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.
Plant
1
2
3
4
5
6
7
8
Height (cm)
15
12
8
20
12
16
11
18
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.
Step-by-step solution
Where the answer comes from: the table lists the eight heights: 15, 12, 8, 20, 12, 16, 11 and 18 cm.
\bar{x} = \dfrac{\text{sum of values}}{\text{number of values}} = \dfrac{112\text{ cm}}{8} = 14\text{ cm}
Correct answer: 14 cm.
Check: a mean should fall between the smallest (8 cm) and largest (20 cm) values. 14 cm does. ✓
Tip: Divide by the number of values (8), not by the largest value or the number of different values.
Question 2 of 36
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.
Plant
1
2
3
4
5
6
7
8
Height (cm)
15
12
8
20
12
16
11
18
What is the median height of the seedlings?
Hint
Put the heights in order first. With an even number of values, the median is halfway between the two middle values.
Step-by-step solution
Where the answer comes from: the table's eight heights: 15, 12, 8, 20, 12, 16, 11, 18 cm.
Step by step:
Order them from least to greatest: 8, 11, 12, 12, 15, 16, 18, 20.
There are 8 values (an even number), so there are two middle values: the 4th (12) and the 5th (15).
The median is their mean: \dfrac{12 + 15}{2} = \dfrac{27}{2} = 13.5 cm.
Correct answer: B (13.5 cm).
Why the others are wrong: 12 cm is the mode (it appears twice). 14 cm is the mean. 16 cm comes from averaging the two middle values of the unsorted list (20 and 12).
Tip: Always sort the data before finding the median. On the GED, one wrong choice is almost always the "middle of the unsorted list."
Question 3 of 36
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.
Plant
1
2
3
4
5
6
7
8
Height (cm)
15
12
8
20
12
16
11
18
What is the range of the seedling heights?
Hint
Range measures spread: largest value minus smallest value.
Step-by-step solution
Where the answer comes from: in the table, the tallest seedling is Plant 4 at 20 cm and the shortest is Plant 3 at 8 cm.
Step by step:
Find the maximum: 20 cm.
Find the minimum: 8 cm.
Subtract: \text{range} = 20 - 8 = 12 cm.
Correct answer: C (12 cm).
Why the others are wrong: 8 cm is the minimum and 20 cm is the maximum; neither alone measures spread. 28 cm adds the two instead of subtracting.
Tip: Mean, median and mode describe the center of a data set. The range describes the spread — how far apart the values are. A big range means the plants grew very differently.
Question 4 of 36
Fill inSeptember Rainfall Records
A farmer in a coastal county is deciding when to harvest. She looks up the total rainfall recorded in September at the nearest weather station for the last seven years.
Year
September rainfall (mm)
Year 1
92
Year 2
105
Year 3
88
Year 4
310
Year 5
97
Year 6
101
Year 7
95
The station's notes say that in Year 4 a hurricane stalled over the county for two days and dropped most of that month's rain.
What is the mean September rainfall for the seven years? Round to the nearest tenth.
mm
Type your answer in the box. Fractions like 3/4 are fine.
Hint
Add all seven values, then divide by how many years there are.
Step-by-step solution
Where the answer comes from: the table lists 92, 105, 88, 310, 97, 101 and 95 mm for the seven Septembers.
Step by step:
Add the values: 92 + 105 + 88 + 310 + 97 + 101 + 95 = 888 mm.
Count the values: 7 years.
Divide: \text{mean} = \dfrac{888}{7} \approx 126.86 mm.
Round to the nearest tenth: 126.9 mm.
Correct answer: 126.9 mm.
Check:126.9 \times 7 \approx 888 ✓. Notice that six of the seven years had less than 110 mm. The mean is pulled far above most of the data by the 310 mm hurricane year.
Tip: The mean uses every value, so one extreme value can move it a lot. After you compute a mean, compare it with the data: if most values sit on one side of it, look for an outlier.
Question 5 of 36
Robin Clutch Sizes
A clutch is the set of eggs a bird lays in one nesting attempt. A wildlife biologist surveyed 20 American robin nests in a city park in May and counted the eggs in each.
Eggs in nest
Number of nests
2
5
3
6
4
7
5
2
Each nest was counted once. The biologist plans to compare these results with nests in a rural area next year.
What is the median number of eggs per nest in the survey?
Hint
List the 20 nests in order by egg count (or count down the table) and find the 10th and 11th nests.
Step-by-step solution
Where the answer comes from: the table shows 5 nests with 2 eggs, 6 with 3 eggs, 7 with 4 eggs and 2 with 5 eggs: 20 nests in all.
Step by step:
With 20 values, the median is the average of the 10th and 11th values in order.
Count down the table: nests 1–5 have 2 eggs; nests 6–11 have 3 eggs; nests 12–18 have 4 eggs; nests 19–20 have 5 eggs.
The 10th and 11th nests both have 3 eggs.
Median = \dfrac{3 + 3}{2} = 3 eggs.
Correct answer: A (3 eggs).
Why the others are wrong: 3.5 is the middle of the egg values 2, 3, 4, 5, ignoring how many nests have each value. 4 is the mode (most common value, 7 nests). 5.5 is the median of the "number of nests" column, which counts nests, not eggs.
Tip: In a frequency table, each row stands for several data points. Count through the frequencies to find the middle position before reading off the value.
Question 6 of 36
Snapdragon Flower Color
In snapdragon plants, flower color shows incomplete dominance: neither allele is dominant, and a plant with one of each allele has a blend of the two colors.
Genotype
Flower color
RR
red
RW
pink
WW
white
A gardener crossed two pink snapdragons (RW \times RW). Each parent passes on one allele, R or W, with equal chance.
R
W
R
RR
RW
W
RW
WW
The gardener planted 120 seeds from this cross and recorded the flower color of every plant that bloomed.
Based on the Punnett square, what is the probability that a plant from this cross will have pink flowers?
Hint
Count the boxes in the Punnett square that show the pink genotype.
Step-by-step solution
Where the answer comes from: the table of genotypes says RW is pink, and the Punnett square shows RR, RW, RW and WW, each equally likely.
Step by step:
Count all boxes: 4.
Count boxes with RW (pink): 2.
Probability = \dfrac{2}{4} = \dfrac{1}{2}, or 50%.
Correct answer: C (\tfrac{1}{2}).
Why the others are wrong:\tfrac{1}{4} is the chance of only one box, such as RR (red) or WW (white). \tfrac{3}{4} applies the familiar 3 : 1 rule for complete dominance, but snapdragon color shows incomplete dominance, so RR is red, not pink. \tfrac{1}{3} treats the three colors as equally likely, but red, pink and white are not equally common.
Check: red \tfrac{1}{4} + pink \tfrac{1}{2} + white \tfrac{1}{4} = 1 ✓
Tip: Always read how the genotypes map to traits in the passage. The 3 : 1 rule works only when one allele is completely dominant.
