Mathbench

Chapter 7 — Statistics and Probability

Statistics and Probability is 12 to 15% of the test — six questions. Very little of it is calculation. Most of it is reading: a table, a chart, or a sentence describing how data was collected.

That makes it the category where careless reading costs most. A two-way table question is arithmetic a ten-year-old could do; the difficulty is entirely in working out which row, which column, and which total the question wants.

Topics covered: mean, median and mode · range and standard deviation · outliers · reading two-way and frequency tables · Venn diagrams · scatterplots and lines of best fit · probability · complements · mutually exclusive events · counting · samples and study design


Section 1 — Centre and Spread

Four words, and the ACT expects all four:

Standard deviation measures how far the values sit from their own mean, on average. You will almost never have to calculate one on the ACT — you will have to compare two, and for that a single idea is enough:

More spread out means a bigger standard deviation. Values bunched near the mean means a smaller one.


Q1Medium

Topic: Comparing standard deviations

Two laboratories each recorded five measurements of the same quantity. Set P is 20,  20,  20,  20,  2020, \; 20, \; 20, \; 20, \; 20 and set Q is 10,  15,  20,  25,  3010, \; 15, \; 20, \; 25, \; 30. Both sets have the same mean. Which statement is true?

A) Set P has the greater standard deviation

B) Set Q has the greater standard deviation

C) The two standard deviations are equal

D) Standard deviation cannot be compared without calculating it

Show the worked solution

Answer: B

Explanation

Both sets have a mean of 20.

Set P is every value identical. Nothing deviates from the mean at all, so its standard deviation is exactly zero — the only time that happens.

Set Q spreads from 10 to 30, so its values sit well away from 20.

Q therefore has the greater standard deviation.

Why each wrong option is wrong:

  • A, Set P has the greater standard deviation — has it backwards. A set of identical values cannot be the more spread out.
  • C, The two standard deviations are equal — equal means do not imply equal spreads. That is exactly what standard deviation exists to distinguish.
  • D, Standard deviation cannot be compared without calculating it — no calculation is needed. One set has zero spread, and no set can have less than that.

Takeaway: standard deviation is about spread, not centre. Identical values give zero. Two sets can share a mean and be nothing alike.


Q2Medium

Topic: What an outlier does to the mean and the median

A set of readings is 12,  14,  15,  13,  1612, \; 14, \; 15, \; 13, \; 16. A sixth reading of 60 is then recorded. Which statement about the effect of that value is true?

F) Neither the mean nor the median changes

G) The median rises far more than the mean does

H) Both rise by the same amount

J) The mean rises far more than the median does

Show the worked solution

Answer: J

Explanation

Work out both, before and after.

Before: the five values total 70, so the mean is 14. Sorted they are 12,13,14,15,1612, 13, 14, 15, 16, so the median is 14 as well.

After: the six values total 130, so the mean is 130÷6=21.67130 \div 6 = 21.67. Sorted they are 12,13,14,15,16,6012, 13, 14, 15, 16, 60, so the median is 14+152=14.5\frac{14 + 15}{2} = 14.5.

The mean moved nearly 8; the median moved a half.

The reason is structural. The mean uses the size of every value, so one huge number drags it a long way. The median only cares about position, and 60 counts for no more than 17 would.

Why each wrong option is wrong:

  • G, The median rises far more than the mean does — exactly backwards.
  • H, Both rise by the same amount — they move by very different amounts, which is the point.
  • F, Neither the mean nor the median changes — both change.

Takeaway: an outlier drags the mean and barely moves the median. That is why incomes and house prices are reported as medians.


Section 2 — Reading Tables

Most ACT statistics questions are a table and a careful sentence. The method is always the same three steps:

  1. Find the group the question restricts you to — that is your denominator.
  2. Find the count it asks about — that is your numerator.
  3. Check the totals add up before you trust the table.

The word to watch is "of". "Of the students who cycle" restricts you to the cyclists' row; the grand total is then irrelevant.


Q3Medium

Topic: Conditional probability from a two-way table

A school recorded which of its 120 students study Spanish and which study history, and set the counts out in the table below.

Studies history Does not study history Total
Studies Spanish 45 25 70
Does not study Spanish 20 30 50
Total 65 55 120

A student who studies Spanish is chosen at random. What is the probability that the student also studies history?

