ACT Math: Statistics and Probability (Drill 1)

Question 1. A student scored 82, 91, 75, 88, and 74 on five tests. What is the mean of the student's test scores?

• A) 82 ✓
• B) 82.5
• C) 83
• D) 85

Explanation: Mean = sum ÷ count = (82 + 91 + 75 + 88 + 74) ÷ 5 = 410 ÷ 5 = 82. Choice B results from an arithmetic error in the sum: computing 82 + 91 + 75 + 88 + 74 = 412 rather than 410, then dividing by 5 to get 82.4 ≈ 82.5. Choice C results from computing the sum as 415 ÷ 5 = 83. Choice D results from averaging only the two highest scores (91 + 88) ÷ 2... or from computing (82 + 91 + 75 + 88 + 74) ÷ 4 = 102.5, then rounding, a common error of dividing by the wrong count.

Question 2. The ages of seven students in a tutoring group are: 14, 16, 14, 17, 15, 16, 16. What are the median and mode of this data set?

• A) Median = 16; Mode = 16 ✓
• B) Median = 15; Mode = 14
• C) Median = 15.5; Mode = 16
• D) Median = 16; Mode = 14

Explanation: First sort the data: 14, 14, 15, 16, 16, 16, 17. With 7 values, the median is the 4th value: 16. The mode is the value that appears most often: 16 appears 3 times (more than any other value). Median = 16, Mode = 16. Choice B gives the wrong median (the 3rd value, 15, instead of the 4th) and names 14 as the mode even though 14 appears only twice versus 16 appearing three times. Choice C gives a median of 15.5, which would be correct for an even number of values but not for 7 (odd). Choice D has the correct median but names 14 as the mode.

Question 3. A bag contains 5 red marbles, 3 blue marbles, and 7 green marbles. If one marble is selected at random, what is the probability that it is NOT green?

• A) 7/15
• B) 8/15 ✓
• C) 1/3
• D) 2/5

Explanation: Total marbles = 5 + 3 + 7 = 15. Non-green marbles = 5 + 3 = 8. P(not green) = 8/15. Choice A is the probability of selecting a green marble (7/15), not the complement. Choice C results from computing only the blue marbles over the total: 3/15 = 1/5... or from misreading 5/15 = 1/3, using only red marbles. Choice D results from computing 8/20 = 2/5, using an incorrect total of 20.

Question 4. The table below shows the results of a survey of 200 students about their preferred lunch option.
Pizza  Salad  Sandwich  Total
Grade 10    38     22       20      80
Grade 11    30     28       32      90
Grade 12    12     10        8      30
Total        80     60       60     200
If a student is selected at random from those who prefer pizza, what is the probability that the student is in Grade 11?

• A) 3/20
• B) 3/8 ✓
• C) 9/20
• D) 1/3

Explanation: This is a conditional probability question: given that the student prefers pizza, what is the probability they are in Grade 11? The condition restricts us to the pizza column only (total = 80). Grade 11 pizza students = 30. P(Grade 11 | pizza) = 30/80 = 3/8. Choice A results from using the overall total (200) as the denominator instead of the pizza column total (80): 30/200 = 3/20. Choice C results from using the Grade 11 row total (90) as the denominator: 30/90 = 1/3... that is Choice D. Choice C (9/20) results from using a denominator of 80 but reading 36 from the table instead of 30: 36/80 = 9/20. Choice D results from computing 30/90, dividing by the Grade 11 total rather than the pizza total.

Question 5. Two classes each took the same 100-point test. Class A had a mean score of 74 with a standard deviation of 3. Class B had a mean score of 74 with a standard deviation of 11. Which of the following best describes the difference between the two classes' results?

• A) Class A performed better overall than Class B
• B) Class B performed better overall than Class A
• C) The scores in Class A were more consistent than the scores in Class B ✓
• D) Class B had a higher median score than Class A

Explanation: Both classes have the same mean (74), so neither performed better overall; Choices A and B are incorrect. Standard deviation measures the spread of scores around the mean. A smaller standard deviation means scores are clustered more tightly around the mean (more consistent). Class A's standard deviation of 3 is much smaller than Class B's 11, so Class A's scores were more consistent. Choice D is incorrect because standard deviation tells us nothing about the median, two distributions can have the same mean and different standard deviations while having the same or different medians.