Question 7 of 36
A Pea Plant Cross
In pea plants, the allele for tall stems (T) is dominant over the allele for short stems (t). A plant with at least one T allele is tall. Only a plant with two t alleles (tt) is short.
A gardener crossed two tall pea plants that were both Tt. The Punnett square shows the possible offspring. Each of the four boxes is equally likely.
The gardener then grew 200 offspring plants from this cross and counted 140 tall plants and 60 short plants.
What is the probability that an offspring from this cross is heterozygous (Tt)?
Hint
Count how many of the four equally likely boxes contain one T and one t.
Step-by-step solution
Where the answer comes from: the Punnett square's four boxes are TT, Tt, Tt and tt, and the passage says "each of the four boxes is equally likely."
Why the others are wrong: 25% is the chance of one particular box, such as TT or tt. 75% is the chance of a tall plant (TT + Tt + Tt). 100% would mean every offspring is Tt, which happens only in a TT \times tt cross.
Tip: Distinguish genotype (the letters, like Tt) from phenotype (the trait you see, like tall). Many wrong answers answer the phenotype question when the genotype was asked, or vice versa.
Question 8 of 36
Fill inA Pea Plant Cross
In pea plants, the allele for tall stems (T) is dominant over the allele for short stems (t). A plant with at least one T allele is tall. Only a plant with two t alleles (tt) is short.
A gardener crossed two tall pea plants that were both Tt. The Punnett square shows the possible offspring. Each of the four boxes is equally likely.
The gardener then grew 200 offspring plants from this cross and counted 140 tall plants and 60 short plants.
Based on the Punnett square, how many of the 200 offspring would be expected to be tall?
plants
Type your answer in the box. Fractions like 3/4 are fine.
Hint
First find the probability of a tall plant from the Punnett square, then multiply by 200.
Step-by-step solution
Where the answer comes from: the Punnett square shows TT, Tt, Tt and tt. The passage says "a plant with at least one T allele is tall," and that the gardener grew 200 offspring.
Step by step:
Tall boxes: TT, Tt, Tt → 3 of 4.
Probability of tall: \dfrac{3}{4} = 0.75.
Expected number = probability \times total: 0.75 \times 200 = 150 plants.
Correct answer: 150 plants.
Check: expected short plants = \tfrac{1}{4} \times 200 = 50, and 150 + 50 = 200 ✓
Tip: Expected number = probability × number of trials. The Punnett square gives the probability; the sample size turns it into a count.
Question 9 of 36
A Pea Plant Cross
In pea plants, the allele for tall stems (T) is dominant over the allele for short stems (t). A plant with at least one T allele is tall. Only a plant with two t alleles (tt) is short.
A gardener crossed two tall pea plants that were both Tt. The Punnett square shows the possible offspring. Each of the four boxes is equally likely.
The gardener then grew 200 offspring plants from this cross and counted 140 tall plants and 60 short plants.
The gardener expected 150 tall plants but counted 140 tall and 60 short. Which statement best explains the difference?
Hint
A Punnett square gives probabilities, not guarantees. Think about flipping a coin 200 times.
Step-by-step solution
Where the answer comes from: the Punnett square predicts a 3 : 1 ratio of tall to short, or 150 tall and 50 short out of 200. The gardener observed 140 tall and 60 short.
Step by step:
The observed ratio is 140 : 60 \approx 2.3 : 1, reasonably close to 3 : 1.
Each offspring receives alleles by chance, like a coin flip for each parent. Probabilities predict what happens on average, not exactly in every sample.
Small differences from the expected numbers are normal. With a larger sample (say, 2,000 plants) the observed ratio would tend to be closer to 3 : 1.
The other choices don't fit: if both parents were TT, there would be no short plants (C). The data still show far more tall than short, so tall is still dominant (D). The Punnett square for Tt \times Tt is correct (A).
Correct answer: B.
Tip: In genetics and probability questions, expect observed results to be close to but not exactly the predicted numbers. Bigger samples reduce the effect of chance.
Question 10 of 36
Fill inEstimating a Milkweed Population
Monarch butterflies lay their eggs only on milkweed plants. A conservation group wanted to estimate how many milkweed plants grow in a 500-square-meter field without counting every plant.
A student used quadrats: square frames that each enclose exactly 1 square meter (m²). She placed 10 quadrats at random locations in the field and counted the milkweed plants inside each one.
Quadrat
Q1
Q2
Q3
Q4
Q5
Q6
Q7
Q8
Q9
Q10
Milkweed plants
3
5
2
4
6
3
4
5
2
6
To estimate the whole population, scientists multiply the mean number of plants per square meter by the total area of the field.
Use the student's quadrat data to estimate the total number of milkweed plants in the 500 m² field.
plants
Type your answer in the box. Fractions like 3/4 are fine.
Hint
Find the mean count per quadrat (per m²), then multiply by the field's area.
Step-by-step solution
Where the answer comes from: the table lists the counts 3, 5, 2, 4, 6, 3, 4, 5, 2 and 6. Each quadrat is 1 m², and the field is 500 m². The passage says to multiply "the mean number of plants per square meter by the total area."
Mean per quadrat: \dfrac{40}{10} = 4 plants per m².
Estimate for the field: 4\ \dfrac{\text{plants}}{\text{m}^2} \times 500\text{ m}^2 = 2{,}000 plants.
Correct answer: 2,000 plants.
Check: 40 plants were counted in 10 m², which is \tfrac{10}{500} = 2\% of the field. 40 \div 0.02 = 2{,}000 ✓
Tip: Sampling lets scientists estimate a large population from a small, random part. The estimate is only as good as the sample is random and large.
Question 11 of 36
Estimating a Milkweed Population
Monarch butterflies lay their eggs only on milkweed plants. A conservation group wanted to estimate how many milkweed plants grow in a 500-square-meter field without counting every plant.
A student used quadrats: square frames that each enclose exactly 1 square meter (m²). She placed 10 quadrats at random locations in the field and counted the milkweed plants inside each one.
Quadrat
Q1
Q2
Q3
Q4
Q5
Q6
Q7
Q8
Q9
Q10
Milkweed plants
3
5
2
4
6
3
4
5
2
6
To estimate the whole population, scientists multiply the mean number of plants per square meter by the total area of the field.
Another student repeated the study but placed all 10 quadrats in the corner of the field where she had noticed milkweed growing most thickly. How would this most likely affect her estimate of the field's milkweed population?
Hint
Would the quadrats in the thickest corner represent the whole field fairly?
Step-by-step solution
Where the answer comes from: the passage says the first student "placed 10 quadrats at random locations." Random placement gives every part of the field an equal chance to be sampled. The second student chose only the thickest patch.
Step by step:
Quadrats in the thickest corner will contain more plants than average — perhaps 8 or 10 instead of 4.
A higher mean per m² multiplied by the same 500 m² gives a larger total.
So the estimate would most likely be too high. This kind of error is called sampling bias.
Keeping the number of quadrats at 10 does not fix the problem; the issue is where they were placed, not how many.
Correct answer: B.