A) 38\frac{3}{8}

B) 712\frac{7}{12}

C) 514\frac{5}{14}

D) 914\frac{9}{14}

Show the worked solution

Answer: D

Explanation

"A student who studies Spanish is chosen" restricts the pool to the Spanish row. That row has 70 students, so 70 is the denominator — not 120.

Of those 70, the number who also study history is 45.

P=4570=914P = \frac{45}{70} = \frac{9}{14}

Why each wrong option is wrong:

  • A, 38\frac{3}{8} — used 120, the whole school. That would be the answer to "a student is chosen at random from the school; what is the probability they study both?" — a different question.
  • B, 712\frac{7}{12} — gave the probability that a random student studies Spanish.
  • C, 514\frac{5}{14} — gave the probability of not studying history within the same row. Note it and the answer add to 1, as they must.

Takeaway: the group the question restricts you to is the denominator. Underline the "of the …" phrase before you look at the table.


Q4Medium

Topic: Reading a frequency table

A tourist survey recorded how many cities each visitor went to during their trip, and the results are summarised below.

Cities visited 1 2 3 4
Number of tourists 30 40 50 20

What is the mean number of cities visited per tourist?

F) 52\frac{5}{2}

G) 3535

H) 177\frac{17}{7}

J) 33

Show the worked solution

Answer: H

Explanation

A frequency table is a shorthand: 30 tourists each visited 1 city, 40 each visited 2, and so on. To find the mean, add up the total cities visited and divide by the total tourists.

total cities=1(30)+2(40)+3(50)+4(20)=30+80+150+80=340\text{total cities} = 1(30) + 2(40) + 3(50) + 4(20) = 30 + 80 + 150 + 80 = 340

total tourists=30+40+50+20=140\text{total tourists} = 30 + 40 + 50 + 20 = 140

mean=340140=1772.43\text{mean} = \frac{340}{140} = \frac{17}{7} \approx 2.43

Sanity check: the answer must lie between 1 and 4, and nearer the middle where most tourists are. It does.

Why each wrong option is wrong:

  • F, 52\frac{5}{2} — averaged 1,2,3,41, 2, 3, 4, ignoring how many tourists gave each answer. That treats one tourist and fifty tourists as equally important.
  • G, 3535 — averaged the frequencies instead, which answers nothing.
  • J, 33 — gave 3, the mode — the most common answer, not the mean.

Takeaway: in a frequency table, multiply each value by its frequency, add, then divide by the total frequency. Never average the values on the top row by themselves.


Q5Hard

Topic: A Venn diagram in words

On a farm survey, 34 farms raise cows, 28 raise sheep, and 12 raise both. Every farm surveyed raises at least one of the two. How many farms were surveyed?

A) 6262

B) 5050

C) 7474

D) 3838

Show the worked solution

Answer: B

Explanation

Draw two overlapping circles. The 12 farms raising both sit in the overlap, and they have been counted twice — once in the 34 and once in the 28. Subtract them once to put that right:

34+2812=5034 + 28 - 12 = 50

Check by parts: cows only is 3412=2234 - 12 = 22; sheep only is 2812=1628 - 12 = 16; both is 12. And 22+16+12=5022 + 16 + 12 = 50. Correct.

Why each wrong option is wrong:

  • A, 6262 — added without removing the double count.
  • C, 7474 — added the overlap a third time.
  • D, 3838 — subtracted it twice, which removes the "both" farms from the survey altogether.

Takeaway: for two overlapping groups, total =A+Bboth= A + B - \text{both}. Subtract the overlap once, because it was counted twice. Then check by adding the three separate regions.


Section 3 — Probability

P(event)=outcomes you wantoutcomes possibleP(\text{event}) = \frac{\text{outcomes you want}}{\text{outcomes possible}}

Every probability is between 0 and 1. An answer outside that range is wrong without further thought.

Three ideas the ACT tests:

  • the complementP(not A)=1P(A)P(\text{not } A) = 1 - P(A). Often far quicker than counting what you want directly.
  • mutually exclusive events cannot both happen, so P(A or B)=P(A)+P(B)P(A \text{ or } B) = P(A) + P(B).
  • independent events do not affect each other, so P(A and B)=P(A)×P(B)P(A \text{ and } B) = P(A) \times P(B).

