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Math problem set Grade 9 Math

Redbank vs. Talbot Ridge: Comparing Chlorine Residual Consistency

A grade 9 statistics problem set comparing the mean and standard deviation of chlorine residual readings from two fictional treatment plants.

Objectives

  1. Students will calculate the mean and standard deviation of a small data set of chlorine residual readings.
  2. Students will compare the spread of two data sets with equal means to judge which is more consistent.

Vocabulary

Key words: mean, standard deviation, range, deviation

Teacher overview

Two fictional treatment plants, Redbank and Talbot Ridge, both report the same average chlorine residual over a five-day period, which might suggest they are performing identically. This problem set uses standard deviation to show why the mean alone does not tell the whole story: one plant holds a tight, consistent residual close to its average every day, while the other swings well above and below its average from day to day, even though both plants land on the same five-day mean. A common misconception in this kind of problem is assuming that two data sets with equal means must be equally consistent; standard deviation is the tool that reveals otherwise.

Redbank vs. Talbot Ridge: Comparing Chlorine Residual Consistency — Grade 9 Math
Name: ____________________________ Date: ______________

Problems

The fictional Redbank treatment plant's five daily chlorine residual readings, in mg/L, were: 1.2, 1.3, 1.1, 1.4, 1.0.

The fictional Talbot Ridge treatment plant's five daily chlorine residual readings, in mg/L, were: 0.8, 1.6, 0.9, 1.7, 1.0.

  1. Calculate the mean of Redbank's five readings.

  2. Calculate the mean of Talbot Ridge's five readings. How does it compare to Redbank's mean from problem 1?

  3. Calculate the range (maximum minus minimum) of each plant's five readings. Based on range alone, which plant's readings appear more spread out?

  4. Complete a deviation table for Redbank: for each reading, find the deviation (reading − mean) and the squared deviation (reading − mean)². Redbank's mean from problem 1 should be the value you subtract.

  5. Using your table from problem 4, calculate Redbank's standard deviation with the formula: standard deviation = square root of [(sum of squared deviations) ÷ number of readings]. Round to the nearest hundredth.

  6. Complete the same deviation table and standard deviation calculation for Talbot Ridge, using the mean from problem 2. Round to the nearest hundredth.

  7. Compare the two standard deviations from problems 5 and 6. Which plant maintains a more consistent chlorine residual, and how does this compare to what the range alone suggested in problem 3?

  8. A third fictional plant, Aldergate, reports a standard deviation of 0.05 mg/L for its five readings, with the same 1.2 mg/L mean as Redbank and Talbot Ridge. Without calculating further, would you expect Aldergate's five readings to be more tightly clustered around 1.2 mg/L or more spread out than Redbank's readings? Explain using the size of the standard deviation.

  9. Redbank's plant manager wants next month's average to stay at 1.2 mg/L but wants to lower the plant's standard deviation. Describe, in terms of the five individual daily readings, what "keeping the same mean while lowering the standard deviation" would look like.

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: 1.2 mg/L.
    (1.2 + 1.3 + 1.1 + 1.4 + 1.0) ÷ 5 = 6.0 ÷ 5 = 1.2 mg/L.

  2. Answer: 1.2 mg/L, the same as Redbank.
    (0.8 + 1.6 + 0.9 + 1.7 + 1.0) ÷ 5 = 6.0 ÷ 5 = 1.2 mg/L, identical to Redbank's mean.

  3. Answer: Redbank's range is 0.4 mg/L; Talbot Ridge's range is 0.9 mg/L. Talbot Ridge appears more spread out.
    Redbank: 1.4 − 1.0 = 0.4 mg/L. Talbot Ridge: 1.7 − 0.8 = 0.9 mg/L. Talbot Ridge's larger range suggests wider variation, before standard deviation confirms it more precisely.

  4. Answer: deviations 0, 0.1, -0.1, 0.2, -0.2; squared deviations 0, 0.01, 0.01, 0.04, 0.04.
    Each reading minus the mean of 1.2: 1.2−1.2=0, 1.3−1.2=0.1, 1.1−1.2=-0.1, 1.4−1.2=0.2, 1.0−1.2=-0.2. Squaring each: 0, 0.01, 0.01, 0.04, 0.04.

  5. Answer: 0.14 mg/L.
    Sum of squared deviations: 0 + 0.01 + 0.01 + 0.04 + 0.04 = 0.10. Divide by 5: 0.10 ÷ 5 = 0.02. Square root: √0.02 ≈ 0.14 mg/L.

  6. Answer: 0.37 mg/L.
    Deviations from the mean of 1.2: 0.8−1.2=-0.4, 1.6−1.2=0.4, 0.9−1.2=-0.3, 1.7−1.2=0.5, 1.0−1.2=-0.2. Squared: 0.16, 0.16, 0.09, 0.25, 0.04. Sum: 0.70. Divide by 5: 0.70 ÷ 5 = 0.14. Square root: √0.14 ≈ 0.37 mg/L.

  7. Answer: Redbank is more consistent.
    Redbank's standard deviation (0.14 mg/L) is much smaller than Talbot Ridge's (0.37 mg/L), confirming what the range comparison in problem 3 already suggested: Talbot Ridge's daily readings swing farther from its average, even though both plants share the same 1.2 mg/L mean. A regulator comparing only the means would miss this difference entirely.

  8. Answer: more tightly clustered.
    A standard deviation of 0.05 mg/L is smaller than Redbank's 0.14 mg/L, so Aldergate's five readings would be expected to sit closer to the 1.2 mg/L mean, with less day-to-day variation than Redbank shows.

  9. Answer: answers vary; must describe readings clustering closer to 1.2 mg/L while still averaging 1.2 mg/L.
    A strong answer explains that the five daily readings would need to move closer together around 1.2 mg/L, for example readings like 1.15, 1.25, 1.20, 1.18, 1.22 instead of the current wider spread, so that the deviations from the mean shrink even though the mean itself stays at 1.2 mg/L.

Differentiation

Support

Provide the deviation tables for both plants with the deviation column already filled in, so students only need to square each value, sum, divide, and take the square root. Work through Redbank's full standard deviation calculation as a class example before releasing problem 6.

Extension

Ask students to calculate the standard deviation of a combined ten-reading data set formed by pooling Redbank's and Talbot Ridge's readings together, and to explain whether this combined value falls between the two individual standard deviations or outside that range.

ELL supports

Pre-teach "deviation" as "how far a number is from the average," paired with a simple number-line diagram showing a few points clustered near a center line versus a few points spread far from it. Provide the standard deviation formula as a labeled step-by-step checklist (find the mean, subtract, square, sum, divide, square root) that students can follow one line at a time.