Tip: A sample must represent the whole population. Non-random sampling (choosing the easiest, closest or most interesting spots) produces biased estimates.
Question 12 of 36
Estimating a Milkweed Population
Monarch butterflies lay their eggs only on milkweed plants. A conservation group wanted to estimate how many milkweed plants grow in a 500-square-meter field without counting every plant.
A student used quadrats: square frames that each enclose exactly 1 square meter (m²). She placed 10 quadrats at random locations in the field and counted the milkweed plants inside each one.
Quadrat
Q1
Q2
Q3
Q4
Q5
Q6
Q7
Q8
Q9
Q10
Milkweed plants
3
5
2
4
6
3
4
5
2
6
To estimate the whole population, scientists multiply the mean number of plants per square meter by the total area of the field.
The conservation group wants a more reliable estimate of the milkweed population. Which change to the student's method would help most?
Hint
Reliability improves when random chance has less effect on the result.
Step-by-step solution
Where the answer comes from: the student used 10 randomly placed quadrats. The counts ranged from 2 to 6 plants per quadrat, so any one quadrat could be quite different from the field's true average.
Step by step:
A larger random sample reduces the effect of chance. With 30 random quadrats, a few unusually high or low counts matter less, and the mean is likely to be closer to the true field average.
B uses only one quadrat — the sample gets smaller and the result depends on luck.
C keeps only the highest count, which deliberately biases the estimate upward.
D replaces random placement with convenience, which introduces bias.
Correct answer: A.
Tip: Two keys to good sampling: random selection (avoids bias) and larger samples (reduce chance error). Repeated trials work the same way in experiments.
Question 13 of 36
Fill inCounting Fish by Mark and Recapture
Biologists cannot count every fish in a pond, so they use mark and recapture.
They net a group of fish, mark each one with a small tag, and release them.
After the marked fish have had time to mix back in, they net a second group and count how many in that group are marked.
The method assumes the fraction of marked fish in the second catch equals the fraction of marked fish in the whole pond:
\dfrac{\text{marked in second catch}}{\text{total in second catch}} = \dfrac{\text{total marked}}{\text{population}}
The method also assumes that tags stay on, that marked fish are no easier or harder to catch, and that no fish are born, die, or move in or out between catches.
In one pond, a team caught, tagged and released 40 bluegill. A week later they caught 50 bluegill, and 8 of them had tags.
Use the mark-and-recapture data to estimate the bluegill population of the pond.
fish
Type your answer in the box. Fractions like 3/4 are fine.
Hint
Set the fraction of tagged fish in the second catch equal to 40 out of the whole population.
Step-by-step solution
Where the answer comes from: the passage gives the proportion \dfrac{\text{marked in second catch}}{\text{total in second catch}} = \dfrac{\text{total marked}}{\text{population}}, with 40 fish tagged, 50 in the second catch, and 8 of those tagged.
Step by step:
Substitute: \dfrac{8}{50} = \dfrac{40}{N}.
Cross-multiply: 8N = 40 \times 50 = 2{,}000.
Divide: N = \dfrac{2{,}000}{8} = 250 fish.
Correct answer: 250 bluegill.
Check: in the second catch, \tfrac{8}{50} = 16\% were tagged. If 40 tagged fish are 16% of the pond, then 40 \div 0.16 = 250 ✓
Common slip: flipping the ratio (\tfrac{8 \times 40}{50} = 6.4) gives fewer fish than were tagged, which is impossible.
Tip: A sample is a small "snapshot" of the population. If 16% of the sample is marked, assume about 16% of the whole population is marked too, as long as the sample is random.
Question 14 of 36
Fill inSnapdragon Flower Color
In snapdragon plants, flower color shows incomplete dominance: neither allele is dominant, and a plant with one of each allele has a blend of the two colors.
Genotype
Flower color
RR
red
RW
pink
WW
white
A gardener crossed two pink snapdragons (RW \times RW). Each parent passes on one allele, R or W, with equal chance.
R
W
R
RR
RW
W
RW
WW
The gardener planted 120 seeds from this cross and recorded the flower color of every plant that bloomed.
Of the 120 seeds the gardener planted, how many plants would be expected to have white flowers?
plants
Type your answer in the box. Fractions like 3/4 are fine.
Hint
Find the probability of white from the Punnett square, then multiply by 120.
Step-by-step solution
Where the answer comes from: the passage says WW plants are white, and the Punnett square has one WW box out of four. The gardener planted 120 seeds.
Step by step:
Probability of white: \dfrac{1}{4}.
Expected number = probability \times number of trials: \dfrac{1}{4} \times 120 = 30.
Common slips: using \tfrac{1}{2} gives 60 (that is the pink count); using \tfrac{3}{4} gives 90, which would only make sense for complete dominance.
Tip: "Expected" means the probability times the number of tries. Real counts will usually be close to, but not exactly, this number.
Question 15 of 36
September Rainfall Records
A farmer in a coastal county is deciding when to harvest. She looks up the total rainfall recorded in September at the nearest weather station for the last seven years.
Year
September rainfall (mm)
Year 1
92
Year 2
105
Year 3
88
Year 4
310
Year 5
97
Year 6
101
Year 7
95
The station's notes say that in Year 4 a hurricane stalled over the county for two days and dropped most of that month's rain.
The farmer wants one value that best describes how much rain falls in a typical September. Which value should she use, and why?
Hint
Is there a value that is far from all the others? Which measure of center is affected less by it?
Step-by-step solution
Where the answer comes from: six Septembers had 88 to 105 mm, but Year 4 had 310 mm, and the station's notes say a hurricane "dropped most of that month's rain."
Step by step:
Sorted data: 88, 92, 95, 97, 101, 105, 310.
Median = middle (4th) value = 97 mm.
Mean = \dfrac{888}{7} \approx 126.9 mm, which is higher than six of the seven values because of the 310 mm outlier.
A "typical" value should look like most years, so the median (97 mm) describes the data better.
Correct answer: B.
Why the others are wrong: A uses all the data, but that is exactly the problem: the one storm year drags the mean far above the usual amount. C is false: no value repeats, so there is no mode. D, the range (222 mm), measures spread, not a typical amount.
Tip: When a data set has an outlier, the median usually describes the "typical" value better than the mean. On the GED, look for words like typical or usual paired with an extreme value.
Question 16 of 36
Robin Clutch Sizes
A clutch is the set of eggs a bird lays in one nesting attempt. A wildlife biologist surveyed 20 American robin nests in a city park in May and counted the eggs in each.
Eggs in nest
Number of nests
2
5
3
6
4
7
5
2
Each nest was counted once. The biologist plans to compare these results with nests in a rural area next year.
Based on the survey, what is the probability that a randomly chosen nest from this park contained at least 4 eggs?
Hint
'At least 4' means 4 or more. Add those nests and compare with the total number of nests.
Step-by-step solution
Where the answer comes from: the table shows 7 nests with 4 eggs and 2 nests with 5 eggs, out of 20 nests surveyed.
Step by step:
"At least 4 eggs" means 4 or 5 eggs: 7 + 2 = 9 nests.