"Or" tends to add; "and" tends to multiply.


Q6Basic

Topic: Probability using the complement

A bag used in a demonstration holds counters identical apart from colour, and the counts are recorded below.

Colour Red Blue Green Total
Number of counters 5 7 8 20

One counter is drawn at random. What is the probability that it is not green?

F) 25\frac{2}{5}

G) 35\frac{3}{5}

H) 23\frac{2}{3}

J) 13\frac{1}{3}

Show the worked solution

Answer: G

Explanation

Two routes, and they agree.

Count directly: not green means red or blue, which is 5+7=125 + 7 = 12 counters out of 20:

1220=35\frac{12}{20} = \frac{3}{5}

Use the complement: green is 820\frac{8}{20}, so not green is

1820=1220=351 - \frac{8}{20} = \frac{12}{20} = \frac{3}{5}

Why each wrong option is wrong:

  • F, 25\frac{2}{5} — answered the opposite question. Note it and the correct answer add to 1.
  • H, 23\frac{2}{3} — compared green with the other two colours rather than with the whole bag. The denominator of a probability is everything possible.
  • J, 13\frac{1}{3} — assumed the three colours are equally likely because there are three of them. They are not; the counts differ.

Takeaway: the denominator is everything possible. And when a question says "not", the complement is usually the faster route: 1P(A)1 - P(A).


Q7Medium

Topic: Mutually exclusive events

Two events AA and BB are mutually exclusive. The probability of AA is 0.30.3 and the probability of BB is 0.450.45. What is the probability that AA or BB occurs?

A) 27200\frac{27}{200}

B) 320\frac{3}{20}

C) 34\frac{3}{4}

D) 11

Show the worked solution

Answer: C

Explanation

Mutually exclusive means the two events never overlap, so nothing gets counted twice and the probabilities simply add:

P(A or B)=0.3+0.45=0.75P(A \text{ or } B) = 0.3 + 0.45 = 0.75

Note the answer is below 1, as it must be. And it does not have to equal 1 — there is a 0.250.25 chance that neither happens.

Why each wrong option is wrong:

  • A, 27200\frac{27}{200} — multiplied. Multiplying is for "and" with independent events, and here "and" is impossible: mutually exclusive events have P(A and B)=0P(A \text{ and } B) = 0.
  • B, 320\frac{3}{20} — subtracted.
  • D, 11 — assumed the two events cover every possibility.

Takeaway: mutually exclusive means add for "or". Independent means multiply for "and". Mutually exclusive is not the same as covering everything.


Q8Hard

Topic: Two draws without replacement

A bag holds 3 red counters and 5 blue counters. One counter is drawn and not replaced, then a second is drawn. What is the probability that both are red?

F) 964\frac{9}{64}

G) 34\frac{3}{4}

H) 328\frac{3}{28}

J) 38\frac{3}{8}

Show the worked solution

Answer: H

Explanation

The first draw: 3 red out of 8 counters.

P(first red)=38P(\text{first red}) = \frac{3}{8}

Now both totals have changed. One red has gone, so there are 2 reds left out of 7 counters.

P(second red)=27P(\text{second red}) = \frac{2}{7}

"And" with these two steps means multiply:

38×27=656=328\frac{3}{8} \times \frac{2}{7} = \frac{6}{56} = \frac{3}{28}

Why each wrong option is wrong:

  • F, 964\frac{9}{64} — used 38\frac{3}{8} twice, which is the answer for drawing with replacement.
  • G, 34\frac{3}{4} — added the counts.
  • J, 38\frac{3}{8} — gave the first draw only.

Takeaway: without replacement, both the numerator and the denominator drop by one for the second draw. Write the two fractions side by side before multiplying.


Section 4 — Samples and Studies

A short but reliable topic. Two rules cover almost every question:

  • A conclusion may only be drawn about the population that was sampled, and only if the sample was random.
  • An observational study can show that two things go together. Only a controlled experiment, with random assignment, can show that one causes the other.

