which statistic tells you so?
  6. Brenmoor's analyzer glitched one day and logged an eighth reading of 2.4 mg/L. What is Brenmoor's mean of all eight readings, in mg/L?
  7. What is the median of Brenmoor's eight readings, in mg/L?
  8. Two plants both average 1.0 mg/L. Plant X has a standard deviation of 0.10 mg/L and Plant Y has 0.35 mg/L. Which plant's residual is more predictable, and why?

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: 1.0 mg/L.
    Aldous mean. The seven readings sum to 7.0 for a mean of 1.0.

  2. Answer: about 0.13 mg/L.
    Aldous standard deviation. The squared deviations sum to 0.10, divided by 6 gives 0.0167, and the square root is about 0.13 mg/L.

  3. Answer: 1.0 mg/L.
    Brenmoor mean. The readings also sum to 7.0 for a mean of 1.0.

  4. Answer: about 0.38 mg/L.
    Brenmoor standard deviation. The squared deviations sum to 0.88, divided by 6 gives 0.1467, and the square root is about 0.38 mg/L.

  5. Answer: Aldous, shown by the standard deviation.
    Aldous is more consistent. Both plants share a mean of 1.0 mg/L, so the mean cannot separate them; the far smaller standard deviation at Aldous (0.13 versus 0.38 mg/L) shows its residual varies much less from day to day.

  6. Answer: 1.175 mg/L.
    With the 2.4 mg/L outlier added, Brenmoor's mean rises to 9.4 ÷ 8 = 1.175 mg/L. One high reading pulls the mean up by about 0.18.

  7. Answer: 1.25 mg/L.
    The median moves only from 1.2 to 1.25 mg/L, so the outlier barely touches it, which is why the median is the more robust measure when a bad reading is suspected.

  8. Answer: Plant X.
    Exit ticket answer: Plant X is more predictable. With equal means, the smaller standard deviation (0.10 versus 0.35 mg/L) means Plant X's readings cluster tightly around the average, so its residual is easier to predict day to day.

Differentiation

Support

Provide the squared-deviation table pre-labeled with columns for reading, deviation, and squared deviation, so students fill values rather than build the structure. Allow the calculator's built-in standard-deviation function once students have done one calculation by hand.

Extension

Ask students to construct a third data set of seven readings that shares the mean of one milligram per liter but has a standard deviation between the Aldous and Brenmoor values, and to prove their set meets both conditions. Then have them describe what operational change might move a plant from the Brenmoor pattern toward the Aldous pattern.

ELL supports

Pre-teach "mean," "spread," "standard deviation," "median," and "outlier" with a simple dot-plot picture showing tight clustering versus wide scatter. Provide sentence frames for the independent practice, such as "I would flag Plant ___ because its ___ is ___."