Probability = \dfrac{9}{20} = 0.45 = 45\%.
Correct answer: B (45%).
Why the others are wrong: 35% counts only the nests with exactly 4 eggs (\tfrac{7}{20}). 55% is the probability of fewer than 4 eggs (\tfrac{11}{20}), the complement. 9% treats the count of 9 nests as if it were a percent without dividing by 20.
Check:45\% + 55\% = 100\% ✓
Tip: A probability from data is a relative frequency: the number of times something happened divided by the total. Read phrases like at least, at most and more than carefully.
Question 17 of 36
Fill inTiming a Toy Car on a Ramp
A physical science student tested how long a toy car takes to roll down a 2-meter ramp. She released the car from the same starting line five times and timed each run with a stopwatch.
Trial
Time (s)
1
2.31
2
2.28
3
2.35
4
3.90
5
2.30
Her lab notebook says: Trial 4: car drifted sideways and scraped along the ramp's side wall before reaching the bottom.
Her teacher reminded the class that repeated trials help reduce the effect of small random errors, and that a trial with a known problem should be noted and left out of the summary instead of being averaged in.
The student then tested a steeper ramp with six trials. The mean time was 1.85 s. Five of the times were 1.82 s, 1.88 s, 1.83 s, 1.86 s and 1.87 s. What was the time for the sixth trial?
s
Type your answer in the box. Fractions like 3/4 are fine.
Hint
Mean × number of trials gives the total of all six times.
Step-by-step solution
Where the answer comes from: the question gives a mean of 1.85 s for six trials and five of the six times. The passage explains that repeated trials are summarized with a mean.
Step by step:
Total of all six times = mean \times count = 1.85 \times 6 = 11.10 s.
Total of the five known times: 1.82 + 1.88 + 1.83 + 1.86 + 1.87 = 9.26 s.
Sixth time = 11.10 - 9.26 = 1.84 s.
Correct answer: 1.84 s.
Check:\dfrac{9.26 + 1.84}{6} = \dfrac{11.10}{6} = 1.85 s ✓
Common slip: averaging only the five known times (9.26 \div 5 = 1.852) does not answer the question; it finds the mean of the five, not the missing value.
Tip: To work backwards from a mean, turn it into a total first (mean × count). Then subtract what you know.
Question 18 of 36
Fill inRobin Clutch Sizes
A clutch is the set of eggs a bird lays in one nesting attempt. A wildlife biologist surveyed 20 American robin nests in a city park in May and counted the eggs in each.
Eggs in nest
Number of nests
2
5
3
6
4
7
5
2
Each nest was counted once. The biologist plans to compare these results with nests in a rural area next year.
What is the mean number of eggs per nest for the 20 nests? Give your answer to the nearest tenth.
eggs
Type your answer in the box. Fractions like 3/4 are fine.
Hint
Multiply each egg count by its number of nests, add the results, then divide by the number of nests.
Step-by-step solution
Where the answer comes from: the table shows 5 nests × 2 eggs, 6 nests × 3 eggs, 7 nests × 4 eggs and 2 nests × 5 eggs, for 20 nests in all.
Step by step:
Total eggs in each row: 2 \times 5 = 10; 3 \times 6 = 18; 4 \times 7 = 28; 5 \times 2 = 10.
Total eggs: 10 + 18 + 28 + 10 = 66.
Number of nests: 5 + 6 + 7 + 2 = 20.
Mean = \dfrac{66}{20} = 3.3 eggs per nest.
Correct answer: 3.3 eggs.
Check: the mean must fall between the smallest (2) and largest (5) values, and close to where most nests are (3 and 4) ✓
Common slips: averaging the egg values \tfrac{2+3+4+5}{4} = 3.5 ignores how many nests have each count; dividing 66 by 4 rows (16.5) divides by the number of rows instead of the number of nests.
Tip: In a frequency table, the mean is a weighted average: multiply, add, then divide by the total frequency.
Question 19 of 36
Inheriting Red-Green Color Blindness
Red-green color blindness is caused by a recessive allele on the X chromosome. Females have two X chromosomes (XX); males have one X and one Y (XY). The Y chromosome does not carry this gene.
X^B = allele for normal color vision (dominant)
X^b = allele for red-green color blindness (recessive)
A female needs two copies (X^bX^b) to be color-blind. A male with just one copy (X^bY) is color-blind.
A couple visits a genetic counselor. The woman has normal vision but is a carrier (X^BX^b). The man has normal vision (X^BY). The counselor draws this Punnett square:
X^B (from father)
Y (from father)
X^B (from mother)
X^BX^B
X^BY
X^b (from mother)
X^BX^b
X^bY
Each box has an equal chance. The counselor explains that each pregnancy is independent: what happened with earlier children does not change the chances for the next one.
The couple learns that their next baby will be a boy. What is the probability that he will be color-blind?
Hint
Look only at the boxes in the Punnett square that are sons (XY).
Step-by-step solution
Where the answer comes from: the Punnett square shows two sons, X^BY and X^bY, and two daughters, X^BX^B and X^BX^b. The passage says "A male with just one copy (X^bY) is color-blind."
Step by step:
We already know the baby is a boy, so only the two son boxes count.
Of those two, one (X^bY) is color-blind.
Probability = \dfrac{1}{2}, or 50%.
Correct answer: B (\tfrac{1}{2}).
Why the others are wrong:\tfrac{1}{4} is the chance that any child (boy or girl) is a color-blind son; the question already tells us it is a boy. 0 assumes the father's normal vision protects his sons, but sons get their X from their mother. 1 assumes every son of a carrier is color-blind, but the mother can pass either X^B or X^b.
Tip: When a question gives extra information ("it will be a boy"), narrow the Punnett square to the boxes that fit, then find the fraction within that group.
Question 20 of 36
Fill inA Weekend Rain Forecast
A meteorologist uses computer models to forecast the weekend weather for a town.
Day
Chance of rain
Saturday
60%
Sunday
50%
A "60% chance of rain" means that on many days with conditions like these, rain fell on about 60 out of 100 of them.
For planning, a community garden club decides to treat the two days as independent events: whether it rains Saturday does not change the chance of rain Sunday. To find the probability that two independent events both happen, multiply their probabilities. The probability that an event does not happen is 1 minus the probability that it does.
Treating the two days as independent, what is the probability that it will rain on both Saturday and Sunday?
%
Type your answer in the box. Fractions like 3/4 are fine.
Hint
Write each chance as a decimal and multiply.
Step-by-step solution
Where the answer comes from: the table gives a 60% chance of rain Saturday and 50% Sunday. The passage says "To find the probability that two independent events both happen, multiply their probabilities."
Step by step:
Convert to decimals: 60% = 0.60; 50% = 0.50.
Multiply: 0.60 \times 0.50 = 0.30.
Convert back to a percent: 0.30 = 30\%.
Correct answer: 30%.
Check: the chance of rain on both days must be less than the chance on either single day (50% and 60%), and 30% is ✓. Adding (110%) would give a probability above 100%, which is impossible.