Q9Hard

Topic: What an observational study can conclude

Researchers recorded the sleep and the test scores of 500 randomly selected students at one university, and found that students who slept longer tended to score higher. No one was told how much to sleep; the researchers only recorded what was already happening. Which conclusion is best supported?

A) Higher scores cause students to sleep longer

B) Sleeping longer causes students to score higher

C) More sleep is associated with higher scores among students at this university

D) More sleep is associated with higher scores among all adults

Show the worked solution

Answer: C

Explanation

Two limits apply, and both matter.

No cause. Nobody was assigned an amount of sleep — the researchers only watched. So this is an observational study, and it can establish that two things move together, not that one produces the other. Something else could explain both: a student who is organised may sleep more and study more.

No wider population. The sample was random, but only within one university. The conclusion can extend to that university's students and no further.

Option A claims association, and only for students at this university. Both limits respected.

Why each wrong option is wrong:

  • B, Sleeping longer causes students to score higher and A, Higher scores cause students to sleep longer — claim causation in one direction or the other. Neither is available from an observational study, and the fact that the two opposite claims are equally consistent with the data is itself the argument.
  • D, More sleep is associated with higher scores among all adults — extends the finding to all adults. The sample was random within one university, so that is as far as it reaches.

Takeaway: observational studies give association; only a randomised experiment gives causation. And a conclusion never travels beyond the population that was actually sampled.


Q10Medium

Topic: Counting arrangements

A café offers 4 kinds of bread, 3 fillings and 2 sauces, and a sandwich is made by choosing exactly one of each. Each choice is made independently of the others. How many different sandwiches are possible?

F) 1414

G) 99

H) 1212

J) 2424

Show the worked solution

Answer: J

Explanation

Each bread can go with each filling, and each of those pairs with each sauce, so the choices multiply:

4×3×2=244 \times 3 \times 2 = 24

Why it multiplies rather than adds: for each of the 4 breads there are 3 fillings, giving 12 combinations; for each of those 12 there are 2 sauces, giving 24. Adding would answer "how many ingredients are there altogether", which is a different question.

Why each wrong option is wrong:

  • G, 99 — added the three counts.
  • H, 1212 — multiplied two of the three and forgot the sauces.
  • F, 1414 — mixed multiplying and adding.

Takeaway: independent choices multiply. If the question says "one of each", multiply the counts together.


Q11Medium

Topic: The range and the spread of a data set

Two sets of five readings are recorded. Set X is 40,  41,  42,  43,  4440, \; 41, \; 42, \; 43, \; 44 and set Y is 10,  25,  42,  60,  7310, \; 25, \; 42, \; 60, \; 73. Both have the same median. Which statement about their spread is true?

A) Neither range nor standard deviation can be compared here

B) Set X has the greater range but set Y the greater standard deviation

C) The two sets have the same range, because they share a median

D) Set Y has both the greater range and the greater standard deviation

Show the worked solution

Answer: D

Explanation

Both medians are 42, which tells you nothing about spread.

Set X runs from 40 to 44, a range of 4. Set Y runs from 10 to 73, a range of 63. Y's values sit far further from their mean, so its standard deviation is greater too.

Why each wrong option is wrong:

  • B, Set X has the greater range but set Y the greater standard deviation — has the range the wrong way round.
  • C, The two sets have the same range, because they share a median — a shared median says nothing about range. That is precisely why both statistics are reported.
  • A, Neither range nor standard deviation can be compared here — both are perfectly comparable from the values given.

Takeaway: median describes the centre; range and standard deviation describe the spread. Two sets can share one and differ wildly in the other.


Q12Hard

Topic: What a random sample supports

A researcher surveys 300 randomly selected students at one university and finds that 42% support a proposed change. The sample was random, but it was drawn from a single university. Which conclusion is best supported?

F) About 42% of students at that university support the change

G) About 42% of students nationwide support the change

H) Exactly 42% of students at that university support the change

J) The survey shows the change will improve student results

Show the worked solution

Answer: F

Explanation

Three limits, and each rules out one option.

The population. The sample was random within one university, so the conclusion reaches that university's students and no further.

Estimate, not certainty. A sample gives an estimate with a margin of error around it. "About 42%" is right; "exactly 42%" is not.

Opinion is not outcome. The survey asked what students support. It says nothing about whether the change would work.

