Statistics & Probability Domain Test

Calculator is allowed on all questions. Figures are not necessarily drawn to scale.

Question 1

What is the mean of the data set \(\{3,\ 7,\ 9,\ 12,\ 14\}\)?




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The mean is the sum of all values divided by the count:

\[\text{Mean} = \frac{3 + 7 + 9 + 12 + 14}{5} = \frac{45}{5} = 9\]

    1. \(7\): This is the second value in the list, not the mean — the student may have reported the median of a different ordering or simply read off a value.
    1. \(10\): Added the five values correctly to get \(45\) but divided by \(4.5\) instead of \(5\).
    1. \(12\): This is the fourth value in the list, not the mean.

Answer: B

Question 2

A bag contains \(4\) red marbles, \(6\) blue marbles, and \(5\) green marbles. If one marble is chosen at random, what is the probability that it is blue?




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Total marbles: \(4 + 6 + 5 = 15\).

\[P(\text{blue}) = \frac{6}{15} = \frac{2}{5}\]

    1. \(\dfrac{2}{15}\): Divided the number of blue marbles by the wrong total — used \(45\) (the product of the three counts) rather than the sum.
    1. \(\dfrac{1}{3}\): Computed \(\dfrac{6}{18}\) — added an extra \(3\) to the total, or confused with \(\dfrac{5}{15}\).
    1. \(\dfrac{3}{5}\): Computed the probability of NOT choosing blue: \(\dfrac{9}{15} = \dfrac{3}{5}\).

Answer: H

Question 3

The two-way table below shows the class period preferences of \(40\) students surveyed.

Morning Afternoon Total
Math 12 8 20
English 9 11 20
Total 21 19 40

How many students prefer Math class?




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The row total for Math gives the number of students who prefer Math class, regardless of period:

\[12 + 8 = 20\]

    1. \(8\): Read only the afternoon cell for Math — ignored the morning Math students.
    1. \(12\): Read only the morning cell for Math — ignored the afternoon Math students.
    1. \(40\): This is the grand total (all students surveyed), not just the Math students.

Answer: C

Question 4

What is the median of the data set \(\{5,\ 8,\ 11,\ 14,\ 19,\ 23\}\)?




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The data set is already ordered. With \(6\) values (an even count), the median is the average of the 3rd and 4th values:

\[\text{Median} = \frac{11 + 14}{2} = \frac{25}{2} = 12.5\]

    1. \(11\): Reported the 3rd value only — did not average the two middle values.
    1. \(13\): Computed the mean of all six values: \((5+8+11+14+19+23)/6 = 80/6 \approx 13.3\), or added the two middle values without dividing: \(11 + 14 = 25\), then… arithmetic slip.
    1. \(14\): Reported the 4th value only — did not average the two middle values.

Answer: G

Question 5

A scatterplot shows the number of hours students studied and their test scores. As the number of study hours increases, the test scores generally increase as well. Which of the following best describes the association shown?




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When both variables increase together, the association is positive. As study hours go up, test scores go up — this is a positive association.

    1. No association: This would mean the data points show no discernible pattern (random scatter). The problem states a clear increasing trend.
    1. Negative association: This would mean one variable increases as the other decreases — the opposite of what is described.
    1. Causation: A scatterplot alone cannot establish causation. A positive association is an observed correlation, not proof that studying directly causes higher scores.

Answer: C

Question 6

A student is getting dressed and can choose from \(4\) shirts, \(3\) pairs of pants, and \(2\) pairs of shoes. Assuming one of each is chosen, how many different outfits are possible?




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By the Fundamental Counting Principle, multiply the number of choices at each step:

\[4 \times 3 \times 2 = 24\]

    1. \(9\): Added all the options: \(4 + 3 + 2 = 9\) — used addition instead of multiplication.
    1. \(12\): Multiplied only two of the three categories: \(4 \times 3 = 12\) — forgot to include the shoes.
    1. \(18\): Multiplied \(3 \times 3 \times 2 = 18\) — misread \(4\) shirts as \(3\).

Answer: J

Question 7

A fair six-sided die is rolled once. What is the probability of NOT rolling a \(3\)?




