6.2 Scatter Plots, Two-Variable Data & Measures of Central Tendency (Mean, Median, Mode, Weighted Average)

Why This Topic Matters for the Exam

This section combines the last piece of the data-display target with the entire measures-of-central-tendency target. Q.6.c ("Represent, display, and interpret data involving two variables in tables and the coordinate plane, including scatter plots and graphs") tests whether you can read a relationship between two variables — for example, hours studied versus test score. Q.7.a ("Calculate the mean, median, mode and range. Calculate a missing data value, given the average and all but one of the values... calculate a weighted average") is the GED's single most computation-heavy statistics indicator, and it is tested with real numeric inputs, not just conceptual recognition.

The GED also uses hot-spot items for scatter plots — you may be asked to click the point on a coordinate plane that matches a given ordered pair, or to identify which plotted point is an outlier. That means scatter-plot skills combine coordinate-plane fluency (see Chapter 10) with data interpretation.

Core Terms: Two-Variable Data

A scatter plot displays two-variable data as points (x, y) on the coordinate plane, where each point represents one paired observation (for example, one student's hours studied and that student's test score). The pattern of points reveals correlation:

A line of best fit (or trend line) is an informal straight line drawn through a scatter plot to approximate the overall trend; the GED expects you to use it to make a prediction — for example, reading off the line's y-value at a given x-value — not to calculate its slope from a regression formula. A two-way table organizes categorical data by two variables at once (rows and columns), such as "Prefers Morning / Prefers Evening" crossed with "Under 30 / 30 and Over"; reading a two-way table means locating the correct row-and-column cell, then often comparing that cell to a row or column total.

Core Terms: Measures of Central Tendency

• Mean (average): the sum of all values divided by the number of values. The GED formula sheet states it plainly: "mean is equal to the total of the values of a data set, divided by the number of elements in the data set."
• Median: the middle value once the data is sorted in order; if there is an even number of values, the median is the mean of the two middle values. The formula sheet also defines this one for you.
• Mode: the value that appears most often. A data set can have no mode, one mode, or several modes (bimodal/multimodal) — and mode is the only measure of center that also works on non-numeric categorical data. Mode is not on the formula sheet, so memorize its definition.
• Range: the maximum value minus the minimum value — a simple measure of spread (different from the interquartile range covered in Section 6.1).
• Weighted average: an average where different values carry different weights (frequencies, group sizes, credit hours) rather than counting equally. Weighted average is also not on the formula sheet.

Worked Example 1: Basic Mean, Median, Mode

A data set of quiz scores is: 78, 85, 78, 92, 67. Find the mean, median, and mode.

• Mean: (78 + 85 + 78 + 92 + 67) ÷ 5 = 400 ÷ 5 = 80
• Median: sort the data — 67, 78, 78, 85, 92 — the middle value is 78
• Mode: 78 appears twice, more than any other value, so the mode is 78

Worked Example 2: Missing Value Given the Mean

Four of five test scores are known: 88, 91, 76, 95. The mean of all five scores is 87. What is the missing fifth score?

First find the total sum: mean × number of values = 87 × 5 = 435. Then subtract the four known scores: 435 − (88 + 91 + 76 + 95) = 435 − 350 = 85. The trap here is trying to average the four known scores directly instead of first rebuilding the total sum — you must multiply by 5 (the total count), not 4, because the mean was defined over all five values.

Worked Example 3: Weighted Average

A class of 30 students takes a test: 10 students score 90, 15 students score 80, and 5 students score 60. What is the weighted average (class mean)?

Multiply each score by how many students earned it, sum those products, then divide by the total number of students: (10 × 90) + (15 × 80) + (5 × 60) = 900 + 1,200 + 300 = 2,400. Divide by 30 total students: 2,400 ÷ 30 = 80. Notice this is not the same as averaging 90, 80, and 60 equally, which would incorrectly give (90 + 80 + 60) ÷ 3 = 76.7 — ignoring how many students are in each group is the classic weighted-average trap.

Common Traps Checklist

• Forgetting to sort data before finding the median — an unsorted "middle" value is meaningless.
• Averaging group scores equally instead of weighting by group size in a weighted-average problem.
• Multiplying the mean by the wrong count (the count before the missing value is added, not after) in missing-value problems.
• Reading a scatter plot's trend direction backward (mistaking a negative correlation for positive, or vice versa).
• Assuming mode and weighted average are given on the formula sheet — they are not; only mean and median are defined there.

Takeaways: sort data before finding the median, remember mode and weighted average must be memorized (they're absent from the formula sheet), rebuild the total sum (mean × count) before solving missing-value problems, and read scatter-plot correlation by the overall direction of the point cloud, not by any single point.