Why each wrong option is wrong:

  • G, About 42% of students nationwide support the change — extends beyond the population sampled.
  • H, Exactly 42% of students at that university support the change — claims exactness a sample cannot give.
  • J, The survey shows the change will improve student results — turns a question about opinion into a claim about results.

Takeaway: a random sample supports an estimate about the population it was drawn from, and nothing else.


Q13Medium

Topic: An experiment versus an observational study

A researcher wants to establish whether a new revision method causes better test results. Which study design would support that conclusion?

A) Survey teachers about which method they believe works better

B) Ask students which method they used and compare their results

C) Compare results at a school that adopted the method with one that did not

D) Randomly assign students to use the new method or the old one, then compare results

Show the worked solution

Answer: D

Explanation

Causation requires a controlled experiment: the researcher assigns the treatment, and assigns it at random.

Random assignment is what makes the two groups alike in every other respect — motivation, prior attainment, hours available — so that any difference in results can be attributed to the method itself.

Why each wrong option is wrong:

  • B, Ask students which method they used and compare their results — students chose their own method, so the groups may differ in ways that also affect results. That is an observational study: it shows association, not cause.
  • C, Compare results at a school that adopted the method with one that did not — the schools differ in many ways besides the method.
  • A, Survey teachers about which method they believe works better — measures belief, not effect.

Takeaway: assigned at random means an experiment, and only an experiment supports a claim about cause. Anything self-selected gives association only.


Q14Medium

Topic: Reading a line of best fit

A clinic recorded the age and height of a group of children and drew a line of best fit through the resulting scatterplot. The line has equation y=6x+78y = 6x + 78, where xx is age in years and yy is height in centimetres. Which is the best interpretation of the number 6?

F) The height of a child at birth, in centimetres

G) The average increase in height, in centimetres, per year of age

H) The age at which a child reaches 6 centimetres

J) The number of children in the study

Show the worked solution

Answer: G

Explanation

In y=mx+cy = mx + c the coefficient of xx is the slope: how much yy changes for each one-unit increase in xx.

Here one extra year of age corresponds to 6 more centimetres of height, on average.

Why each wrong option is wrong:

  • F, The height of a child at birth, in centimetres — that is the 78, the value where the line meets the vertical axis.
  • H, The age at which a child reaches 6 centimetres and J, The number of children in the study — neither quantity appears in the model.

Takeaway: in a line of best fit, the slope is a rate of change and the intercept is the value at zero. Say the units out loud: "6 centimetres per year".


Q15Medium

Topic: Comparing an actual value with a predicted one

A line of best fit through a scatterplot is y=2x+5y = 2x + 5. One of the data points used to draw it is (4,15)(4, 15). How does the actual value compare with the predicted one?

A) 2-2

B) 22

C) 1313

D) 00

Show the worked solution

Answer: B

Explanation

The line predicts

y=2(4)+5=13y = 2(4) + 5 = 13

The actual value is 15, so the point sits 2 above the line:

1513=215 - 13 = 2

That difference is called the residual. A positive residual means the line under-predicted.

Why each wrong option is wrong:

  • A, 2-2 — subtracted actual from predicted, giving the residual's negative.
  • C, 1313 — gave the prediction rather than the difference.
  • D, 00 — assumed the point lies exactly on the line. A line of best fit rarely passes through any particular point.

Takeaway: residual == actual - predicted. Positive means the point is above the line.


Q16Hard

Topic: What a scatterplot's association does and does not show

A scatterplot of 40 towns shows that towns with more bookshops tend to have higher average incomes, and a line of best fit slopes upward. Which conclusion is best supported by the scatterplot alone?

F) Among these towns, more bookshops is associated with higher average income

G) Opening more bookshops would raise a town's average income

H) Higher average income causes towns to open more bookshops

J) Bookshop numbers and income are unrelated in these towns

Show the worked solution

Answer: F

Explanation

An upward-sloping line of best fit shows association: the two quantities tend to rise together across these towns.

It cannot show cause, in either direction. Nobody opened bookshops experimentally to see what happened; the data was only observed. A third factor — population size, say — could easily drive both.