Tip: For "both A and B" with independent events, multiply. Multiplying two probabilities (each less than 1) always gives a smaller number.
Question 21 of 36
Measuring Reaction Time
A student measured her reaction time with a ruler-drop test. A partner held a ruler above her open hand and let it fall without warning; she caught it as fast as she could. The distance the ruler fell was converted to a reaction time in seconds. She did five trials.
Trial
1
2
3
4
5
Reaction time (s)
0.31
0.29
0.33
0.30
0.92
In her notes she wrote: "During trial 5, a fire alarm test started ringing in the hallway just before the ruler was dropped."
The student wants to report one value that best represents her typical reaction time. Which choice is best, and why?
Hint
Compute both the mean and the median. Which one is pulled away from most of the data by trial 5?
Step-by-step solution
Where the answer comes from: the table gives 0.31, 0.29, 0.33, 0.30 and 0.92 s, and her notes say that during trial 5 "a fire alarm test started ringing."
Median: sort the values → 0.29, 0.30, 0.31, 0.33, 0.92. The middle value is 0.31 s.
Four of the five trials are between 0.29 and 0.33 s. The value 0.92 s is an outlier, and the notes give a reason: she was distracted by the alarm.
The mean (0.43 s) is higher than every one of the four normal trials, so it does not represent her typical time. The median (0.31 s) sits right among them.
Correct answer: B.
Why the others are wrong: A uses all the data, but an outlier pulls the mean toward itself. C reports the outlier itself. D's 0.63 s is the range (0.92 - 0.29), a measure of spread, not a typical value.
Tip: When a data set has an outlier, the median usually represents the typical value better than the mean. Look for a note in the passage that explains an unusual trial.
Question 22 of 36
Hot spotFertilizer Trials with Error Bars
An agricultural researcher tested three fertilizers on tomato plants. She grew five plots for each treatment, including a control group that got no fertilizer, and recorded the mean yield in kilograms per plot.
On the graph, each bar shows the mean yield. The error bar on each bar shows the range of values within which the true mean is likely to fall. A large error bar means the plots in that group gave very different results, so the mean is less certain.
How to read error bars: if the error bars of two groups overlap, the difference between their means may be due to chance. If they do not overlap, the difference is more likely to be real.
Click the bar for the fertilizer whose mean yield is most clearly different from the control group's mean yield.
Click on the figure to mark your answer (or focus it and use the arrow keys).
Hint
Compare each fertilizer's error bar with the control's error bar (19 to 21 kg). Which one does not overlap at all?
Step-by-step solution
Where the answer comes from: the graph shows Control 20 kg (error bar 19 to 21 kg), Fertilizer A 22 kg (19 to 25 kg), Fertilizer B 25 kg (24 to 26 kg) and Fertilizer C 26 kg (20 to 32 kg). The passage says that if error bars "do not overlap, the difference is more likely to be real."
Step by step:
Control's error bar: 19 to 21 kg.
Fertilizer A: 19 to 25 kg. Its lower end (19) is inside the control's range → overlap.
Fertilizer B: 24 to 26 kg. Its lowest value (24) is above the control's highest (21) → no overlap. ✓
Fertilizer C: 20 to 32 kg. Its lower end (20) is inside the control's range → overlap, even though its mean is highest.
Correct answer: the Fertilizer B bar.
Tip: The highest bar is not automatically the "real" winner. A difference is convincing when it is large compared with the uncertainty — that is what non-overlapping error bars show.
Question 23 of 36
Hot spotAntacid Tablets and Water Temperature
A chemistry student timed how long one antacid tablet took to dissolve completely in 200 mL of water at four temperatures. She repeated each temperature six times.
On the graph, each bar shows the mean dissolving time. The error bar shows the range within which the true mean is likely to fall. A longer error bar means the six trials at that temperature were less consistent.
How to read error bars: if the error bars of two groups overlap, the difference between their means may be due to chance. If they do not overlap, the difference is more likely to be real.
Click the bar for the temperature whose mean dissolving time is not clearly different from the mean time at 25 °C.
Click on the figure to mark your answer (or focus it and use the arrow keys).
Hint
Write down the 25 °C error bar's range (41 to 49 s). Which other error bar overlaps it?
Step-by-step solution
Where the answer comes from: the graph shows 10 °C: 62 s (56 to 68 s); 25 °C: 45 s (41 to 49 s); 40 °C: 38 s (30 to 46 s); 55 °C: 24 s (22 to 26 s). The passage says that if error bars "overlap, the difference between their means may be due to chance."
Step by step:
The 25 °C error bar runs from 41 to 49 s.
10 °C: lowest value 56 s is above 49 s → no overlap → clearly different.
40 °C: 30 to 46 s. The top part (41 to 46 s) lies inside the 25 °C range → overlap → not clearly different.
55 °C: highest value 26 s is below 41 s → no overlap → clearly different.
Correct answer: the 40 °C bar.
Check: the 40 °C bar also has the longest error bar (16 s), showing its six trials were the least consistent, which is why its mean is uncertain.
Tip: Compare the ends of the error bars, not the tops of the bars. Two means can look different and still overlap once their uncertainty is included.
Question 24 of 36
Counting Fish by Mark and Recapture
Biologists cannot count every fish in a pond, so they use mark and recapture.
They net a group of fish, mark each one with a small tag, and release them.
After the marked fish have had time to mix back in, they net a second group and count how many in that group are marked.
The method assumes the fraction of marked fish in the second catch equals the fraction of marked fish in the whole pond:
\dfrac{\text{marked in second catch}}{\text{total in second catch}} = \dfrac{\text{total marked}}{\text{population}}
The method also assumes that tags stay on, that marked fish are no easier or harder to catch, and that no fish are born, die, or move in or out between catches.
In one pond, a team caught, tagged and released 40 bluegill. A week later they caught 50 bluegill, and 8 of them had tags.
The team later learns that about half of the tags they attached had fallen off before the second catch. How does this most likely affect their population estimate?
Hint
With fewer fish still tagged, would the second catch contain more or fewer tagged fish? Then follow the proportion.
Step-by-step solution
Where the answer comes from: the passage says the method "assumes that tags stay on" and uses \dfrac{\text{marked in second catch}}{\text{total in second catch}} = \dfrac{\text{total marked}}{\text{population}}.
Step by step:
If tags fell off, fewer fish in the pond actually carry tags than the 40 the team counts in the formula.
So the second catch contains fewer tagged fish than it should.
A smaller number of tagged fish on the left side makes the fraction smaller, so the calculated population N = \dfrac{40 \times 50}{\text{tagged recaught}} gets larger.
The estimate is therefore too high.
Correct answer: A.
Example: if only 4 tagged fish had been recaught instead of 8, the estimate would be \dfrac{40 \times 50}{4} = 500, double the 250 from the original data.
Why the others are wrong: B reverses the direction of the bias. C ignores that the method depends on the tag count. D is wrong because the error is predictable: losing tags always pushes the estimate up.
