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Probability and Statistics in Science

Find and choose measures of center and spread, use Punnett squares and probability, and judge samples and error bars.

Practice this skill 18 min Probability and statistics in science Leer en español
Key idea

What you'll learn

Living things vary, measurements are never perfect, and many natural events happen by chance. Scientists use statistics to summarize data and probability to predict outcomes. The GED calls this skill probability and statistics in science.

In this lesson you will learn to:

  • find and choose between the mean, median and mode, and measure spread with the range;
  • recognize outliers and how they affect each measure;
  • calculate probability, including from a Punnett square, and turn a probability into an expected number;
  • judge samples: random vs. biased, small vs. large;
  • read error bars and decide whether a difference between groups is convincing.

How it shows up on the GED: You will see tables of repeated trials, Punnett squares, survey data and bar graphs with error bars. Questions ask you to calculate a mean or median (often fill-in-the-blank), give the probability of an offspring's trait, estimate a population from a sample, explain why observed results differ from predictions, or choose the conclusion best supported by uncertain data.

Words to know

Key terms

  • mean (average) — the sum of the values divided by how many there are (112 ÷ 8 = 14).
  • median — the middle value when the data are in order (8, 11, 12, 12, 15, 16, 18, 20 → 13.5).
  • mode — the value that occurs most often (12 appears twice).
  • range — largest value minus smallest value; a measure of spread (20 − 8 = 12).
  • outlier — a value far from the rest of the data (0.92 s among times near 0.30 s).
  • probability — how likely an event is, from 0 (impossible) to 1 (certain); written as a fraction, decimal or percent (\tfrac{1}{4} = 0.25 = 25%).
  • allele — one version of a gene (T for tall, t for short).
  • dominant / recessive — a dominant allele shows its trait with just one copy; a recessive allele shows only with two copies.
  • genotype / phenotype — the allele letters (Tt) vs. the visible trait (tall).
  • homozygous / heterozygous — two identical alleles (TT, tt) vs. two different alleles (Tt).
  • Punnett square — a grid that shows all equally likely allele combinations from a cross.
  • expected value — probability × number of trials (\tfrac{3}{4} \times 200 = 150).
  • sample — the part of a population that is actually measured.
  • random sample — every member has an equal chance of being chosen.
  • bias — a systematic error that pushes results in one direction.
  • error bar — a line on a graph showing the uncertainty in a value.

All Science words in the glossary

1. Mean, median and mode

These three numbers each describe the center of a data set.

Mean. Add all the values; divide by how many values there are. Seedling heights 15, 12, 8, 20, 12, 16, 11, 18: sum = 112, count = 8, mean = 14 cm.

Median. Sort the values, then find the middle.

  • Odd number of values: the median is the single middle value. (0.29, 0.30, 0.31, 0.33, 0.92 → 0.31)
  • Even number of values: the median is the mean of the two middle values. (8, 11, 12, 12, 15, 16, 18, 20 → \dfrac{12 + 15}{2} = 13.5)

Mode. The most frequent value. A data set can have one mode, more than one, or none (if every value appears once). In the seedling data, 12 appears twice, so the mode is 12 cm.

Dot plot of 8 seedling heights in centimeters: 8, 11, 12, 12, 15, 16, 18, 20, with two dots stacked at 12. The mode, 12, is the tallest stack. The median, 13.5, is halfway between the two middle values 12 and 15. The mean, 14, is the sum 112 divided by 8. 6810121416182022mode 12median 13.5mean 14Seedling height (cm)Sorted: 8, 11, 12, 12, 15, 16, 18, 20 (sum 112, count 8)

Which measure should you report?

SituationBest measureWhy
Values are fairly even, no outliersMeanUses every value
There is an outlier or the data are skewedMedianNot pulled by extreme values
You want the most common category or valueModeWorks even for non-numbers (eye color)

2. Spread, outliers and repeated trials

Two data sets can have the same mean but very different spreads. The range (maximum − minimum) is the simplest measure of spread.

Outliers. An outlier is a value far from the others. It might come from a mistake (misread instrument, distraction) or from a real but rare event. Outliers affect the mean and the range strongly, but barely affect the median.

Reaction times (s)MeanMedianRange
0.31, 0.29, 0.33, 0.300.30750.3050.04
0.31, 0.29, 0.33, 0.30, 0.920.430.310.63

Adding one outlier raised the mean by about 0.12 s, but the median changed by only 0.005 s.