Why each wrong option is wrong:

  • G, Opening more bookshops would raise a town's average income and H, Higher average income causes towns to open more bookshops — claim causation, in opposite directions. That both are equally consistent with the same scatterplot is the argument against either.
  • J, Bookshop numbers and income are unrelated in these towns — contradicts the upward slope.

Takeaway: a scatterplot shows association. Only a randomised experiment shows cause, and "these two rise together" never says which one moves the other.


Q17Hard

Topic: Counting arrangements in order

Four runners finish a race and no two finish together. In how many different orders can they finish?

A) 2424

B) 1616

C) 1010

D) 44

Show the worked solution

Answer: A

Explanation

Four choices for first place, then three left for second, two for third, and the last runner takes fourth:

4×3×2×1=244 \times 3 \times 2 \times 1 = 24

The counts shrink by one each time because a runner cannot finish in two places at once — unlike the sandwich question, where each choice was independent and the counts stayed the same.

Why each wrong option is wrong:

  • B, 1616 — used 424^2, which would allow repeats.
  • C, 1010 — added the choices.
  • D, 44 — gave the number of runners.

Takeaway: arrangements of nn distinct things in order come to n×(n1)××1n \times (n-1) \times \dots \times 1. When the same item cannot be reused, the count drops by one at every step.


Q18Medium

Topic: Reading a two-way table for a total

A cinema recorded which of 300 customers bought a snack, split by ticket type.

Bought a snack No snack Total
Adult 84 96 180
Child 78 42 120
Total 162 138 300

What fraction of all customers bought a snack?

F) 715\frac{7}{15}

G) 1320\frac{13}{20}

H) 2750\frac{27}{50}

J) 2723\frac{27}{23}

Show the worked solution

Answer: H

Explanation

"All customers" means the grand total, 300, goes underneath. The snack column totals 162.

162300=2750\frac{162}{300} = \frac{27}{50}

Why each wrong option is wrong:

  • F, 715\frac{7}{15} and G, 1320\frac{13}{20} — restricted the question to one row. Neither was asked for.
  • J, 2723\frac{27}{23} — compared the two column totals with each other rather than with the whole.

Takeaway: the denominator is whatever group the question names. "All" means the grand total; "of the adults" would mean 180.


Q19Medium

Topic: Reading a frequency table for a median

A shop recorded how many items each customer bought during one hour.

Items bought 1 2 3 4
Number of customers 6 9 4 1

What is the median number of items bought?

A) 52\frac{5}{2}

B) 22

C) 99

D) 32\frac{3}{2}

Show the worked solution

Answer: B

Explanation

There are 6+9+4+1=206 + 9 + 4 + 1 = 20 customers, so the median sits between the 10th and the 11th when they are lined up in order.

Counting along: customers 1–6 bought 1 item, customers 7–15 bought 2. Both the 10th and the 11th fall in that second group, so both bought 2 items, and the median is 2.

Why each wrong option is wrong:

  • A, 52\frac{5}{2} — averaged the numbers on the top row, ignoring how many customers gave each.
  • C, 99 — gave the largest frequency, which counts customers, not items.
  • D, 32\frac{3}{2} — took a middle value of the wrong list.

Takeaway: in a frequency table, the median is found by counting along the frequencies to the middle position, not by looking at the top row.


Q20Hard

Topic: Reading a bar chart described in words

A survey of 50 households recorded the number of cars owned. Twelve households owned no car, twenty owned one, fourteen owned two, and four owned three. What is the mean number of cars per household?

F) 11

G) 65\frac{6}{5}

H) 32\frac{3}{2}

J) 5050

Show the worked solution

Answer: G

Explanation

Total cars owned:

0(12)+1(20)+2(14)+3(4)=0+20+28+12=600(12) + 1(20) + 2(14) + 3(4) = 0 + 20 + 28 + 12 = 60

Households: 50.

mean=6050=1.2\text{mean} = \frac{60}{50} = 1.2

Why each wrong option is wrong:

  • F, 11 — gave the mode, the most common answer, which is 1.
  • H, 32\frac{3}{2} — averaged 0,1,2,30, 1, 2, 3 without weighting by how many households gave each.
  • J, 5050 — gave the number of households.

Takeaway: multiply each value by its frequency, add, then divide by the total frequency. A mean need not be a whole number even when every value is.


