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G07 • Lesson 48 of 105

Data Displays & Measures of Center and Spread

Builds on basic graphing to introduce histograms, dot plots, and box plots for displaying distributions of numerical data. Students calculate and interpret mean, median, mode, range, interquartile range (IQR), and mean absolute deviation (MAD), choosing appropriate measures for symmetric versus skewed distributions.

Middle School Geometry & Data • 6-8

Prerequisites: E12

Key Concepts

  • histograms and dot plots for numerical data
  • mean, median, and mode as measures of center
  • range, IQR, and mean absolute deviation as measures of spread
  • effect of outliers and skewness on center and spread

Data Displays & Measures of Center and Spread

Data is everywhere—test scores, temperatures, game statistics, survey results. But a list of numbers by itself is hard to interpret. This lesson teaches you how to organize data visually and calculate numbers that describe what is "typical" and how spread out the data is.

Data Displays

DisplayBest ForKey Feature
Dot PlotSmall data sets, seeing individual valuesA dot above a number line for each data point; clusters and gaps are visible
HistogramLarger data sets, grouped intervalsBars show frequency of data in ranges (bins); no gaps between bars
Box Plot (Box-and-Whisker)Comparing distributions, showing spreadShows minimum, Q1, median, Q3, maximum; the box spans the middle 50%

Measures of Center

MeasureHow to Find ItWhen to Use
MeanAdd all values, divide by the countWhen data is symmetric with no extreme outliers
MedianMiddle value when data is orderedWhen data is skewed or has outliers
ModeMost frequently occurring valueWhen you want the most common response

Measures of Spread

MeasureHow to Find It
RangeMaximum − Minimum
IQR (Interquartile Range)Q3 − Q1 (spread of the middle 50%)
MAD (Mean Absolute Deviation)Average distance of each data point from the mean

Symmetric vs. Skewed

In a symmetric distribution, the mean and median are close together and the data looks balanced. In a skewed distribution, a tail stretches to one side. Skewed right = tail goes right (mean pulled higher than median). Skewed left = tail goes left (mean pulled lower than median). Use the median for skewed data.

Effect of Outliers

An outlier is a data point far from the rest. Outliers strongly affect the mean and range but have little effect on the median and IQR. This is why the median is often preferred for skewed data.

Example 1 — Mean, Median, Mode

Test scores: 78, 85, 90, 85, 92, 88, 85.

  • Ordered: 78, 85, 85, 85, 88, 90, 92.
  • Mean = (78 + 85 + 85 + 85 + 88 + 90 + 92) / 7 = 603 / 7 = 86.1 (rounded).
  • Median = middle value (4th of 7) = 85.
  • Mode = most frequent = 85 (appears 3 times).
  • Example 2 — Range, IQR, and MAD

    Data: 3, 5, 7, 8, 10, 12, 15.

  • Range = 15 − 3 = 12.
  • Q1 = median of lower half (3, 5, 7) = 5. Q3 = median of upper half (10, 12, 15) = 12.
  • IQR = 12 − 5 = 7.
  • Mean = 60 / 7 ≈ 8.57. Distances from mean: |3 − 8.57| = 5.57, |5 − 8.57| = 3.57, |7 − 8.57| = 1.57, |8 − 8.57| = 0.57, |10 − 8.57| = 1.43, |12 − 8.57| = 3.43, |15 − 8.57| = 6.43.
  • MAD = (5.57 + 3.57 + 1.57 + 0.57 + 1.43 + 3.43 + 6.43) / 7 = 22.57 / 7 ≈ 3.22.
  • Example 3 — Outlier Impact

    Salaries (in thousands): 40, 42, 45, 47, 50, 200. Compare mean and median.

  • Mean = (40 + 42 + 45 + 47 + 50 + 200) / 6 = 424 / 6 ≈ 70.7.
  • Median = average of 3rd and 4th values = (45 + 47) / 2 = 46.
  • The outlier (200) pulled the mean far above the typical salary. The median (46) better represents the group. This data is skewed right.
  • Common Mistake

    Students forget to order the data before finding the median. The median is the middle of the sorted list, not the middle of the original list. Always sort first.

    Practice Problems

    1. Find the mean, median, and mode: 12, 15, 18, 15, 20, 15, 22.

    Show Solution

    Ordered: 12, 15, 15, 15, 18, 20, 22. Mean = 117/7 ≈ 16.7. Median = 15 (4th value). Mode = 15.

    2. Data: 4, 6, 8, 10, 12, 14, 16, 18, 20. Find the IQR.

    Show Solution

    Median = 12 (5th value). Lower half: 4, 6, 8, 10 → Q1 = (6 + 8)/2 = 7. Upper half: 14, 16, 18, 20 → Q3 = (16 + 18)/2 = 17. IQR = 17 − 7 = 10.

    3. Scores: 70, 75, 80, 85, 90. Add an outlier of 10. Does the mean or median change more?

    Show Solution

    Original mean = 80, median = 80. New data (sorted): 10, 70, 75, 80, 85, 90. New mean = 410/6 ≈ 68.3, new median = (75 + 80)/2 = 77.5. The mean dropped by 11.7; the median only dropped by 2.5. The mean changed more.

    4. A histogram shows bins 0–10, 10–20, 20–30 with frequencies 2, 8, 5. Which bin contains the median?

    Show Solution

    Total data points = 2 + 8 + 5 = 15. Median is the 8th value. First bin has values 1–2, second bin has values 3–10. The 8th value falls in the 10–20 bin.

    5. Would you use mean or median to describe the typical home price in a neighborhood where most homes cost around $300,000 but one mansion sold for $5,000,000? Explain.

    Show Solution

    Use the median. The $5,000,000 mansion is an extreme outlier that would pull the mean far above the typical price. The median is resistant to outliers and better represents the typical home.

    Lesson Summary

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