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Tables, Scatter Plots, and Line Graphs

Read two-variable data in tables, scatter plots, and line graphs; describe trends and correlation; and use trend lines and rates of change to predict.

Practice this skill 30 min Scatter plots and two-variable data Leer en español
Key idea

What you'll learn

Many real questions involve two quantities at once: hours worked and pay, age of a car and its value, time of day and temperature. This lesson teaches you to read and use the three ways such data are shown: tables, scatter plots, and line graphs.

You will learn to:

  • read values from a table, a scatter plot, and a line graph,
  • describe the association (correlation) in a scatter plot: positive, negative, or none, and strong or weak,
  • find outliers and explain why they don't fit,
  • use a trend line to make predictions and work backwards,
  • find and explain a rate of change (slope) with its units, and
  • avoid the classic mistake of treating correlation as cause.

Why it matters on the GED: Two-variable data connect the data skills to algebra. The GED shows a graph or table and asks you to read a value, describe a trend, predict with a trend line, or find a rate. Many of these questions are word problems about money, time, or measurement. In daily life the same skills help you read a car's depreciation, a pay table, or a chart of your utility bills.

Words to know

  • Variable — a quantity that can change, such as hours or temperature.
  • Two-variable data — pairs of numbers that go together, like (4 hours, $62).
  • Table — rows or columns that list each pair, such as hours 4, 6, 9 with pay 62, 93, 139.50.
  • Scatter plot — a graph with one dot for each pair (x, y). The dots are not connected.
  • Line graph — a graph that connects points in order, usually over time, to show how something changes.
  • Association (correlation) — a relationship between two variables.
  • Positive association — as x increases, y tends to increase (more study, higher score).
  • Negative association — as x increases, y tends to decrease (older car, lower value).
  • No association — no up or down pattern (shoe size and test score).
  • Strong / weak — points lie close to a line (strong) or loosely around it (weak).
  • Trend line (line of best fit) — a straight line drawn through the middle of the points to show the pattern.
  • Outlier — a point far from the pattern of the others, such as (6, 58) when similar students scored 85.
  • Rate of change (slope) — how much y changes for each 1-unit change in x, such as $2,000 per year.
  • Causation — one thing directly causes another. Correlation alone never proves causation.

All Math words in the glossary

Tables and line graphs

Tables. A table pairs each x-value with a y-value. To find a rate, divide the change in y by the change in x:

Hours worked (x)46912
Pay in dollars (y)6293139.50186

Example: From 4 to 6 hours, pay rises $31 for 2 hours: 31 \div 2 = \$15.50 per hour. Check another pair: \tfrac{186 - 139.50}{12 - 9} = \tfrac{46.50}{3} = 15.50 ✓. Because the rate is the same for every pair, the relationship is linear, and you can predict: 20 hours → 20 \times 15.50 = \$310.

Dana's pay table graphed: hours worked on the x-axis (0 to 13) and pay in dollars on the y-axis (0 to 200). The points (4, 62), (6, 93), (9, 139.50) and (12, 186) lie on one straight line through the origin. A slope triangle from 4 to 6 hours shows a run of 2 hours and a rise of 31 dollars, so the rate is 31 ÷ 2 = 15.50 dollars per hour.0255075100125150175200012345678910111213Hours worked (x)Pay in dollars (y)+2 h+$31(4, 62)(6, 93)(9, 139.50)(12, 186)rate = 31 ÷ 2 = $15.50/h

Careful: if the x-values jump by 2s (weeks 0, 2, 4, 6), the change in y between columns is for 2 units of x. Divide by 2 for the rate per unit.

Line graphs. A line graph connects points in order, so you can see how a quantity changes over time.