Q21Medium

Topic: Predicting from a line of best fit

A scatterplot of temperature against ice-cream sales has the line of best fit y=12x40y = 12x - 40, where xx is the temperature in degrees and yy is the number of sales. What does the line predict for a day at 15 degrees?

A) 180180

B) 40-40

C) 140140

D) 1212

Show the worked solution

Answer: C

Explanation

Substitute x=15x = 15 into the line:

y=12(15)40=18040=140y = 12(15) - 40 = 180 - 40 = 140

Why each wrong option is wrong:

  • A, 180180 — multiplied and forgot to subtract.
  • B, 40-40 — gave the intercept, which is the prediction at zero degrees — and note it is negative, which is why a line of best fit should not be trusted far outside the data it came from.
  • D, 1212 — gave the slope.

Takeaway: a line of best fit predicts by substitution. The intercept is the prediction at zero, which is often physically meaningless.


Q22Medium

Topic: Interpreting the sign of a slope in a scatterplot

A scatterplot of a car's age against its value has a line of best fit with a negative slope. Which statement does that support?

F) Older cars tend to be worth less

G) Older cars tend to be worth more

H) Age and value are unrelated

J) Every older car is worth less than every newer one

Show the worked solution

Answer: F

Explanation

A negative slope means that as xx increases, yy decreases. Here xx is age and yy is value, so older cars tend to be worth less.

The word "tend" is doing real work. A line of best fit describes a trend across the whole scatterplot; individual points sit above and below it.

Why each wrong option is wrong:

  • G, Older cars tend to be worth more — describes a positive slope.
  • H, Age and value are unrelated — a slope of zero would mean no relationship; this one is negative.
  • J, Every older car is worth less than every newer one — far too strong. A trend is not a rule about every pair of cars, and a scatterplot with a clear negative slope will still contain an old car worth more than some newer one.

Takeaway: the sign of the slope gives the direction of the trend. "Tends to" is right; "always" is not.


Q23Hard

Topic: The residual of a data point

A line of best fit is y=4x+2y = 4x + 2. One observed data point is (3,11)(3, 11). What is the residual for that point?

A) 3-3

B) 33

C) 1414

D) 1111

Show the worked solution

Answer: A

Explanation

The line predicts

y=4(3)+2=14y = 4(3) + 2 = 14

The actual value is 11, so

residual=1114=3\text{residual} = 11 - 14 = -3

A negative residual means the point lies below the line — the model over-predicted.

Why each wrong option is wrong:

  • B, 33 — subtracted predicted from actual the wrong way round. The sign carries the meaning, so the order matters.
  • C, 1414 and D, 1111 — gave one of the two values rather than their difference.

Takeaway: residual = actual − predicted. Negative means the point sits below the line.


Q24Medium

Topic: Counting with a restriction

A four-digit code is made from the digits 1 to 9, and no digit may be repeated. How many different codes are possible?

F) 65616561

G) 3636

H) 2424

J) 30243024

Show the worked solution

Answer: J

Explanation

Nine digits for the first position; once one is used, eight remain for the second, then seven, then six:

9×8×7×6=30249 \times 8 \times 7 \times 6 = 3024

Why each wrong option is wrong:

  • F, 65616561949^4 allows repeats, which the question forbids.
  • G, 3636 — multiplied the digit count by the number of positions.
  • H, 2424 — counted the orderings of four fixed items, ignoring the nine digits to choose from.

Takeaway: when items cannot repeat, the count drops by one at each step. When they can, it stays the same.


Q25Medium

Topic: Counting choices that combine

A meal deal offers one of 5 mains, one of 4 sides and one of 3 drinks. Each choice is made independently of the others. How many different meals are possible?

A) 1212

B) 2323

C) 6060

D) 1515

Show the worked solution

Answer: C

Explanation

Every main can go with every side, and each of those pairs with every drink, so the counts multiply:

5×4×3=605 \times 4 \times 3 = 60

The counts stay the same at each step here, because choosing a main does not use up any side.

Why each wrong option is wrong:

  • A, 1212 — added, which answers "how many items are on the menu".
  • B, 2323 — mixed multiplying and adding.
  • D, 1515 — left out one of the three choices.