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The complement rule: \(P(\text{not } A) = 1 - P(A)\).

\[P(\text{not rolling a 3}) = 1 - \frac{1}{6} = \frac{5}{6}\]

Alternatively, five of the six outcomes (1, 2, 4, 5, 6) are not a 3.

    1. \(\dfrac{1}{6}\): This is the probability of rolling a 3 — the student found \(P(A)\) instead of \(P(\text{not }A)\).
    1. \(\dfrac{1}{3}\): Computed \(\dfrac{2}{6}\) — counted two “not 3” outcomes instead of five.
    1. \(\dfrac{1}{2}\): Treated the event as a 50/50 — incorrectly assumed the die has only two equally likely outcomes.

Answer: D

Question 8

Cameron scored \(85\), \(90\), and \(78\) on three equally weighted tests. He then took a final exam that counts as two test grades. His final exam score was \(92\). What is Cameron’s overall average across all five test grades?




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Since the final exam counts as two test grades, it is entered twice in the average:

\[\text{Average} = \frac{85 + 90 + 78 + 92 + 92}{5} = \frac{437}{5} = 87.4\]

    1. \(85.4\): Averaged only the four distinct scores without doubling the final: \((85+90+78+92)/4 = 345/4 = 86.25\)… this matches G, not F. F = averaged all four and divided by something else — likely a computational error summing to \(427\) then dividing by \(5\).
    1. \(86.25\): Averaged the four scores without doubling the final: \((85+90+78+92)/4 = 86.25\) — did not account for the final counting twice.
    1. \(88.2\): Doubled the final exam score in the numerator but also added \(1\) to the denominator incorrectly, or used a different weighting scheme.

Answer: H

Question 9

A standard deck of \(52\) cards is shuffled. One card is drawn, recorded, and returned to the deck before a second card is drawn. What is the probability that both cards drawn are aces?




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Since the card is replaced, the two draws are independent events. There are \(4\) aces in \(52\) cards:

\[P(\text{ace}) = \frac{4}{52} = \frac{1}{13}\]

\[P(\text{both aces}) = \frac{1}{13} \times \frac{1}{13} = \frac{1}{169}\]

    1. \(\dfrac{1}{52}\): Computed \(\dfrac{4}{52} \times \dfrac{1}{4}\) — only partially multiplied the probabilities.
    1. \(\dfrac{2}{52}\): Added the two probabilities instead of multiplying: \(\dfrac{1}{13} + \dfrac{1}{13} = \dfrac{2}{13}\), then reduced incorrectly.
    1. \(\dfrac{4}{169}\): Multiplied \(\dfrac{4}{52} \times \dfrac{4}{52} = \dfrac{16}{2704} = \dfrac{1}{169}\) but wrote \(\dfrac{4}{169}\) — forgot to simplify \(\dfrac{16}{2704}\) correctly, or multiplied numerators incorrectly as \(4 \times 1\).

Answer: A

Question 10

The two-way table below shows the reading preferences of \(50\) students, broken down by gender.

Fiction Nonfiction Total
Female 18 7 25
Male 12 13 25
Total 30 20 50

Given that a randomly selected student is female, what is the probability that she prefers fiction?




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This is a conditional probability question. The condition is “given female,” so the sample space is restricted to the \(25\) female students only:

\[P(\text{fiction} \mid \text{female}) = \frac{18}{25}\]

    1. \(\dfrac{9}{25}\): Used the nonfiction count for females (\(7\)) and misread it as \(9\), or confused with another cell — not based on the correct cell value.
    1. \(\dfrac{18}{50}\): Used the grand total (\(50\)) as the denominator instead of restricting to female students (\(25\)).
    1. \(\dfrac{30}{50}\): Gives the overall probability of preferring fiction across all students — did not apply the given condition.

Answer: H

Question 11

A line of best fit for a set of data is given by the equation \(y = 2x + 3\). According to this equation, what is the predicted value of \(y\) when \(x = 8\)?




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Substitute \(x = 8\) into the equation:

\[y = 2(8) + 3 = 16 + 3 = 19\]

    1. \(11\): Computed \(2 + 8 + 3 = 13\)… or substituted \(x = 4\) instead of \(x = 8\): \(2(4) + 3 = 11\).
    1. \(16\): Computed \(2(8) = 16\) but forgot to add the \(y\)-intercept of \(3\).
    1. \(22\): Computed \(2(8) + 3 \times 2 = 16 + 6 = 22\) — doubled the intercept.