Tip: To predict the direction of a bias, trace the error through the formula: which number gets bigger or smaller, and is it on top or bottom of the fraction?
Question 25 of 36
Drop-downSeptember Rainfall Records
A farmer in a coastal county is deciding when to harvest. She looks up the total rainfall recorded in September at the nearest weather station for the last seven years.
Year
September rainfall (mm)
Year 1
92
Year 2
105
Year 3
88
Year 4
310
Year 5
97
Year 6
101
Year 7
95
The station's notes say that in Year 4 a hurricane stalled over the county for two days and dropped most of that month's rain.
The farmer decides to remove the Year 4 hurricane value and look at the other six years only.
Compared with the seven-year values, the mean would .
Compared with the seven-year values, the median would .
Choose an answer from each menu.
Hint
Recompute both measures for the six remaining values: 88, 92, 95, 97, 101, 105.
Step-by-step solution
Where the answer comes from: the seven-year data are 88, 92, 95, 97, 101, 105 and 310 mm. With all seven, the mean is \dfrac{888}{7} \approx 126.9 mm and the median (4th value) is 97 mm.
Blank 1 (mean):
Remove 310: the total becomes 888 - 310 = 578 mm for 6 years.
New mean = \dfrac{578}{6} \approx 96.3 mm.
Change: 126.9 - 96.3 \approx 30.6 mm, so the mean would decrease by about 31 mm.
Blank 2 (median):
Six values in order: 88, 92, 95, 97, 101, 105.
New median = \dfrac{95 + 97}{2} = 96 mm.
Change: 97 - 96 = 1 mm, so the median would decrease by about 1 mm.
Correct answers: mean: decrease by about 31 mm; median: decrease by about 1 mm.
Tip: An outlier changes the mean a lot but the median very little. That is why the median is called resistant to outliers.
Question 26 of 36
Fill inEstimating Craters on the Moon
A planetary science class studied a high-resolution photograph of a flat region of the Moon covering 2,000 km². Counting every crater would take too long, so the students laid a grid of equal squares over the photo. Each square covered 100 km².
The class picked 5 squares using a random number generator and counted every crater wider than 1 km in each.
Square
Craters counted
A
12
B
9
C
15
D
11
E
13
To estimate the total, they planned to find the mean number of craters per square and multiply by the number of squares that fit in the whole region.
Use the class's sample to estimate the total number of craters wider than 1 km in the whole 2,000 km² region.
craters
Type your answer in the box. Fractions like 3/4 are fine.
Hint
Find the mean count per square, then find how many 100 km² squares fit in 2,000 km².
Step-by-step solution
Where the answer comes from: the table lists 12, 9, 15, 11 and 13 craters in five squares of 100 km² each. The region covers 2,000 km², and the passage says to "find the mean number of craters per square and multiply by the number of squares that fit in the whole region."
Number of squares in the region: \dfrac{2{,}000\text{ km}^2}{100\text{ km}^2} = 20 squares.
Estimate: 12 \times 20 = 240 craters.
Correct answer: about 240 craters.
Check: the five squares cover 500 km², a quarter of the region, and hold 60 craters; 60 \times 4 = 240 ✓
Common slips: multiplying the total 60 by 20 (1,200) forgets to find the mean first; multiplying 12 by 2,000 (24,000) uses the area instead of the number of squares.
Tip: Scale a sample up by asking "how many samples would fill the whole area?" and multiply the average sample by that number.
Question 27 of 36
Snapdragon Flower Color
In snapdragon plants, flower color shows incomplete dominance: neither allele is dominant, and a plant with one of each allele has a blend of the two colors.
Genotype
Flower color
RR
red
RW
pink
WW
white
A gardener crossed two pink snapdragons (RW \times RW). Each parent passes on one allele, R or W, with equal chance.
R
W
R
RR
RW
W
RW
WW
The gardener planted 120 seeds from this cross and recorded the flower color of every plant that bloomed.
Of the 120 plants, the gardener counted 34 red, 55 pink and 31 white. Which statement best describes these results?
Hint
Compare each count with the expected 30 red, 60 pink, 30 white. Are the differences big or small?
Step-by-step solution
Where the answer comes from: the Punnett square predicts \tfrac{1}{4} red, \tfrac{1}{2} pink and \tfrac{1}{4} white. For 120 plants, that is 30 red, 60 pink, 30 white.
Step by step:
Compare: red 34 vs 30 (+4); pink 55 vs 60 (−5); white 31 vs 30 (+1).
Each difference is small, at most about 5 plants, or roughly 4% of 120.
Each seed's genotype depends on chance, like a coin toss, so real counts rarely match the prediction exactly.
The results fit the predicted 1 : 2 : 1 ratio, with small differences due to chance.
Correct answer: C.
Why the others are wrong: A demands an exact match; probability predicts what to expect on average, not exact counts. B is not supported: pink is the blended RW color, not an allele, so it cannot be dominant; pink being most common is exactly what the 1 : 2 : 1 prediction expects. D is wrong because the prediction itself is 60 pink and 60 red-plus-white; 55 versus 65 is a small difference that chance can easily explain.
Tip: Small differences between observed and expected results are normal. Ask whether the overall pattern matches the prediction.
Question 28 of 36
Fill inA Week of High Temperatures
An earth science class kept a weather station on the school roof. The station records the daily high temperature. During one week, the sensor lost power for one day, so that day's reading is missing from the class log.
Day
High temperature (°C)
Monday
16.5
Tuesday
19.5
Wednesday
17.5
Thursday
20.5
Friday
18.0
Saturday
15.5
Sunday
missing
A nearby official weather station, which matched the school station's readings closely on every other day, reported that the mean daily high for the full seven days was 18.0 °C.
Use the official station's seven-day mean to estimate the high temperature on Sunday.
°C
Type your answer in the box. Fractions like 3/4 are fine.
Hint
Turn the mean into a seven-day total, then subtract the six known days.
Step-by-step solution
Where the answer comes from: the table lists six highs: 16.5, 19.5, 17.5, 20.5, 18.0 and 15.5 °C. The passage says the official station "reported that the mean daily high for the full seven days was 18.0 °C" and matched the school station closely.
Step by step:
Seven-day total = mean \times number of days = 18.0 \times 7 = 126.0 °C.
Sum of the six known days: 16.5 + 19.5 + 17.5 + 20.5 + 18.0 + 15.5 = 107.5 °C.
Sunday = 126.0 - 107.5 = 18.5 °C.
Correct answer: 18.5 °C.
Check:\dfrac{107.5 + 18.5}{7} = \dfrac{126.0}{7} = 18.0 °C ✓
Common slip: taking the mean of the six known days (107.5 \div 6 \approx 17.9) does not use the official mean and does not find Sunday's value.
Tip: Mean × count = total. This one fact lets you find a missing value, add a new value or remove a value from any data set.
Question 29 of 36
Fill inRadioactive Decay and Half-Life
Hospitals use a radioactive tracer, technetium-99m, to make images of organs. It has a half-life of about 6 hours: in each 6-hour period, every radioactive atom has a 50% chance of decaying.
