Drill 2 · Math · Statistics and Probability
ACT Math: Statistics and Probability (Drill 2) is a Math practice drill covering Statistics and Probability. It contains 5 original questions created by Brian Stewart, a Barron's test prep author with over 20 years of tutoring experience.
This drill covers range and the effect of outliers on measures of center, the fundamental counting principle, compound probability using AND and OR rules, expected value, and counting with permutations and combinations.
Question 1. The daily high temperatures (°F) for a week were: 72, 75, 71, 73, 74, 70, 98. Which of the following statements is true?
Explanation: Sorted data: 70, 71, 72, 73, 74, 75, 98. Mean = (70 + 71 + 72 + 73 + 74 + 75 + 98) ÷ 7 = 533 ÷ 7 = 76.1. Median = the 4th value (middle of 7) = 73. Since 76.1 > 73, the mean is greater than the median. The outlier (98) pulls the mean up but does not affect the median. Choice A is false: mean ≈ 76.1 ≠ 73. Choice C is false: the median (73) is less than the mean (76.1), not greater. Choice D: there is no mode since all values appear exactly once, so this statement cannot be evaluated as true.
Question 2. A restaurant offers a fixed-price dinner with 4 choices of appetizer, 6 choices of entrée, and 3 choices of dessert. How many different complete dinners are possible?
Explanation: By the fundamental counting principle, multiply the number of choices at each stage: 4 × 6 × 3 = 72. Choice A results from adding the choices instead of multiplying: 4 + 6 + 3 = 13. Choice B results from multiplying only two of the three categories: 6 × 3 = 18... or 4 × 3 × 3 = 36, forgetting one category. Choice D results from computing 4 × 6 × 6 = 144, doubling the dessert count by mistake.
Question 3. A fair coin is flipped and a standard six-sided die is rolled. What is the probability of getting heads on the coin AND rolling a number greater than 4 on the die?
Explanation: The coin flip and die roll are independent events. P(heads) = 1/2. Numbers greater than 4 on a six-sided die are 5 and 6, so P(greater than 4) = 2/6 = 1/3. For independent events, P(A and B) = P(A) × P(B) = (1/2)(1/3) = 1/6. Choice A gives only P(die > 4) = 1/3, ignoring the coin entirely. Choice C results from computing P(heads) + P(die > 4) = 1/2 + 1/3 = 5/6... or from computing 1 − P(not heads) − P(die ≤ 4): 1 − 1/2 + 1/6 = 2/3, misapplying the OR rule. Choice D results from computing P(heads) × P(die > 4) using P(die > 4) = 1/2 instead of 1/3: (1/2)(1/2) = 1/4.
Question 4. A carnival game costs $2 to play. A player wins $10 with probability 1/10, wins $3 with probability 1/5, and wins nothing otherwise. What is the expected net gain (or loss) per play?
Explanation: Expected gross winnings = (10)(1/10) + (3)(1/5) + (0)(remaining) = 1 + 0.60 + 0 = $1.60. The cost to play is $2, so the expected net gain = $1.60 − $2.00 = −$0.40. On average, a player loses $0.40 per play. Choice B results from computing the expected gross winnings ($1.60) and forgetting to subtract the $2 cost. Choice C results from computing only the $10 prize component: (10)(1/10) − 2 = 1 − 2 = −$1, then adjusting incorrectly to −$1.40. Choice D is the expected gross winnings ($1.60) reported as the net gain without subtracting the cost.
Question 5. A school club has 10 members. The club needs to select a committee of 3 members. How many different committees are possible?
Explanation: Order does not matter for a committee, so use combinations: C(10, 3) = 10! / (3! × 7!) = (10 × 9 × 8) / (3 × 2 × 1) = 720 / 6 = 120. Choice A results from computing 10 × 3 = 30, incorrectly multiplying the number of members by the committee size. Choice C gives the permutation P(10, 3) = 10 × 9 × 8 = 720, the number of ordered arrangements, without dividing by 3! to account for the fact that order does not matter in a committee. Choice D results from computing 103 = 1,000, treating each slot as an independent choice with replacement.