Dot plot of five reaction times: 0.29, 0.30, 0.31 and 0.33 seconds clustered together, and an outlier at 0.92 seconds. The median, 0.31 seconds, stays with the cluster. The mean, 0.43 seconds, is pulled up toward the outlier. 0.30.40.50.60.70.80.9outlier 0.92median 0.31mean 0.43the outlier pulls the mean upReaction time (s)

Why scientists repeat trials. Every measurement has some random error. Repeating trials and averaging reduces the effect of that error, and it reveals outliers. A result found in many trials is more reliable than a result from one trial. A small range across trials means the measurements are precise (close to each other).

3. Probability and Punnett squares

Probability of an event = \dfrac{\text{number of favorable outcomes}}{\text{total number of equally likely outcomes}}.

A fair coin: P(heads) = \tfrac{1}{2}. A six-sided die: P(rolling a 4) = \tfrac{1}{6}.

Punnett squares. Each parent passes one allele to each offspring, chosen by chance. A Punnett square lists all combinations. Each box is equally likely, so each box is \tfrac{1}{4} (25%) in a 2 × 2 square.

CrossBoxesGenotype ratioPhenotype ratio (dominant : recessive)
Tt \times TtTT, Tt, Tt, tt1 TT : 2 Tt : 1 tt3 : 1 (75% : 25%)
Tt \times ttTt, Tt, tt, tt2 Tt : 2 tt1 : 1 (50% : 50%)
TT \times ttTt, Tt, Tt, Ttall Ttall dominant (100%)
TT \times TtTT, TT, Tt, Tt2 TT : 2 Ttall dominant (100%)

Read the question carefully: is it asking about a genotype (TT, Tt, tt) or a phenotype (tall, short)? For Tt \times Tt: P(Tt) = 50%, but P(tall) = 75%.

Expected numbers. Multiply the probability by the number of offspring. Out of 200 offspring of Tt \times Tt: tall = \tfrac{3}{4} \times 200 = 150; short = \tfrac{1}{4} \times 200 = 50.

Observed vs. expected. Real counts rarely match exactly. 140 tall and 60 short is close to 150 and 50; the difference comes from chance. Larger samples usually come closer to the predicted ratio — the same way 1,000 coin flips come closer to half heads than 10 flips do.

Bar graph for 200 offspring of a Tt by Tt cross. Tall: expected 150 (three-fourths of 200), observed 140. Short: expected 50 (one-fourth of 200), observed 60. Chance makes observed counts differ a little from expected. 0255075100125150175Number of offspring150140tall5060shortexpectedobservedExpected:¾ × 200 = 150¼ × 200 = 50Chance makesobserved countsdiffer a little.

4. Sampling, bias and error bars

Scientists rarely measure every member of a population. They measure a sample and use it to estimate the whole.

Estimating a population. Find the mean per unit (per m², per liter, per hour), then multiply by the total. Example: 10 random 1 m² quadrats contain 40 milkweed plants → mean 4 per m². Field = 500 m². Estimate: 4 \times 500 = 2{,}000 plants.

A 25 meter by 20 meter field, 500 square meters, with 10 randomly placed 1 square meter quadrats. The plant counts in the quadrats total 40, a mean of 4 plants per square meter. Estimate for the field: 4 times 500 equals 2,000 plants. Field: 25 m × 20 m = 500 m²435462453410 random 1 m²quadratsplants counted:4+3+5+4+6+2+4+5+3+4= 40mean = 40 ÷ 10= 4 per m²estimate:4 × 500 = 2,000

Good samples are:

  • Random — every part of the population has an equal chance of being chosen. This prevents bias.
  • Large enough — bigger samples reduce the effect of chance (sampling error).

Bias pushes results in one direction. Sampling only the thickest patch of plants overestimates; surveying only people at a gym about exercise habits overestimates how much people exercise. Increasing the sample size does not fix bias — only better sampling methods do.

Error bars. On a graph, an error bar shows the uncertainty in a mean.

  • A short error bar: results were consistent; the mean is fairly certain.
  • A long error bar: results varied a lot; the mean is less certain.
  • Overlapping error bars: the difference between the groups may be due to chance.
  • Non-overlapping error bars: the difference is more likely real.
Formula

Formulas

\text{mean} = \bar{x} = \dfrac{\text{sum of all values}}{\text{number of values}} \text{range} = \text{maximum} - \text{minimum} \text{median} = \text{middle value of the sorted data (mean of the two middle values if the count is even)} P(\text{event}) = \dfrac{\text{favorable outcomes}}{\text{total equally likely outcomes}} \text{expected number} = P(\text{event}) \times \text{number of trials} \text{population estimate} = \text{mean per sample unit} \times \text{total number of units}

The mode has no formula: it is simply the most frequent value.