Answer: C

Question 12

Which of the following data sets has the greater standard deviation?

Set P: \(\{10,\ 10,\ 10,\ 10,\ 10\}\)

Set Q: \(\{6,\ 9,\ 11,\ 14,\ 15\}\)




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Standard deviation measures how spread out values are from the mean.

  • Set P: All values equal \(10\), so there is zero deviation from the mean. Standard deviation \(= 0\).
  • Set Q: Values range from \(6\) to \(15\), spread around a mean of \(11\). Standard deviation \(> 0\).

Set Q has the greater standard deviation because its values vary from the mean, while Set P has no variation at all.

    1. Set P: All identical values give a standard deviation of \(0\) — the smallest possible standard deviation, not the largest.
    1. Both equal: Set P has \(SD = 0\); Set Q has \(SD > 0\). They cannot be equal.
    1. Cannot be determined: Standard deviation is a conceptual property of spread. Set P’s \(SD = 0\) is certain without calculation; Set Q’s \(SD > 0\) is certain because its values differ.

Answer: G

Question 13

Events \(A\) and \(B\) are independent. The probability of event \(A\) occurring is \(0.4\), and the probability of event \(B\) occurring is \(0.3\). What is the probability that both \(A\) and \(B\) occur?




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For independent events, \(P(A \text{ and } B) = P(A) \times P(B)\):

\[P(A \text{ and } B) = 0.4 \times 0.3 = 0.12\]

    1. \(0.10\): Computed \(0.4 \times 0.3\) incorrectly as \(0.10\) — decimal placement error.
    1. \(0.58\): Used the addition rule for mutually exclusive events: \(P(A) + P(B) - P(A \cap B) \approx 0.4 + 0.3 - 0.12 = 0.58\) — applied the “or” rule for the “and” question.
    1. \(0.70\): Added the two probabilities: \(0.4 + 0.3 = 0.7\) — used addition instead of multiplication.

Answer: B

Question 14

A boxplot for a data set shows the following five-number summary: minimum \(= 5\), \(Q_1 = 14\), median \(= 22\), \(Q_3 = 30\), maximum \(= 41\). What is the interquartile range (IQR) of the data?




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The interquartile range is the difference between the third and first quartiles:

\[\text{IQR} = Q_3 - Q_1 = 30 - 14 = 16\]

    1. \(8\): Computed the distance from the median to \(Q_3\): \(30 - 22 = 8\).
    1. \(14\): Reported \(Q_1\) itself as the IQR rather than computing \(Q_3 - Q_1\).
    1. \(36\): Computed the range (max \(-\) min): \(41 - 5 = 36\) — confused range with IQR.

Answer: H

Question 15

A jar contains \(5\) red marbles and \(3\) blue marbles. Two marbles are drawn at random, one after the other, without replacement. What is the probability that both marbles drawn are red?




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Because the marbles are drawn without replacement, the events are dependent. After one red marble is removed, only \(4\) red marbles remain in a jar of \(7\):

\[P(\text{both red}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}\]

    1. \(\dfrac{5}{32}\): Computed \(\dfrac{5}{8} \times \dfrac{1}{4}\) — used \(\dfrac{1}{4}\) for the second draw instead of \(\dfrac{4}{7}\).
    1. \(\dfrac{15}{56}\): Computed \(\dfrac{5}{8} \times \dfrac{3}{7}\) — used the number of blue marbles (\(3\)) in the numerator of the second draw instead of the remaining red marbles (\(4\)).
    1. \(\dfrac{25}{64}\): Treated the draws as independent (with replacement): \(\dfrac{5}{8} \times \dfrac{5}{8} = \dfrac{25}{64}\) — ignored the “without replacement” condition.

Answer: B

Question 16

A teacher wants to select a committee of \(3\) students from a class of \(8\). The order in which students are chosen does not matter. How many different committees are possible?




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Since order does not matter, use the combination formula:

\[\binom{8}{3} = \frac{8!}{3!\,(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56\]

    1. \(24\): Computed \(8 \times 3 = 24\) — multiplied instead of using the combination formula.
    1. \(40\): An arithmetic error in applying the formula — possibly computed \(\frac{8 \times 7 \times 6}{3 \times 2}\) without dividing by the final \(1\).
    1. \(336\): Computed the permutation \(P(8,3) = 8 \times 7 \times 6 = 336\) — did not divide by \(3!\) to account for the fact that order does not matter.

