ACT Math Mean, Median, Mode, and Range: 16 Practice Problems with Step-by-Step Explanations
Statistics questions on mean, median, mode, and range appear regularly on the ACT Math section, and while the definitions themselves are simple, the ACT tests them through layered scenarios: frequency tables, weighted averages, missing-value problems, and questions about how a single changed data point affects each statistic differently. Mean is the sum of all values divided by how many values there are; median is the middle value when the data is ordered from least to greatest (or the average of the two middle values when there's an even count); mode is the value that appears most often; and range is the difference between the greatest and least value.
Frequency tables show up often in this chapter and can trip students up if they treat the "frequency" column as data values instead of as a count of how many times each value occurred. To find the mean from a frequency table, multiply each value by its frequency, add those products together, and divide by the total frequency (not by the number of rows in the table). To find the median from a frequency table, first find the total number of data points, then count through the frequencies in order to locate the middle position, rather than assuming the median is one of the values listed in the middle row of the table.
A recurring theme in harder ACT problems is understanding how mean, median, and range respond differently when a data set changes. Adding a constant to every value shifts the mean, median, and range predictably, but changing just one value (especially an extreme one) can shift the mean and range while leaving the median completely unaffected. Work through the 16 problems below, including two that use frequency tables, then click to reveal each step-by-step explanation.
On this page:
- Problem 1 — Weighted mean
- Problem 2 — Average of a data set
- Problem 3 — Finding a missing value from the mean
- Problem 4 — Mean of consecutive integers
- Problem 5 — Weighted average, algebraic
- Problem 6 — Mean of a group after a change
- Problem 7 — Mean from a frequency table
- Problem 8 — Median from a frequency table
- Problem 9 — Effect of an error on statistics
- Problem 10 — Median from a frequency table
- Problem 11 — Weighted average rate
- Problem 12 — Weighted average pace
- Problem 13 — Average of consecutive even integers (algebraic)
- Problem 14 — Mean of a modified data set
- Problem 15 — Solving for a needed score
- Problem 16 — Effect on the sum from a mean change
Practice Problems
A certain group consists of 7 girls, 5 of whom are 15 and 2 of whom are 8. What is the mean age of the girls in the group?
- A) 8
- B) 10
- C) 11.5
- D) 13
Mean is the sum of all ages divided by the number of girls. The total age is (5 × 15) + (2 × 8) = 75 + 16 = 91.
Divide by the total number of girls: 91 / 7 = 13.
The tips that a barista earned for 1 week were $25 on Monday, $45 on Tuesday, and $30 on Wednesday, Thursday, and Friday. What was the barista's average daily tips for these 5 days?
- A) $31
- B) $32
- C) $33
- D) $35
The five daily amounts are $25, $45, $30, $30, and $30 (Wednesday, Thursday, and Friday were each $30). The total is 25 + 45 + 30 + 30 + 30 = 160.
Divide by 5 days: 160 / 5 = 32.
The average of 6 numbers is 93. What is the 6th number if the first 5 are 99, 86, 93, 89, 92?
- A) 88
- B) 93
- C) 95
- D) 99
Since the average of 6 numbers is 93, their total sum is 6 × 93 = 558. The first 5 numbers sum to 99 + 86 + 93 + 89 + 92 = 459.
The 6th number is the difference: 558 − 459 = 99.
What is the sum of three consecutive even integers whose mean is 44?
- A) 88
- B) 120
- C) 126
- D) 132
For any evenly spaced set with an odd number of terms, the mean equals the middle value. So the middle of the three consecutive even integers is 44 (meaning the integers are 42, 44, and 46).
The sum can be found directly by multiplying the mean by the count of numbers: 44 × 3 = 132.
John and Karen work at the bakery. John works 7 hours per day, and Karen works 6 hours per day. John produces x brownies per hour and Karen produces y brownies per hour. Which of the following expressions gives the average number of brownies John and Karen produce per hour?
- A) (x+y)/13
- B) (7x+6y)/13
- C) xy/13
- D) (6x+7y)/13
The total brownies produced is John's hours times his rate, plus Karen's hours times her rate: 7x + 6y. The total hours worked combined is 7 + 6 = 13.
Average brownies per hour is total brownies divided by total hours: (7x + 6y) / 13.
The mean age of 6 people at a graduation dinner is 40. When the oldest person, who is 95 years old, leaves the table, what is the mean age of the 5 people remaining at the table?
- A) 29
- B) 32
- C) 33
- D) 35
Since the mean age of 6 people is 40, the total combined age is 6 × 40 = 240. Removing the 95-year-old leaves a total of 240 − 95 = 145 among the remaining 5 people.
The new mean is 145 / 5 = 29.
Questions 7–8 refer to the table below. A student playing Skee-Ball tracked the points he received from each of his 10 throws.
| Points | Frequency |
|---|---|
| 0 | 3 |
| 10 | 0 |
| 20 | 1 |
| 30 | 2 |
| 50 | 3 |
| 100 | 1 |
Which of the following is the mean score of the student's 10 throws?
- A) 20
- B) 28
- C) 33
- D) 35
To find the mean from a frequency table, multiply each point value by its frequency, then add: (0×3) + (10×0) + (20×1) + (30×2) + (50×3) + (100×1) = 0 + 0 + 20 + 60 + 150 + 100 = 330.