Line graph of temperature in degrees Fahrenheit during one day: 6 am 48, 8 am 53, 10 am 62, 12 pm 68, 2 pm 72, 4 pm 70, 6 pm 61. Vertical axis from 44 to 76 in steps of 4.4448525660646872766 am8 am10 am12 pm2 pm4 pm6 pmTime of dayTemperature (°F)
  • Read a value: find the time, go up to the point, then across to the scale. At 10 am it was 62^\circ\text{F}.
  • Increasing / decreasing: a segment going up means an increase; going down means a decrease. Here the temperature rises until 2 pm, then falls.
  • Steepness shows speed of change. The steepest upward segment (8 am to 10 am, +9^\circ) is the fastest rise — even though the highest point is at 2 pm.
  • Average rate of change between two points = \dfrac{\text{change in } y}{\text{change in } x}. From 6 am to 2 pm: \dfrac{72 - 48}{8\text{ h}} = 3^\circ\text{F} per hour.
  • Broken axis: a zig-zag at the bottom of the axis means it doesn't start at 0. Don't judge size by bar or line height alone.
The temperature line graph (6 am 48, 8 am 53, 10 am 62, 12 pm 68, 2 pm 72, 4 pm 70, 6 pm 61 degrees F). The steepest segment, 8 am to 10 am, is red and labelled plus 9 degrees: the fastest rise. The highest point is at 2 pm. A dashed green line from 6 am (48) to 2 pm (72) shows the average rate of change, 24 degrees over 8 hours = 3 degrees F per hour. The zig-zag on the vertical axis shows it starts at 44, not 0.4448525660646872766 am8 am10 am12 pm2 pm4 pm6 pmTime of dayTemperature (°F)steepest: +9°highest 72°average 6 am → 2 pm:(72 − 48) ÷ 8 = 3°F per hour

Scatter plots and association

A scatter plot shows each pair as a dot. You look at the whole cloud of dots to see the pattern.

Three scatter plots. Graph A: points rise from lower left to upper right (positive association). Graph B: points fall from upper left to lower right (negative association). Graph C: points are scattered with no pattern (no association).Graph AxyGraph BxyGraph Cxy

Direction. Read from left to right:

  • Graph A — dots go up: positive association (height and shoe size; hours worked and pay).
  • Graph B — dots go down: negative association (car age and value; hours of exercise and resting heart rate).
  • Graph C — no direction: no association (birth month and test score).

Strength. If the dots lie close to a straight line, the association is strong. If they are spread far from any line, it is weak. Strength and direction are separate: a strong association can be negative.

Linear or not. If the dots follow a curve (for example, rising and then leveling off), the association is nonlinear.

Three scatter plots. Strong positive: dots rise from left to right and lie close to a line. Weak positive: dots still rise but are spread far from any line. Nonlinear: dots rise quickly and then level off, following a curve.Strong positiveclose to a lineWeak positivespread outNonlinearfollows a curve

Outliers. An outlier is a point far from the pattern. In the study scatter plot, (6, 58) is an outlier: other students who studied 5 to 8 hours scored 80 to 93. An outlier is not simply the highest or lowest point — the points (1, 62) and (8, 93) are extremes, but they fit the pattern.

Study scatter plot: hours studied (0 to 9) versus test score (50 to 100). Scores rise as hours increase. The point (6, 58) is circled in red as an outlier because it lies far below the pattern. The points (1, 62) and (8, 93) are labelled as extremes that still fit the pattern.50607080901000123456789Hours studiedTest scoreoutlier (6, 58)(1, 62) lowest, fits(8, 93) highest, fits

Correlation is not causation. Ice cream sales and sunburns both rise in summer: they are associated, but ice cream doesn't cause sunburn — hot, sunny weather drives both. On the GED, a statement that a scatter plot proves one thing causes another is not supported.

A tendency, not a rule. A positive association means y tends to increase. It does not mean every single point is higher than every point to its left.

Trend lines: predicting and working backwards

A trend line is a straight line through the middle of the dots, with about as many points above it as below.

Scatter plot of car age in years (0 to 10) versus value in thousands of dollars (0 to 26), with a dashed trend line from (0, 24) to (10, 4). Points: (1, 21), (2, 21), (3, 17), (4, 17), (5, 13), (6, 13), (7, 11), (8, 7), (9, 7), (10, 3).02468101214161820222426012345678910Age of car (years)Value (thousands of $)

Find its equation. Pick two points on the line that cross gridlines exactly — here (0, 24) and (10, 4).

