Worksheet Grade 11 Math
Harmon Ridge vs. Cutler Bend: Comparing Chlorine Residual Consistency
A grade 11 worksheet where students calculate the mean and standard deviation of two fictional treatment plants' chlorine residual data to compare consistency.
Objectives
- Students will calculate the mean and standard deviation of a data set of chlorine residual readings.
- Students will compare the center and spread of two data sets to judge which shows more consistent operation.
Vocabulary
Key words: mean, standard deviation, spread, disinfectant residual
Teacher overview
Harmon Ridge and Cutler Bend (two fictional water treatment plants used for this worksheet) both report a mean chlorine residual of 1.0 mg/L across eight days of testing, yet their daily readings vary by very different amounts. A common misconception is that two data sets with the same average are equally consistent; this worksheet asks students to calculate the standard deviation of each plant's readings to show that center alone does not describe a data set. Utilities must keep enough chlorine residual to protect water as it travels through the distribution system while staying below the maximum residual disinfectant level, so a plant that swings widely around its average carries more operational risk than one that holds steady.
Student Worksheet
Both fictional plants tested their chlorine residual once per day for eight days. Readings are in milligrams per liter (mg/L).
Harmon Ridge: 1.0, 1.1, 0.9, 1.0, 1.2, 0.8, 1.0, 1.0
Cutler Bend: 0.4, 1.6, 0.3, 1.7, 0.5, 1.5, 0.4, 1.6
Use the population standard deviation formula: standard deviation = the square root of (the sum of each reading's squared distance from the mean, divided by the number of readings).
- Calculate the mean chlorine residual for Harmon Ridge.
- Calculate the mean chlorine residual for Cutler Bend.
- Compare the two means. What does this comparison alone tell you, or fail to tell you, about how each plant is running?
- Calculate the standard deviation of Harmon Ridge's eight readings. Show the deviation from the mean, the squared deviation, and the final result for each step.
- Calculate the standard deviation of Cutler Bend's eight readings, using the same method.
- Which plant's chlorine residual is more consistent? Use the two standard deviations to justify your answer.
- Both plants must stay below the 4.0 mg/L maximum residual disinfectant level. Based on Cutler Bend's data, does any single reading come close to exceeding that limit?
- Suppose a new internal guideline required a plant's standard deviation to stay below 0.5 mg/L to count as "consistent." Would Harmon Ridge and Cutler Bend each meet that guideline?
- In one or two sentences, explain a reason a plant's chlorine residual might swing more widely than another's, even with the same target dose.
- A third fictional plant, Dellwood, reports a mean residual of 1.0 mg/L with a standard deviation of 0.05 mg/L. Rank Harmon Ridge, Cutler Bend, and Dellwood from most to least consistent.
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: 1.0 mg/L.
Sum of Harmon Ridge's readings (1.0 + 1.1 + 0.9 + 1.0 + 1.2 + 0.8 + 1.0 + 1.0) = 8.0; 8.0 ÷ 8 = 1.0 mg/L.Answer: 1.0 mg/L.
Sum of Cutler Bend's readings (0.4 + 1.6 + 0.3 + 1.7 + 0.5 + 1.5 + 0.4 + 1.6) = 8.0; 8.0 ÷ 8 = 1.0 mg/L.Answer: Both plants dose to the same average target.
The identical means show both plants center their dosing at the same target, but a mean says nothing about how far individual days stray from that target, so it cannot tell us which plant is more reliable day to day.Answer: 0.11 mg/L.
Deviations from 1.0: 0, 0.1, -0.1, 0, 0.2, -0.2, 0, 0. Squared deviations: 0, 0.01, 0.01, 0, 0.04, 0.04, 0, 0, which sum to 0.10. Variance = 0.10 ÷ 8 = 0.0125. Standard deviation = the square root of 0.0125 ≈ 0.11 mg/L.Answer: 0.60 mg/L.
Deviations from 1.0: -0.6, 0.6, -0.7, 0.7, -0.5, 0.5, -0.6, 0.6. Squared deviations: 0.36, 0.36, 0.49, 0.49, 0.25, 0.25, 0.36, 0.36, which sum to 2.92. Variance = 2.92 ÷ 8 = 0.365. Standard deviation = the square root of 0.365 ≈ 0.60 mg/L.Answer: Harmon Ridge.
Harmon Ridge's standard deviation (0.11 mg/L) is much smaller than Cutler Bend's (0.60 mg/L), meaning Harmon Ridge's daily readings cluster far more tightly around its mean.Answer: No.
Cutler Bend's highest single reading is 1.7 mg/L, which stays well below the 4.0 mg/L maximum residual disinfectant level.Answer: Harmon Ridge yes, Cutler Bend no.
Harmon Ridge's standard deviation (0.11 mg/L) is below 0.5 mg/L, so it meets the guideline. Cutler Bend's standard deviation (0.60 mg/L) exceeds 0.5 mg/L, so it does not.Answer: Answers will vary.
A plant might see wider swings because of variable flow through the day, sensor lag between dose changes and readings, or an operator adjusting the feed less frequently than conditions change.Answer: Dellwood, then Harmon Ridge, then Cutler Bend.
Dellwood's standard deviation (0.05 mg/L) is the smallest, followed by Harmon Ridge's (0.11 mg/L), with Cutler Bend's (0.60 mg/L) the largest, so that is the order from most to least consistent.
Differentiation
Support (below grade level)
Provide a partially completed table with the deviation column already filled in for Harmon Ridge, so students only need to square the values, sum them, divide, and take the square root. Allow a calculator with a square-root key for every item.
Extension (above grade level)
Ask students to calculate what the fictional Cutler Bend plant's standard deviation would become if its highest reading (1.7 mg/L) were corrected to 1.0 mg/L as a data-entry fix, and to compare how much the correction changed the mean and the standard deviation.
ELL supports
Pre-teach "mean," "deviation," and "consistent" with a simple visual: two dot plots on a number line, one tightly clustered and one spread out, both centered on the same point. Allow students to write numeric answers and short phrases rather than full sentences for items 3 and 9.