ACT Math Statistics: 14 Practice Problems with Step-by-Step Explanations
Statistics on the ACT covers a broader range than most other chapters: sampling methods (how data gets collected), reading data displays (box plots, stem-and-leaf plots, frequency tables), describing distribution shapes, and applying the normal distribution's empirical rule. There's less pure calculation here than in most ACT Math chapters — the bigger skill is correctly interpreting what a data display or sampling description is actually telling you.
For sampling problems, the key distinction is between a census (surveying an entire population) and a sample (surveying a randomly selected subset), and between a randomized method (using random selection) and a nonrandomized one. For distribution shape problems, picture the data as a histogram: normal means roughly symmetric with one peak in the middle, uniform means roughly flat across all values, bimodal means two separate peaks, and skewed left or right means a long tail stretching in that direction. Box plots show five key values — minimum, first quartile, median, third quartile, and maximum — but never show the mean directly, which matters on "must be true" questions comparing two data sets.
For the normal distribution, the empirical rule (68-95-99.7 rule) is essential: about 68% of data falls within 1 standard deviation of the mean, about 95% falls within 2 standard deviations, and about 99.7% falls within 3. Since the normal distribution is symmetric, "at least 1 standard deviation below the mean" always captures 50% (everything above the mean) plus half of the 68% within 1 standard deviation (34%), for a combined 84%. Work through the 14 problems below, including two that reference a box plot, then click to reveal each step-by-step explanation.
On this page:
- Problem 1 — Identifying a sampling method
- Problem 2 — Sample proportion
- Problem 3 — Census vs. sample
- Problem 4 — Probability from a stem-and-leaf plot
- Problem 5 — Range from a box plot
- Problem 6 — Identifying a distribution shape
- Problem 7 — Identifying a distribution shape
- Problem 8 — Percentage from a stem-and-leaf plot
- Problem 9 — Predicting a distribution shape
- Problem 10 — Comparing box plots
- Problem 11 — Effect of a constant shift on statistics
- Problem 12 — Predicting a distribution shape
- Problem 13 — Applying the empirical rule
- Problem 14 — Applying the empirical rule
Practice Problems
A school librarian wants to predict how often the 500 students at Walker High School will use a new 3D printer. To find out, she randomly selects 60 students to complete an online survey. Which of the following best describes the method the librarian used?
- A) Randomized study
- B) Nonrandomized study
- C) Randomized survey
- D) Nonrandomized survey
The librarian selected students randomly (not based on any bias or convenience), and she is asking them to complete a survey rather than conducting an experiment.
This is a randomized survey.
A research team in Botswana tagged 50 elephants in a preserve. One month later, the research team returned to the same preserve and observed a random sample of 80 elephants, 15 of which were tagged. Let p be the proportion of elephants in the preserve that are tagged. What is p̂, the sample proportion, for this sample?
- A) 3/16
- B) 3/10
- C) 7/20
- D) 5/8
The sample proportion is the number tagged in the sample divided by the total sample size: 15/80.
Simplify: 15/80 = 3/16.
Monica is assigned to find out how far seniors at her school drive each day as part of a statistics project. She decides to stand out front of the school and ask every senior how far he or she drives each day. The method that Monica is using can be best described as a:
- A) Census
- B) Nonrandomized study
- C) Randomized study
- D) Nonrandomized survey
Monica is asking every senior, not a sample of seniors — she's collecting data from the entire population she's interested in.
Surveying an entire population (rather than a sample) is called a census.
18 balls are placed into a box. Each ball is numbered, and the numbers are shown in the stem and leaf plot shown below. What is the probability that a randomly selected ball has a number less than 30?
| Stem | Leaf |
|---|---|
| 1 | 5, 8 |
| 2 | 3, 5, 9 |
| 3 | 0, 0, 1, 3, 4, 7, 7, 9 |
| 4 | 3, 6, 7, 9 |
| 5 | 6 |
- A) 5/18
- B) 8/18
- C) 13/18
- D) 5/23
Numbers less than 30 come from stems 1 and 2 only: stem 1 has 2 values (15, 18), and stem 2 has 3 values (23, 25, 29), for a total of 5 values.
Probability: 5/18 (out of 18 total balls).
The box plot above displays information about the heights of plants in a garden. Which of the following values is closest to the range for the heights of the plants in the garden?
- A) 13
- B) 18
- C) 20
- D) 24
The range is the difference between the maximum and minimum values shown by the box plot's whiskers: approximately 35−15=20.
Use the following information to answer questions 6–8. In 2015, a poll asked 1,800 households, "On average, how much does your family spend on groceries each week?" The households were split into 2 groups, each including 900 households. Group 1 households have children, and group 2 households have no children. The table below lists the percent of each sample that gave each response.
| Average amount spent | Group 1 | Group 2 |
|---|---|---|
| Less than $100 | 3% | 3% |
| $101 to $150 | 15% | 39% |
| $151 to $200 | 32% | 9% |
| $201 to $250 | 33% | 11% |
| $251 to $300 | 13% | 37% |
| $301 or more | 4% | 1% |
Which of the following best characterizes the distribution of the group 1 responses?