Answer: H

Question 17

Events \(A\) and \(B\) are mutually exclusive. The probability of event \(A\) is \(0.35\), and the probability of event \(B\) is \(0.45\). What is the probability that event \(A\) or event \(B\) occurs?




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For mutually exclusive events (they cannot both occur at the same time), \(P(A \cap B) = 0\), so the addition rule simplifies to:

\[P(A \text{ or } B) = P(A) + P(B) = 0.35 + 0.45 = 0.80\]

    1. \(0.10\): Subtracted the probabilities: \(0.45 - 0.35 = 0.10\) — applied subtraction with no basis.
    1. \(0.45\): Reported only \(P(B)\), the larger of the two probabilities.
    1. \(1.00\): Assumed mutually exclusive events must cover the entire sample space — that would only be true if \(A\) and \(B\) were also complementary events.

Answer: C

Question 18

A game involves rolling a fair six-sided die once. If a \(6\) is rolled, the player wins \(\$5\). If a \(1\) is rolled, the player loses \(\$2\). For any other outcome, the player wins or loses nothing. What is the expected value of the player’s winnings per roll?




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Expected value = sum of (outcome \(\times\) probability) for all outcomes:

\[E = \left(\frac{1}{6}\right)(5) + \left(\frac{1}{6}\right)(-2) + \left(\frac{4}{6}\right)(0)\]

\[E = \frac{5}{6} - \frac{2}{6} + 0 = \frac{3}{6} = \frac{1}{2} = \$0.50\]

    1. \(\$0.25\): Computed \(\dfrac{5-2}{6 \times 2} = \dfrac{3}{12}\) — divided by \(12\) instead of \(6\).
    1. \(\$1.00\): Computed \(\dfrac{5-2}{3} = 1\) — used \(3\) as the denominator (number of non-zero outcomes) instead of \(6\) (total outcomes).
    1. \(\$3.00\): Computed \(5 - 2 = 3\) without weighting by probability — did not divide by the number of outcomes.

Answer: G

Question 19

A researcher finds a strong positive correlation between the number of fire stations in a city and the number of fires in that city. Which of the following is the most appropriate conclusion?




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Correlation does not imply causation. The most likely explanation is a lurking (confounding) variable: larger cities have more people, which leads to both more fires and more fire stations. The positive correlation is a result of population size driving both variables simultaneously.

    1. Fire stations cause fires: This reverses the likely causal direction and is logically unsound — fire stations exist to suppress fires, not cause them.
    1. Fires cause cities to build stations: While plausible as a partial story, this alone doesn’t explain a sustained positive correlation across all cities at all times. It also mistakes correlation for direct causation.
    1. The correlation coefficient proves a causal relationship: Correlation statistics, regardless of their strength, cannot establish causation — they only describe the degree of linear association.

Answer: C

Question 20

A bar chart displays the test scores of two groups. The \(y\)-axis begins at \(80\) instead of \(0\). Group A scored \(84\) and Group B scored \(88\). In the chart, the bar for Group B appears to be \(3\) times as tall as the bar for Group A. What is the actual ratio of Group B’s score to Group A’s score?




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The visual distortion occurs because the \(y\)-axis starts at \(80\), making the bars represent only the portion above \(80\): Group A’s bar height \(= 84 - 80 = 4\) and Group B’s bar height \(= 88 - 80 = 8\), giving a visual \(2:1\) ratio (the problem states it appears \(3:1\) — the key point is the visual ratio is exaggerated).

Regardless of what the chart looks like, the actual ratio of the scores is:

\[\frac{88}{84} = \frac{22}{21}\]

    1. \(3:1\): This is the misleading visual ratio read off the distorted chart — not the ratio of the actual scores.
    1. \(11:10\): A rough approximation — \(88/84\) does not simplify to \(11/10 = 1.1\); the actual value is \(88/84 = 22/21 \approx 1.048\).
    1. \(88:80\): Compared Group B’s score to the \(y\)-axis starting value rather than to Group A’s score.

Answer: G