No one can predict which atoms will decay in a given period, just as no one can predict which coins will land heads in a large pile of flipped coins. But with huge numbers of atoms, the expected result is very reliable: after each half-life, about half of the remaining radioactive atoms are left.
A science textbook illustrates this with a simple model: a sample that starts with 1,600 radioactive atoms.
In the textbook model, how many radioactive atoms would be expected to remain after 24 hours?
atoms
Type your answer in the box. Fractions like 3/4 are fine.
Hint
How many 6-hour half-lives fit into 24 hours? Halve the number once for each one.
Step-by-step solution
Where the answer comes from: the passage says the tracer's half-life is "about 6 hours," that "after each half-life, about half of the remaining radioactive atoms are left," and that the model starts with 1,600 atoms.
Step by step:
Number of half-lives: \dfrac{24\text{ h}}{6\text{ h}} = 4.
Halve four times:
after 6 h: 1{,}600 \times \tfrac{1}{2} = 800
after 12 h: 400
after 18 h: 200
after 24 h: 100
In one step: 1{,}600 \times \left(\tfrac{1}{2}\right)^4 = \dfrac{1{,}600}{16} = 100.
Correct answer: 100 atoms.
Common slips: halving only once (800) or dividing by 4 instead of 2^4 = 16 (400).
Tip: An expected value comes from probability × number of chances. Here each atom has a \tfrac{1}{2} chance of surviving each half-life, so the chances multiply: \left(\tfrac{1}{2}\right)^n.
Question 30 of 36
Timing a Toy Car on a Ramp
A physical science student tested how long a toy car takes to roll down a 2-meter ramp. She released the car from the same starting line five times and timed each run with a stopwatch.
Trial
Time (s)
1
2.31
2
2.28
3
2.35
4
3.90
5
2.30
Her lab notebook says: Trial 4: car drifted sideways and scraped along the ramp's side wall before reaching the bottom.
Her teacher reminded the class that repeated trials help reduce the effect of small random errors, and that a trial with a known problem should be noted and left out of the summary instead of being averaged in.
Which value should the student report as the car's typical time to roll down the ramp?
Hint
Read the notebook entry and the teacher's reminder before averaging.
Step-by-step solution
Where the answer comes from: the notebook says Trial 4's car "scraped along the ramp's side wall," and the teacher said "a trial with a known problem should be noted and left out of the summary instead of being averaged in."
Step by step:
Trial 4 (3.90 s) had a known problem, so leave it out.
Mean of the four valid trials: \dfrac{2.31 + 2.28 + 2.35 + 2.30}{4} = \dfrac{9.24}{4} = 2.31 s.
Report 2.31 s and note why Trial 4 was excluded.
Correct answer: B.
Why the others are wrong: A averages in Trial 4: \dfrac{13.14}{5} \approx 2.63 s, which is slower than every valid trial. C reports one trial only, which ignores the purpose of repeated trials. D reports the faulty trial.
Check: 2.31 s sits in the middle of the valid times (2.28 to 2.35 s) ✓
Tip: Repeated trials reduce random error, but a trial with a known mistake is not random error. Exclude it, say why, and average the rest.
Question 31 of 36
Fertilizer Trials with Error Bars
An agricultural researcher tested three fertilizers on tomato plants. She grew five plots for each treatment, including a control group that got no fertilizer, and recorded the mean yield in kilograms per plot.
On the graph, each bar shows the mean yield. The error bar on each bar shows the range of values within which the true mean is likely to fall. A large error bar means the plots in that group gave very different results, so the mean is less certain.
How to read error bars: if the error bars of two groups overlap, the difference between their means may be due to chance. If they do not overlap, the difference is more likely to be real.
Fertilizer C produced the highest mean yield. Which conclusion about Fertilizer C is best supported by the data?
Hint
Look at how wide Fertilizer C's error bar is and where its bottom end falls.
Step-by-step solution
Where the answer comes from: the graph shows Fertilizer C with a mean of 26 kg and an error bar from 20 to 32 kg; the control's error bar runs from 19 to 21 kg. The passage says "a large error bar means the plots in that group gave very different results, so the mean is less certain," and overlapping error bars mean "the difference … may be due to chance."
Step by step:
C's error bar is 12 kg wide (20 to 32) — the widest on the graph — so its five plots varied a lot.
The bottom of C's error bar (20 kg) overlaps the control's error bar (19 to 21 kg).
So the data cannot rule out that C's true effect is about the same as no fertilizer. The high mean is uncertain.
The best conclusion is cautious: C might raise yield, but more trials (more plots) are needed to reduce the uncertainty.
Correct answer: C.
Why the others are wrong: A ignores the large uncertainty. B is not supported — C's mean is above the control's, not below. D claims certainty that error bars can never give; overlapping bars mean the difference is unclear, not that the values are identical.
Tip: On the GED, the best-supported conclusion about uncertain data is usually the most cautious one that still matches the numbers. Watch for words like "best," "proves" and "certainly."
Question 32 of 36
Select allTesting a Germination Claim
A seed company prints on its tomato seed packets: "90% germination rate." This means that, on average, 90 out of every 100 seeds should sprout under good conditions.
Student test: A student planted 10 seeds from one packet in a tray at home. 7 sprouted (70%).
Lab test: An agricultural extension lab planted 500 seeds taken from many packets of the same seed batch, using controlled warmth and moisture. 448 sprouted (89.6%).
The lab's report notes that when only a few seeds are tested, the percentage that sprout can vary a lot by chance from one small group to the next. With hundreds of seeds, chance differences tend to even out.
Select all of the statements that are supported by the information in the passage.
Select all that apply.
Hint
Think about how much a 10-seed test can vary by chance compared with a 500-seed test.
Step-by-step solution
Where the answer comes from: the student's test had 10 seeds (70% sprouted); the lab's test had 500 seeds (89.6% sprouted). The lab notes that "when only a few seeds are tested, the percentage that sprout can vary a lot by chance," and "With hundreds of seeds, chance differences tend to even out."
Evaluate each statement:
A — supported. The lab result (89.6%) comes from a much larger sample and is very close to 90%.
B — supported. With only 10 seeds, one or two seeds make a 10–20 point difference, so 70% can happen by chance even if the true rate is 90%.
C — not supported. One small test cannot prove the claim false; the lab's larger test agrees with the claim.
D — supported. The lab used seeds from many packets and controlled conditions, so its result is the more reliable evidence.
E — not supported. Large samples get closer to the true rate, but not exactly: the lab got 89.6%, not 90.0%.
Correct answers: A, B and D.
Tip: Larger samples give more reliable estimates, but "more reliable" never means "exact." Small samples can easily differ from the true value by chance.
Question 33 of 36
A Weekend Rain Forecast
A meteorologist uses computer models to forecast the weekend weather for a town.
Day
Chance of rain
Saturday
60%
Sunday
50%
A "60% chance of rain" means that on many days with conditions like these, rain fell on about 60 out of 100 of them.