  • Slope: m = \dfrac{4 - 24}{10 - 0} = -2.
  • y-intercept: the line crosses the y-axis at 24, so b = 24.
  • Equation: y = -2x + 24 (value in thousands of dollars, age in years).

Interpret the slope and intercept with units.

  • Slope -2: the car loses about $2,000 per year.
  • Intercept 24: a brand-new car (age 0) is worth about $24,000.

Predict (x known, find y): age 7 → y = -2(7) + 24 = 10 → $10,000. You can also read the graph: go up from 7 to the line, then across.

Work backwards (y known, find x): worth $9,000 → 9 = -2x + 24 → -15 = -2x → x = 7.5 years.

Car trend line y = −2x + 24 through (0, 24) and (10, 4). Predict: from age 7 go up to the line and across to a value of 10 (10,000 dollars), shown with red arrows. Work backwards: from a value of 9 go across to the line and down to age 7.5, shown with green arrows.0246810121416182022242601234567891011Age of car (years)Value (thousands of $)(0, 24)(10, 4)predict: age 7 → $10,000backwards: $9,000 → 7.5 yry = −2x + 24

Interpolation vs. extrapolation. Predicting inside the range of the data (between 1 and 10 years) is fairly reliable. Predicting far outside it is risky: the line says a 12-year-old car is worth -2(12) + 24 = 0 dollars, and a 15-year-old car -\$6{,}000 — impossible. Trend lines only describe the data near where they were measured.

Interpolation versus extrapolation for the car trend line y = −2x + 24. The data cover ages 1 to 10 (shaded band); predictions there are fairly reliable. Extended past the data (dashed), the line gives 0 dollars at age 12 and −6,000 dollars at age 15, which is impossible.-8-404812162024012345678910111213141516Age of car (years)Value (thousands of $)age 12: $0age 15: −$6,000?data range (interpolate)extrapolate: risky
Formula

Formulas

m = \frac{y_2 - y_1}{x_2 - x_1} y = mx + b \text{Average rate of change} = \frac{\text{change in } y}{\text{change in } x}
  • (x_1, y_1) and (x_2, y_2) — two points (use points on the trend line or on the line graph)
  • m — slope, the rate of change of y per one unit of x (with units like $ per year or °F per hour)
  • b — the y-intercept, the value of y when x = 0

The slope formula and slope-intercept form y = mx + b ARE on the GED formula sheet. The "average rate of change" is the same as the slope formula, so you don't need to memorize anything extra — but you do need to know when to use it.

Worked example

Example 1 (easy): Read a table

Using Dana's pay table ($62 for 4 hours, $93 for 6, $139.50 for 9, $186 for 12), how much would she earn for 20 hours?

  1. Rate: 62 \div 4 = \$15.50 per hour.
  2. Pay for 20 hours: 15.50 \times 20 = 310.
Example 1. Dana's pay line extended to 20 hours. The table points (4, 62), (6, 93), (9, 139.50), (12, 186) are blue; the predicted point (20, 310) is red: 15.50 × 20 = 310 dollars.0501001502002503003500246810121416182022Hours workedPay in dollars(20, 310)$15.50 × 20 = $310

Answer: $310.

Check: 20 hours = 12 + 4 + 4 hours, so pay = 186 + 62 + 62 = \$310 ✓

Worked example

Example 2 (medium): Describe a scatter plot — real world

A gym records each member's hours of exercise per week (x) and resting heart rate (y). The points fall from the upper left to the lower right and lie close to a line. Describe the association.

  1. Direction: as x increases, y tends to decrease → negative.
  2. Strength: the points lie close to a line → strong.
  3. Form: a straight pattern → linear.
Example 2. An illustrative scatter plot of hours of exercise per week (0 to 10) versus resting heart rate in beats per minute (50 to 85). The points fall from the upper left to the lower right and lie close to a dashed trend line: a strong, negative, linear association.5055606570758085012345678910Exercise (hours per week)Resting heart rate (bpm)strong, negative, linear

Answer: a strong, negative, linear association — people who exercise more tend to have lower resting heart rates.