- A) Uniform
- B) Normal
- C) Bimodal
- D) Skewed left
Group 1's percentages (3%, 15%, 32%, 33%, 13%, 4%) rise to a peak in the middle categories ($151–$250) and taper off roughly symmetrically on both sides.
This single-peaked, roughly symmetric shape is characteristic of a normal distribution.
Using the same table, which of the following best characterizes the distribution of the group 2 responses?
- A) Normal
- B) Bimodal
- C) Skewed left
- D) Skewed right
Group 2's percentages (3%, 39%, 9%, 11%, 37%, 1%) show two separate high points — 39% at $101–$150 and 37% at $251–$300 — with lower percentages in between.
Two distinct peaks is the defining feature of a bimodal distribution.
The scores of 25 students on a fitness test are shown in the stem and leaf plot below.
| Stem | Leaf |
|---|---|
| 2 | 1, 1, 2, 6, 8 |
| 3 | 0, 1, 3, 3, 6, 8, 9 |
| 4 | 1, 2, 2, 4, 4, 5, 8, 9 |
| 5 | 3, 6 |
| 6 | 1, 3, 3 |
What percentage of students scored above a 45 on the fitness test?
- A) 52%
- B) 32%
- C) 28%
- D) 20%
Scores above 45 (not including 45 itself): from stem 4, only 48 and 49 qualify (2 scores). All of stem 5 (2 scores) and stem 6 (3 scores) also qualify.
Total above 45: 2+2+3=7 out of 25 students. Percentage: 7/25=0.28=28%.
Mike has a theory that when asked to pick a random number from 1 to 100, people are more likely to pick values with a single digit than any other number. If Mike's theory is correct, which of the following descriptors will most likely characterize the distribution of the numbers that 10,000 randomly selected people pick if they are asked to pick a number from 1 to 100?
- A) Uniform
- B) Normal
- C) Skewed left
- D) Skewed right
If single-digit numbers (1–9) are picked more often, the data would cluster heavily at the low end of the range, with a long tail of less-frequently picked values extending toward 100.
A cluster at low values with a tail stretching toward high values describes a distribution skewed right.
The box plots above summarize data set A and data set B. Which of the following statements must be true?
I. The mean of data set A must be greater than the mean of data set B.
II. The median of data set A must be greater than the median of data set B.
- A) I only
- B) II only
- C) I and II
- D) Neither I nor II
A box plot displays the median directly (the line inside the box), so if data set A's median line sits to the right of data set B's, that comparison is confirmed by the plot itself — statement II must be true.
A box plot never displays the mean — only the minimum, quartiles, median, and maximum. Since the mean isn't shown, there's no way to confirm statement I from a box plot alone, even if the medians differ.
Let m, r, and s be the mean, range, and standard deviation of the ages of players on a professional soccer team. Which of the following gives the mean, range, and standard deviation of the ages of the players in 8 years?
| Mean | Range | Standard Deviation | |
|---|---|---|---|
| A) | m | r | s+8 |
| B) | m+8 | r | s |
| C) | m+8 | r | s+8 |
| D) | m+8 | r+8 | s |
In 8 years, every player's age increases by exactly 8, so the mean shifts up by 8 as well: m+8.
Adding the same constant to every value doesn't change how spread out the data is — the range (the gap between the max and min) and the standard deviation (which measures spread around the mean) both stay the same: r and s.
A pair of 6-sided dice are rolled 4,000 times in an experiment. The sum of the numbers rolled is collect after each roll. Which of the following will most likely characterize the distribution of the 4,000 numbers?
- A) Uniform
- B) Normal
- C) Bimodal
- D) Skewed left
The sum of two dice is most likely to be 7 (the most common combination), with sums further from 7 (like 2 or 12) becoming progressively less likely on both sides.
This single-peaked, roughly symmetric pattern is best characterized as approximately normal among the given choices.
The lengths of 10,000 salmon during the annual salmon run are normally distributed. If the mean length of a salmon is 25 inches and the standard deviation is 1.75 inches, how many of the 10,000 salmon are expected to have lengths from 21.5 inches to 28.5 inches?
- A) 3,400
- B) 5,000
- C) 6,800
- D) 9,500
Check how many standard deviations 21.5 and 28.5 are from the mean of 25: 25−21.5=3.5, and 28.5−25=3.5. Since 3.5/1.75=2, this range covers exactly 2 standard deviations in each direction.
By the empirical rule, about 95% of data falls within 2 standard deviations of the mean: 0.95×10,000=9,500.
The heights of 18,000 students are normally distributed. If the mean height is 66 inches and the standard deviation is 3 inches, what percentage of students are at least 63 inches tall?
- A) 50%
- B) 68%
- C) 84%
- D) 95%
63 inches is exactly 1 standard deviation below the mean (66−3=63). "At least 63 inches" includes everyone above the mean (50%) plus everyone between 63 inches and the mean.
By the empirical rule, 68% of data falls within 1 standard deviation of the mean, so half of that (34%) lies between 63 inches and the mean. Total: 50%+34%=84%.
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