Takeaway: independent choices multiply. Compare with the previous question: there the counts shrank, because a digit once used was gone.


Q26Hard

Topic: Counting arrangements with a fixed position

Five books are placed in a row on a shelf, and one particular book must go on the far left. The remaining four can go anywhere among the other positions. In how many orders can the books be arranged?

F) 120120

G) 2020

H) 2424

J) 55

Show the worked solution

Answer: H

Explanation

The far-left position is fixed, so there is only one choice for it. That leaves four books for four positions:

1×4×3×2×1=241 \times 4 \times 3 \times 2 \times 1 = 24

Why each wrong option is wrong:

  • F, 1201205!=1205! = 120 counts every arrangement, including the ones with the wrong book on the left.
  • G, 2020 — counted only the first two positions.
  • J, 55 — gave the number of books.

Takeaway: handle the restricted position first, then count the free ones. A fixed position contributes a factor of 1.


Q27Medium

Topic: What makes a sample representative

A school wants to estimate how much time its 900 students spend on homework. Four methods of choosing 100 students are proposed. Which is most likely to give a representative sample?

A) Pick 100 students at random from the full list of all 900

B) Ask the first 100 students who arrive at school one morning

C) Ask 100 students in the library after school

D) Ask for 100 volunteers to fill in a form

Show the worked solution

Answer: A

Explanation

A sample is representative when every member of the population has the same chance of being chosen. Only a random draw from the full list of 900 does that.

The other three are self-selected, and each in a way that is plausibly linked to the very thing being measured — which is what makes the bias serious rather than merely untidy.

Why each wrong option is wrong:

  • B, Ask the first 100 students who arrive at school one morning — students who arrive early may be more organised, and so may do more homework.
  • C, Ask 100 students in the library after school — students in the library after school are almost certainly doing homework.
  • D, Ask for 100 volunteers to fill in a form — volunteers choose themselves, and people with strong feelings volunteer more.

Takeaway: representative means randomly selected from the whole population. Ask whether the way people ended up in the sample is connected to what is being measured.


Q28Hard

Topic: How far a conclusion may travel

A researcher randomly samples 200 adults in one city and finds that 38% cycle to work. Which conclusion is best supported?

F) About 38% of adults in that city cycle to work

G) About 38% of adults in the country cycle to work

H) Exactly 38% of adults in that city cycle to work

J) Cycling to work makes adults healthier

Show the worked solution

Answer: F

Explanation

Three limits, one per wrong option.

The population. The sample was random within one city, so the conclusion reaches that city and no further.

Estimate, not certainty. A sample gives an estimate surrounded by a margin of error. "About 38%" is supportable; "exactly 38%" is not.

Behaviour is not effect. The survey recorded what people do, not what it does to them.

Why each wrong option is wrong:

  • G, About 38% of adults in the country cycle to work — extends beyond the population sampled.
  • H, Exactly 38% of adults in that city cycle to work — claims a precision a sample cannot give.
  • J, Cycling to work makes adults healthier — a claim about health from a survey about travel.

Takeaway: a random sample supports an estimate about the population it was drawn from. Check both of those before accepting a conclusion.


Q29Medium

Topic: Why a control group is needed

A company tests a new training course by measuring employees' scores before and after it, and finds the scores went up. To claim the course caused the rise, what does the study most need?

A) A comparison group of similar employees who did not take the course

B) A larger number of employees taking the course

C) A second test of the same employees after the course

D) A survey asking employees whether they enjoyed the course

Show the worked solution

Answer: A

Explanation

Scores rising after a course is consistent with the course working — and equally consistent with employees simply getting better at their jobs over the same period, or getting better at the test itself.

A comparison group of similar employees who did not take the course separates those explanations: if the course group improved more, the course is the plausible reason.

Why each wrong option is wrong:

  • B, A larger number of employees taking the course — a bigger group makes the same flawed comparison more precisely. Size fixes noise, not bias.
  • C, A second test of the same employees after the course — retesting the same people adds no comparison.
  • D, A survey asking employees whether they enjoyed the course — enjoyment is not attainment.

Takeaway: to claim cause, you need something to compare against. "Before and after" on its own cannot rule out everything else that happened in between.