Check: Pick two points: a member with more hours should usually have a lower heart rate. Note: this does not prove exercise causes the lower rate; it shows a tendency ✓

Worked example

Example 3 (medium): Rate from a line graph

Using the temperature line graph, what was the average rate of change from 2 pm to 6 pm?

  1. 2 pm: 72^\circ\text{F}. 6 pm: 61^\circ\text{F}.
  2. Change in temperature: 61 - 72 = -11^\circ\text{F}.
  3. Change in time: 2 pm to 6 pm = 4 hours.
  4. Rate: \dfrac{-11}{4} = -2.75^\circ\text{F} per hour.
Example 3. The temperature line graph with the part from 2 pm (72 degrees F) to 6 pm (61 degrees F) highlighted in red. A triangle shows the change in time, 4 hours, and the change in temperature, −11 degrees, so the average rate is −11 ÷ 4 = −2.75 degrees F per hour. The zig-zag on the vertical axis shows it starts at 44, not 0.4448525660646872766 am8 am10 am12 pm2 pm4 pm6 pmTime of dayTemperature (°F)−11°4 hours72°61°rate = −11 ÷ 4 = −2.75°F per hour

Answer: -2.75°F per hour (falling about 2.75 degrees each hour).

Check: 72 - 2.75 \times 4 = 72 - 11 = 61 ✓

Worked example

Example 4 (GED-hard): Trend-line prediction — money

Using the car trend line y = -2x + 24 (y in thousands of dollars), how much more is a 3-year-old car predicted to be worth than a 9-year-old car?

  1. Age 3: y = -2(3) + 24 = 18 → $18,000.
  2. Age 9: y = -2(9) + 24 = 6 → $6,000.
  3. Difference: 18{,}000 - 6{,}000 = 12{,}000.
Example 4. The trend line y = −2x + 24. At age 3 the line gives 18 (18,000 dollars); at age 9 it gives 6 (6,000 dollars). A red bracket shows the difference: 18 − 6 = 12, so 12,000 dollars. The ages differ by 6 years at 2,000 dollars per year.0246810121416182022242601234567891011Age of car (years)Value (thousands of $)(3, 18)(9, 6)12y = −2x + 24

Answer: $12,000 more.

Check (shortcut): the ages differ by 6 years, and the car loses $2,000 per year: 6 \times 2{,}000 = \$12{,}000 ✓

Worked example

Example 5 (GED-hard): Work backwards from a table

A phone plan charges a fixed monthly fee plus a cost per gigabyte (GB) of data. Two bills: 3 GB cost $41; 8 GB cost $61. What is the fixed fee, and how many GB would make a $73 bill?

  1. Rate: \dfrac{61 - 41}{8 - 3} = \dfrac{20}{5} = \$4 per GB.
  2. Fixed fee: 41 = 4(3) + b → b = 41 - 12 = 29. So y = 4x + 29.
  3. Work backwards for $73: 73 = 4x + 29 → 4x = 44 → x = 11.
Example 5. Phone plan line y = 4x + 29, data in GB on the x-axis (0 to 12) and bill in dollars on the y-axis (0 to 80). The line starts at the fixed fee, 29 dollars, on the y-axis. The bills (3, 41) and (8, 61) are marked; a slope triangle between them shows a run of 5 GB and a rise of 20 dollars, so 4 dollars per GB. The point (11, 73) shows that a 73 dollar bill means 11 GB.010203040506070800123456789101112Data used (GB)Monthly bill ($)+5 GB+$20(3, 41)(8, 61)fixed fee $29(11, 73)y = 4x + 29

Answer: fixed fee $29; a $73 bill means 11 GB.

Check: 4(8) + 29 = 32 + 29 = 61 ✓ and 4(11) + 29 = 44 + 29 = 73 ✓

Calculator tip

Using the TI-30XS MultiView

Reading graphs is visual, but the calculator speeds up rates and predictions:

  • Slope / rate: use parentheses so the subtraction happens first: ( 4 − 24 ) ÷ ( 10 − 0 ) enter → -2.
  • Negative numbers: use the (−) key, not the subtraction key, to type -2 at the start of an expression: (−) 2 × 7 + 24 enter → 10.
  • Fraction results: if you get -11/4, press toggle to see -2.75.