Worked example

Worked example 1: Choosing the right measure of center

Stimulus. Five students record how long it takes an ice cube to melt in a cup of water: 182 s, 175 s, 190 s, 178 s and 410 s. The student with 410 s notes that her water had just come from the refrigerator and was much colder than the others'.

Question. Which value best represents the typical melting time?

A. Mean, 227 s B. Median, 182 s C. Range, 235 s D. 410 s

Think-aloud.

  1. Mean: \dfrac{182 + 175 + 190 + 178 + 410}{5} = \dfrac{1135}{5} = 227 s.
  2. Sorted: 175, 178, 182, 190, 410 → median 182 s.
  3. 410 s is an outlier with an explanation (colder water). It pulls the mean to 227, higher than all four normal trials.
  4. The median, 182 s, sits among the typical values. ✓ C is a spread, not a center. D is the outlier.
Dot plot of ice cube melting times: 175, 178, 182 and 190 seconds clustered together, and an outlier at 410 seconds from colder water. The median, 182 seconds, sits among the typical values. The mean, 1135 divided by 5, is 227 seconds, higher than all four normal trials. 150180210240270300330360390420outlier 410 s(colder water)median 182 smean 227 sMelting time (s)Mean = 1135 ÷ 5 = 227 s: higher than all four normal trials

Answer: B.

Worked example

Worked example 2: A Punnett square probability

Stimulus. In rabbits, black fur (B) is dominant over brown fur (b). A heterozygous black rabbit (Bb) is crossed with a brown rabbit (bb).

Question. What percent of the offspring are expected to have brown fur? Out of 12 offspring, how many are expected to be brown?

Think-aloud.

  1. Parent 1 gives B or b; parent 2 gives b or b.
  2. Boxes: Bb, Bb, bb, bb.
  3. Brown requires bb: 2 of 4 boxes → \tfrac{2}{4} = 50\%.
  4. Expected brown out of 12: 0.50 \times 12 = 6.
  5. Trap check: "25%" comes from assuming every cross gives 3 : 1. That is only true for Bb \times Bb.
Punnett square for a heterozygous black rabbit, Bb, crossed with a brown rabbit, bb. Parent 1 gives B or b; parent 2 gives b or b. The four boxes are Bb, Bb, bb, bb. Black Bb is 2 of 4 boxes and brown bb is 2 of 4, so the probability of brown is 50 percent, and 50 percent of 12 offspring is 6 brown. Parent 1: Bb (black)Parent 2: bb (brown)BbbbBbbbBbbbBb = black: 2 of 4bb = brown: 2 of 4P(brown) = 2/4 = 50%50% × 12 = 6 brown

Answer: 50%; about 6 brown offspring.

Worked example

Worked example 3: Spotting a biased sample

Stimulus. A researcher wants to know what fraction of fish in a large lake carry a certain parasite. She catches 50 fish, all from a shallow, warm cove near a sewage outflow, and finds that 40% carry the parasite.

Question. Which statement best evaluates her estimate?

A. The estimate is reliable because 50 is a large sample. B. The estimate may be biased because all the fish came from one unusual part of the lake. C. Exactly 40% of the fish in the lake carry the parasite. D. She should have caught only 10 fish to save time.

Think-aloud.

  1. Is the sample random? No — all from one warm cove near sewage, where parasites may be more common.
  2. So the sample may not represent the whole lake → bias. ✓ B
  3. A — sample size does not fix bias. C — an estimate is never "exactly" certain, and this one may be biased. D — a smaller sample would make the estimate less reliable.
Two maps of the same lake. Left: all of the researcher’s fish came from one shallow, warm cove near a sewage outflow, so the sample may be biased and not representative. Right: a random sample spreads catches across the whole lake, so every part has an equal chance of being chosen. Her sample: one warm coveRandom sample: whole lakesewage outflowMay be biased: not representativeEvery part has an equal chance

Answer: B.

Worked example

Worked example 4: Reading error bars

Stimulus. A study compares the mean resting heart rate of two groups. Group 1 (daily walkers): 68 beats per minute, error bar 66–70. Group 2 (non-walkers): 72 beats per minute, error bar 69–75.

Question. Which conclusion is best supported?

A. Daily walking definitely lowers heart rate by exactly 4 beats per minute. B. The error bars overlap, so the data do not clearly show a difference between the groups. C. Non-walkers have lower heart rates. D. The study proves walking has no effect.

Think-aloud.

