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Math problem set Grade 11 Math

Deer Hollow Monitoring Wells: Box Plots, Outliers, and Nitrate

A grade 11 problem set where students build a box plot from fictional groundwater monitoring wells' nitrate data and decide how an outlier should be handled.

Objectives

  1. Students will calculate a five-number summary and identify outliers using the 1.5 times IQR rule.
  2. Students will compare how the mean and median each respond to an outlier in a data set.

Vocabulary

Key words: five-number summary, interquartile range, outlier, median

Teacher overview

Deer Hollow, a fictional groundwater monitoring network used for this problem set, tracks nitrate levels across twelve wells, and nitrate from fertilizer runoff is a common contaminant that groundwater programs watch closely. A common misconception with data like these is assuming every reading should be averaged in without question; one well's unusually high reading can pull a mean far from where most of the data actually sits, so this problem set has students calculate a five-number summary, test for an outlier using the 1.5 times IQR rule, and compare how the mean and the median each respond before deciding which measure best represents a "typical" well.

Deer Hollow Monitoring Wells: Box Plots, Outliers, and Nitrate — Grade 11 Math
Name: ____________________________ Date: ______________

Problems

The fictional Deer Hollow monitoring network's twelve wells reported the following nitrate levels, in milligrams per liter (mg/L), during this quarter's sampling round: 2, 3, 3, 4, 4, 4, 5, 5, 6, 6, 7, 15.

  1. Sort the data if it is not already sorted, and identify the five-number summary: minimum, first quartile (Q1), median, third quartile (Q3), and maximum.
  2. Calculate the interquartile range (IQR), using IQR = Q3 − Q1.
  3. Use the 1.5 × IQR rule to calculate the lower fence and the upper fence for outliers (lower fence = Q1 − 1.5 × IQR; upper fence = Q3 + 1.5 × IQR).
  4. Which well reading, if any, falls outside the fences from problem 3? Identify it as an outlier or state that none exists. Then draw a box plot of the data on a number line from 0 to 16 mg/L, ending the upper whisker at the largest value that is not an outlier and marking any outlier as a separate point.
  5. Calculate the mean nitrate reading for all twelve wells, including the reading identified in problem 4.
  6. Calculate the mean nitrate reading for the eleven wells that remain if that reading is excluded.
  7. Calculate the median nitrate reading with the outlier included and with it excluded. Compare how much each measure changed.
  8. A well technician needs to report a single "typical" nitrate level for Deer Hollow to the town council. Based on your answers to problems 5 through 7, should the technician use the mean or the median, 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: Min 2, Q1 3.5, median 4.5, Q3 6, max 15 mg/L.
    Sorted data: 2, 3, 3, 4, 4, 4, 5, 5, 6, 6, 7, 15. Median (average of the 6th and 7th values, 4 and 5) = 4.5. Q1 (median of the lower six values 2, 3, 3, 4, 4, 4) = 3.5. Q3 (median of the upper six values 5, 5, 6, 6, 7, 15) = 6.

  2. Answer: 2.5 mg/L.
    IQR = Q3 − Q1 = 6 − 3.5 = 2.5 mg/L.

  3. Answer: -0.25 mg/L; lower fence -0.25 mg/L and upper fence 9.75 mg/L.
    Lower fence = 3.5 − (1.5 × 2.5) = 3.5 − 3.75 = -0.25 mg/L. Upper fence = 6 + (1.5 × 2.5) = 6 + 3.75 = 9.75 mg/L.

  4. Answer: the 15 mg/L reading.
    15 mg/L is greater than the upper fence of 9.75 mg/L, so it counts as an outlier under the 1.5 × IQR rule; no reading falls below the lower fence. Box plot: the box runs from 3.5 to 6 with a median line at 4.5; the whiskers run from 2 to 7; 15 is plotted as a separate point.

  5. Answer: 5.33 mg/L.
    Sum of all twelve readings (2 + 3 + 3 + 4 + 4 + 4 + 5 + 5 + 6 + 6 + 7 + 15) = 64; 64 ÷ 12 ≈ 5.33 mg/L.

  6. Answer: 4.45 mg/L.
    Sum of the eleven readings without the outlier (64 − 15) = 49; 49 ÷ 11 ≈ 4.45 mg/L.

  7. Answer: median 4.5 mg/L with outlier, 4 mg/L without.
    With the outlier, the median is 4.5 mg/L (as found in problem 1). Without it, the eleven remaining values (2, 3, 3, 4, 4, 4, 5, 5, 6, 6, 7) have a median of 4 mg/L (the 6th of 11 values). The median changed by only 0.5 mg/L, far less than the mean's change of about 0.88 mg/L (5.33 minus 4.45).

  8. Answer: the median.
    The median barely moved when the outlier well was removed, while the mean dropped by nearly a full mg/L, which means the median better represents the typical well in the Deer Hollow network and is less distorted by one unusual reading.

Differentiation

Support (below grade level)

Provide the sorted data list already written out, and give students a labeled diagram of a box plot with blanks for the minimum, Q1, median, Q3, and maximum so they can see where each value belongs before calculating the fences.

Extension (above grade level)

Ask students to invent a thirteenth well reading that would become a new outlier under the same fences from problem 3, then recalculate the mean with that added value and compare how much further it shifts.

ELL supports

Pre-teach "outlier," "median," and "quartile" with a simple visual: a box-and-whisker diagram drawn on a number line with the twelve well readings marked as dots above it, showing where the "whisker" ends and the lone outlier sits beyond it. Allow students to answer problem 8 with a short phrase and the two numbers rather than a full paragraph.