When not to use it: describing an association, spotting an outlier, and deciding whether a statement is supported are reading and reasoning tasks. The calculator can't help, so spend your time looking carefully at the graph.

Common mistake

Common mistakes

  • Correlation as causation. Wrong: "the graph proves studying causes higher scores." Right: it shows the two are associated; other factors may play a part.
  • Using a data point instead of the trend line. Wrong: the 7-year-old car in the data is worth $11,000, so the prediction is $11,000. Right: the trend line predicts $10,000.
  • Forgetting to divide by the change in x. Wrong: the plant grows 3 inches per week. Right: 3 inches per 2 weeks = 1.5 inches per week.
  • Highest point vs. steepest segment. Wrong: the temperature rose fastest just before 2 pm because 2 pm is the highest. Right: the steepest rise is 8 am–10 am (+9^\circ).
  • Sign errors with negative slope. Wrong: y = 2x + 24 for a falling line. Right: a line that falls left to right has a negative slope.
  • Ignoring axis units and breaks. If the axis says thousands of dollars, 10 means $10,000; if it has a zig-zag, it doesn't start at zero.
Test strategy

GED test strategy

  • Label first. Before reading points, check what x and y stand for and their units.
  • Estimate the direction of a scatter plot in two seconds: up, down, or neither. Eliminate answer choices with the wrong sign or direction.
  • Use points on gridlines. For a trend line's slope, choose two points where the line crosses gridlines exactly — never guess values between gridlines.
  • Plug back in. For an equation or a work-backwards answer, substitute your answer to make sure it gives the right value.
  • Read the question twice for words like "based on the trend line," "increase," or "consecutive." They decide which number is right.
  • Watch for causation words ("proves," "causes"); these statements are usually not supported.
  • Time: most two-variable questions take 1–2 minutes. Write the equation of a trend line once and reuse it for several questions about the same graph.

Check yourself

Scatter plots and two-variable dataHot spot

The scatter plot shows the number of hours 11 students studied for a test and the score each student earned.

Click the point that is an outlier — the one that does not fit the pattern of the other data.

Click on the figure to mark your answer (or focus it and use the arrow keys).

Scatter plot of hours studied (0 to 9) versus test score (50 to 100) for 11 students. Points: (1, 62), (2, 65), (2, 70), (3, 72), (4, 74), (4, 79), (5, 80), (6, 85), (6, 58), (7, 88), (8, 93). Scores generally rise as hours increase; the point (6, 58) lies far below the others.505560657075808590951000123456789Hours studiedTest score
Hint

First describe the overall pattern: as hours go up, what happens to scores? Then look for the point that breaks that pattern.

Check yourself

Scatter plots and two-variable dataFill in

The line graph shows the outdoor temperature at different times on one day.

Line graph of temperature in degrees Fahrenheit during one day: 6 am 48, 8 am 53, 10 am 62, 12 pm 68, 2 pm 72, 4 pm 70, 6 pm 61. Vertical axis from 44 to 76 in steps of 4.4448525660646872766 am8 am10 am12 pm2 pm4 pm6 pmTime of dayTemperature (°F)

By how many degrees Fahrenheit did the temperature rise from 6 am to 12 pm?

°F

Type your answer in the box. Fractions like 3/4 are fine.

Hint

Read the temperature at each of the two times, then find the difference. Notice that the vertical axis does not start at 0.

Summary

  • Tables, scatter plots, and line graphs all show pairs (x, y); always read the labels and units first.
  • Rate of change = \dfrac{\text{change in } y}{\text{change in } x} — the slope — with units like $ per hour.
  • Scatter plots show positive, negative, or no association; close to a line = strong.
  • An outlier is far from the pattern, not just the biggest or smallest point.
  • Use the trend line y = mx + b to predict (x \to y) and to work backwards (y \to x).
  • Correlation does not prove causation.
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