Divide by the total number of throws (the sum of the frequencies, which is 3+0+1+2+3+1 = 10): 330 / 10 = 33.
Using the same table from Problem 7, what was the median value of the points scored from the 10 throws?
| Points | Frequency |
|---|---|
| 0 | 3 |
| 10 | 0 |
| 20 | 1 |
| 30 | 2 |
| 50 | 3 |
| 100 | 1 |
- A) 20
- B) 25
- C) 30
- D) 40
List all 10 throws in order using the frequencies: 0, 0, 0, 20, 30, 30, 50, 50, 50, 100.
With 10 values (an even count), the median is the average of the 5th and 6th values. Both the 5th and 6th values are 30, so the median is (30+30)/2 = 30.
The scores for the 43 students in AP Biology were reported, and the mean, median, range, and standard deviation were found. The teacher made an error in grading, and the student with the highest score actually scored 8 points higher. Which of the following will not change after the student's score is corrected?
- A) Range
- B) Mean
- C) Standard Deviation
- D) Median
Increasing the highest score changes the total sum (so the mean changes), increases the spread between the highest and lowest score (so the range changes), and changes how spread out the scores are (so the standard deviation changes).
The median, however, is based on position in the ordered list, not value. Since the highest score was already the highest and stays the highest after increasing by 8, its position in the ordered list doesn't move, so the value in the middle position — the median — is unaffected.
A recent survey asked 21 households how many iPhones they owned, shown in the table below. Based on the table, what was the median number of iPhones?
| iPhones | Frequency |
|---|---|
| 0 | 6 |
| 1 | 3 |
| 2 | 1 |
| 3 | 5 |
| 4 | 4 |
| 5 | 2 |
Number of iPhones per household (21 households surveyed)
- A) 1
- B) 2
- C) 3
- D) 4
With 21 households (an odd count), the median is the 11th value once all responses are listed in order. Track the running (cumulative) totals: 6 households own 0 (positions 1–6), 3 own 1 (positions 7–9), 1 owns 2 (position 10), and 5 own 3 (positions 11–15).
The 11th position falls within the group of households owning 3 iPhones, so the median is 3.
John drove 40 miles per hour for 15 minutes and then drove 20 miles per hour for 10 minutes. Which of the following gives the average rate, in miles per hour, that he drove during the 25 minutes?
- A) 28
- B) 30
- C) 32
- D) 34
Average rate is total distance divided by total time, not a simple average of the two speeds. First find each distance: 40 mph for 15 minutes (1/4 hour) is 40 × 1/4 = 10 miles. 20 mph for 10 minutes (1/6 hour) is 20 × 1/6 = 10/3 miles.
Total distance is 10 + 10/3 = 40/3 miles, and total time is 25 minutes = 25/60 = 5/12 hour. Average rate = (40/3) / (5/12) = (40/3) × (12/5) = 32 mph.
James is running in a 50-mile ultramarathon. For the first 15 miles, James runs at an average pace of 8 minutes per mile. For the next 25 miles, James runs at an average pace of 6.5 minutes per mile. For the last 10 miles, James runs at an average pace of 5 minutes per mile. What was James' average pace over the entire race?
- A) 6
- B) 6.50
- C) 6.65
- D) 7
Average pace over the whole race is total time divided by total distance — not a simple average of the three paces. Find the time for each segment: 15 miles × 8 min/mile = 120 minutes; 25 miles × 6.5 min/mile = 162.5 minutes; 10 miles × 5 min/mile = 50 minutes.
Total time is 120 + 162.5 + 50 = 332.5 minutes, over the full 50 miles. Average pace = 332.5 / 50 = 6.65 minutes per mile.
Which of the following is equivalent to the average of 5 consecutive even integers a, b, c, d, and e such that a < b < c < d < e?
- A) (a+d)/2
- B) bc/2
- C) bd/2
- D) c
For any evenly spaced set of numbers with an odd count of terms, the average equals the middle term. With 5 consecutive even integers a, b, c, d, e listed in increasing order, c is the middle (3rd) term.
So the average of the 5 integers is simply c — no calculation with the other variables is needed.
The average score of a team of 4 golfers is 86. A second team has the same first three scores, but the fourth score is 16 points higher than the fourth score of the first team. What is the average score of the second team?
- A) 68.5
- B) 90
- C) 92
- D) 94
The first team's total score is 4 × 86 = 344. Since the second team has the exact same first three scores, and its fourth score is just 16 points higher than the first team's fourth score, the second team's total is simply the first team's total plus 16: 344 + 16 = 360.
The second team's average is 360 / 4 = 90.
Ben has an average score of 72 points on 7 equally weighted tests. How many points higher than his average must Ben score on his 8th equally weighted test to raise his average on all 8 tests by 3 points?
- A) 6
- B) 18
- C) 21
- D) 24
Ben's current total after 7 tests is 7 × 72 = 504. To raise his average on 8 tests to 72 + 3 = 75, his new total must be 8 × 75 = 600.
His 8th test score must be 600 − 504 = 96. Since his average is 72, that score is 96 − 72 = 24 points higher than his average.
To increase the mean of 6 numbers by 5, by how much would the sum of the 6 numbers have to increase?
- A) 6/5
- B) 5
- C) 25
- D) 30
Mean is sum divided by count, so sum equals mean times count. Raising the mean by 5 across 6 numbers requires the sum to increase by 5 × 6 = 30.
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