For planning, a community garden club decides to treat the two days as independent events: whether it rains Saturday does not change the chance of rain Sunday. To find the probability that two independent events both happen, multiply their probabilities. The probability that an event does not happen is 1 minus the probability that it does.
The garden club will cancel its outdoor work day if it rains on at least one of the two days. Treating the days as independent, what is the probability that the club cancels?
Hint
It is easier to find the chance of NO rain on either day first, then subtract from 1.
Step-by-step solution
Where the answer comes from: the table gives 60% (Saturday) and 50% (Sunday). The passage says to multiply probabilities for two independent events that both happen, and "The probability that an event does not happen is 1 minus the probability that it does."
Step by step:
Chance of no rain Saturday: 1 - 0.60 = 0.40.
Chance of no rain Sunday: 1 - 0.50 = 0.50.
Chance of no rain on either day (both dry): 0.40 \times 0.50 = 0.20.
"At least one rainy day" is the opposite of "both days dry": 1 - 0.20 = 0.80 = 80\%.
Correct answer: D (80%).
Why the others are wrong: 110% comes from adding 60% + 50%, which counts the "rain both days" case twice; a probability can never be above 100%. 30% is the chance of rain on both days. 20% is the chance of rain on neither day.
Check by listing cases: rain both 0.30 + rain Sat only 0.60 \times 0.50 = 0.30 + rain Sun only 0.40 \times 0.50 = 0.20= 0.80 ✓
Tip: For "at least one," use the complement: 1 - P(\text{none}). It avoids double-counting.
Question 34 of 36
Antacid Tablets and Water Temperature
A chemistry student timed how long one antacid tablet took to dissolve completely in 200 mL of water at four temperatures. She repeated each temperature six times.
On the graph, each bar shows the mean dissolving time. The error bar shows the range within which the true mean is likely to fall. A longer error bar means the six trials at that temperature were less consistent.
How to read error bars: if the error bars of two groups overlap, the difference between their means may be due to chance. If they do not overlap, the difference is more likely to be real.
Which conclusion is best supported by the graph?
Hint
Check each claim against the error bars. Which pairs overlap, and which error bar is longest?
Step-by-step solution
Where the answer comes from: the graph shows 10 °C: 62 s (56 to 68 s); 25 °C: 45 s (41 to 49 s); 40 °C: 38 s (30 to 46 s); 55 °C: 24 s (22 to 26 s). The passage says a longer error bar "means the six trials at that temperature were less consistent," and overlapping error bars mean a difference "may be due to chance."
Evaluate each option:
A: the 10 °C range (56 to 68 s) and the 55 °C range (22 to 26 s) are far apart, so raising the temperature from 10 °C to 55 °C clearly shortened the time. Supported.
B: 40 °C (30 to 46 s) overlaps 25 °C (41 to 49 s), so we cannot be sure 40 °C is faster.
C: 40 °C has the longest error bar (16 s), so its trials were the least consistent, not the most.
D: the means fall steadily as temperature rises, and two pairs are clearly different, so temperature does matter.
Correct answer: A.
Tip: A well-supported conclusion uses only the differences the error bars say are real. Be cautious about any claim that depends on overlapping bars.
Question 35 of 36
Inheriting Red-Green Color Blindness
Red-green color blindness is caused by a recessive allele on the X chromosome. Females have two X chromosomes (XX); males have one X and one Y (XY). The Y chromosome does not carry this gene.
X^B = allele for normal color vision (dominant)
X^b = allele for red-green color blindness (recessive)
A female needs two copies (X^bX^b) to be color-blind. A male with just one copy (X^bY) is color-blind.
A couple visits a genetic counselor. The woman has normal vision but is a carrier (X^BX^b). The man has normal vision (X^BY). The counselor draws this Punnett square:
X^B (from father)
Y (from father)
X^B (from mother)
X^BX^B
X^BY
X^b (from mother)
X^BX^b
X^bY
Each box has an equal chance. The counselor explains that each pregnancy is independent: what happened with earlier children does not change the chances for the next one.
The couple plans to have two children. What is the probability that both children will be color-blind sons?
Hint
First find the chance that one child is a color-blind son using all four boxes. Then use the counselor's point about independence.
Step-by-step solution
Where the answer comes from: the Punnett square has four equally likely boxes, and only one, X^bY, is a color-blind son. The counselor says "each pregnancy is independent."
Step by step:
Chance one child is a color-blind son: \dfrac{1}{4} (one box out of four).
The two births are independent, so multiply: \dfrac{1}{4} \times \dfrac{1}{4} = \dfrac{1}{16}.
Correct answer: A (\tfrac{1}{16}).
Why the others are wrong:\tfrac{1}{4} multiplies \tfrac{1}{2} \times \tfrac{1}{2}, using the chance that a son is color-blind; but the children are not known to be sons, so the chance of being a son must be included. \tfrac{1}{8} adds \tfrac{1}{16} + \tfrac{1}{16} or halves \tfrac{1}{4}, neither of which matches the rule for independent events. \tfrac{1}{2} adds \tfrac{1}{4} + \tfrac{1}{4}; adding is for "either/or," not "both."
Check:\tfrac{1}{16} is smaller than the \tfrac{1}{4} chance for one child, as it must be when two things must both happen ✓
Tip: Read who the probability is about. "A son is color-blind" (\tfrac{1}{2}) and "a child is a color-blind son" (\tfrac{1}{4}) are different questions.
Question 36 of 36
Resting Heart Rates in a Fitness Study
A community health center measured the resting heart rate of 10 adults at the start of a walking program. Resting heart rate is measured after a person has been sitting quietly for at least 5 minutes.
The center's first report gave a mean resting heart rate of 72 beats per minute (bpm) for the 10 adults.
Later, a staff member found that one participant's reading, 90 bpm, had been taken right after the participant hurried up two flights of stairs, not after resting. The center decided to remove that reading from the data set rather than guess a replacement value.
After the 90 bpm reading is removed, what is the mean resting heart rate of the remaining participants?
Hint
Use the original mean to find the total of all 10 readings. Remember how many readings are left.
Step-by-step solution
Where the answer comes from: the passage gives a mean of 72 bpm for 10 adults and says one reading of 90 bpm was removed.
Step by step:
Total of all 10 readings = mean \times count = 72 \times 10 = 720 bpm.
Remove the 90 bpm reading: 720 - 90 = 630 bpm.
Readings left: 10 - 1 = 9.
New mean = \dfrac{630}{9} = 70 bpm.
Correct answer: C (70 bpm).
Why the others are wrong: 63 bpm divides 630 by 10, keeping the old count after removing a reading. 72 bpm assumes removing a value cannot change the mean. 81 bpm averages 72 and 90, which mixes a mean with a single value.
Check: removing a value above the mean must lower the mean, and 70 < 72 ✓. Also 70 \times 9 + 90 = 720 ✓
Tip: You never need the individual data to add or remove a value: total = mean × count, adjust the total and the count, then divide again.