  1. Group 1's range is 66–70; Group 2's is 69–75. They overlap between 69 and 70.
  2. Overlap means the 4-beat difference in means could be due to chance. ✓ B
  3. A claims certainty. C reverses the data. D — "no clear difference" is not the same as "proves no effect."
Means with error bars for resting heart rate. Group 1, daily walkers: mean 68 beats per minute, error bar from 66 to 70. Group 2, non-walkers: mean 72, error bar from 69 to 75. The error bars overlap between 69 and 70, so the difference may be due to chance. 6062646668707274767880Resting heart rate (beats/min)mean 687066Group 1: walkersmean 727569Group 2: non-walkersoverlap 69–70

Answer: B.

Calculator tip

Calculator tip (TI-30XS MultiView)

  • Mean: type the whole sum in parentheses, then divide: ( 15 + 12 + 8 + 20 + 12 + 16 + 11 + 18 ) ÷ 8 enter → 14. Without the parentheses, the calculator would divide only the last number by 8.
  • Expected number: 3 ÷ 4 × 200 enter → 150.
  • Percent to decimal: divide by 100 (25% = 0.25) before multiplying.
  • The median, mode and range are usually faster by hand: sort the values on your erasable note board first.
Common mistake

Common mistakes

  • Finding the median without sorting. Wrong: the middle of 15, 12, 8, 20, 12, 16, 11, 18 is "20 and 12 → 16." Right: sort first: 8, 11, 12, 12, 15, 16, 18, 20 → median 13.5.
  • Mixing up mean, median and mode. Wrong: reporting the most frequent value as the "average." Right: mean = sum ÷ count; median = middle; mode = most frequent.
  • Using the mean when there is an outlier. Wrong: "Typical reaction time is 0.43 s" when four of five trials are near 0.30 s. Right: the median (0.31 s) better represents typical data with an outlier.
  • Confusing genotype and phenotype probabilities. Wrong: P(Tt) = 75% for Tt \times Tt. Right: P(Tt) = 50%; P(tall) = 75%.
  • Expecting exact ratios. Wrong: "140 tall instead of 150 means the Punnett square is wrong." Right: chance causes variation; larger samples come closer to the prediction.
  • Choosing the highest bar regardless of error bars. Wrong: "Fertilizer C is best because its bar is tallest." Right: check whether its error bar overlaps the control's.
Test strategy

GED strategy for probability and statistics

  1. Identify the measure asked for. Underline mean, median, mode, range, probability or expected number in the question; the wrong answers are often the other measures.
  2. Sort before you find a median. Write the sorted list on your erasable note board and cross off values from both ends.
  3. Estimate first. A mean must fall between the smallest and largest values. A probability must be between 0 and 1 (0% and 100%).
  4. Look for outliers and notes. If a passage mentions something unusual in one trial, expect a question about how it affects the mean or the conclusion.
  5. Judge the sample: random? large enough? from the right population?
  6. With error bars, check overlap before trusting a difference. Prefer cautious conclusions ("may," "more trials needed") over "proves" or "certainly."

Check yourself

Probability and statistics in scienceFill in
Heights of Sunflower Seedlings

A student planted eight sunflower seeds from the same seed packet in identical pots. Each pot had the same amount of the same potting soil, and all the pots sat side by side on a sunny windowsill. She gave every pot the same amount of water each morning. After 21 days she measured the height of each seedling, from the soil surface to the tip of the top leaf, to the nearest centimeter. Even under identical conditions, the seedlings grew to different heights.

Plant12345678
Height (cm)151282012161118

What is the mean height of the eight seedlings?

cm

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

Hint

Add all eight heights, then divide by the number of plants.

Check yourself

Probability and statistics in science
Heights of Sunflower Seedlings

A student planted eight sunflower seeds from the same seed packet in identical pots. Each pot had the same amount of the same potting soil, and all the pots sat side by side on a sunny windowsill. She gave every pot the same amount of water each morning. After 21 days she measured the height of each seedling, from the soil surface to the tip of the top leaf, to the nearest centimeter. Even under identical conditions, the seedlings grew to different heights.

Plant12345678
Height (cm)151282012161118

What is the median height of the seedlings?

Answer choices
Hint

Put the heights in order first. With an even number of values, the median is halfway between the two middle values.

Summary

  • Mean = sum ÷ count; median = middle of sorted data; mode = most frequent; range = max − min.
  • Outliers pull the mean and range but barely move the median — use the median when an outlier is present.
  • Probability = favorable ÷ total equally likely outcomes; in a 2 × 2 Punnett square each box is 25%. Tt \times Tt gives 3 : 1 dominant : recessive.
  • Expected number = probability × number of trials; observed counts vary by chance, less so in large samples.
  • Good samples are random (no bias) and large (less chance error).
  • Overlapping error bars mean a difference may be due to chance; non-overlapping bars suggest a real difference.
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