Statistics and Probability

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Practice Statistics and Probability with tests designed for the University Practice. Each question includes a full explanation so you can learn from every mistake. Mastering this Mathematics topic is key to improving your score on the entrance exam.

Statistics and Probability accounts for about 5-10% of the ACT Math section. Though it is the smallest content area, the questions are often straightforward if you know the formulas and concepts — making these some of the easiest points to earn.

Mean, Median, and Mode

  • Mean (average): sum of all values ÷ number of values
  • Median: the middle value when data is ordered. For an even count, average the two middle values
  • Mode: the most frequently occurring value (there can be more than one, or none)
Mean, Median, and Mode — Worked Example Data set: 3, 5, 5, 7, 8, 12 Mean (3+5+5+7+8+12) / 6 = 40/6 ≈ 6.67 Median Middle of 5 and 7 = (5+7)/2 = 6 Mode Most frequent value = 5 Key ACT Insight: Total = Mean × Count If the mean of 5 numbers is 20, their total is 5 × 20 = 100. Use this to find a missing value: if 4 numbers sum to 78, the 5th = 100 − 78 = 22. Example 1: After 4 tests, a student has a mean score of 82. What score is needed on the 5th test to raise the mean to 85?
Total for 4 tests = 82 × 4 = 328. Needed total for 5 tests = 85 × 5 = 425. Fifth test = 425 − 328 = 97.

Example 2: Data set: 14, 18, 18, 22, 25, 30, 35. Find the mean, median, and mode.
Mean: (14+18+18+22+25+30+35)/7 = 162/7 ≈ 23.1
Median: 4th value = 22 (7 values, middle is the 4th)
Mode: 18 (appears twice)

Example 3: The mean of 6 numbers is 15. If the number 3 is removed, what is the new mean?
Total = 15 × 6 = 90. Remove 3: new total = 87. New mean = 87/5 = 17.4.

Range and Standard Deviation

Range = maximum − minimum. This is the simplest measure of spread.

Standard deviation measures how spread out data is around the mean. A larger standard deviation means more variability. You will not need to calculate it by hand on the ACT, but you should understand that:
  • If all values are the same, standard deviation = 0.
  • Adding the same constant to every value does not change the standard deviation.
  • Multiplying every value by k multiplies the standard deviation by |k|.
Example: Data set A: {10, 10, 10, 10}. Data set B: {1, 5, 15, 19}. Both have a mean of 10, but B has a much larger standard deviation because the values are more spread out.

Probability Basics

P(event) = favorable outcomes / total outcomes. Probability is always between 0 and 1. P(not A) = 1 − P(A).

Example 1: A bag contains 4 red, 3 blue, and 5 green marbles. What is the probability of drawing a blue marble?
P(blue) = 3/12 = 1/4 = 25%

Example 2: A card is drawn from a standard 52-card deck. What is the probability it is a face card (J, Q, K)?
There are 12 face cards. P = 12/52 = 3/13 ≈ 23.1%.

Example 3: Two dice are rolled. What is the probability the sum is 7?
Favorable outcomes: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) = 6 outcomes. Total = 36.
P(sum of 7) = 6/36 = 1/6 ≈ 16.7%.

Compound Probability

  • Independent events: P(A and B) = P(A) × P(B)
  • Mutually exclusive: P(A or B) = P(A) + P(B)
  • General "or": P(A or B) = P(A) + P(B) − P(A and B)
Example 1: A coin is flipped 3 times. What is the probability of getting exactly 2 heads?
Favorable outcomes: HHT, HTH, THH = 3 outcomes. Total outcomes: 2³ = 8.
P(exactly 2 heads) = 3/8 = 37.5%

Example 2: What is the probability of NOT rolling a 6 on three consecutive rolls of a fair die?
P(not 6 per roll) = 5/6. For all three: (5/6)³ = 125/216 ≈ 57.9%.

Probability Decision Tree A AND B? Multiply P(A) × P(B) A OR B? Add (then subtract overlap) P(A)+P(B)−P(A and B) NOT A? Complement 1 − P(A) "At Least One" Shortcut P(at least one) = 1 − P(none). Almost always faster than direct calculation.

Counting Principles

Permutations vs. Combinations Permutations (Order Matters) nPr = n! / (n−r)! Example: Arrange 3 of 5 books 5P3 = 5!/(5−3)! = 120/2 = 60 Think: "How many orders?" Passwords, rankings, seating order, president/VP elections Combinations (Order Doesn't Matter) nCr = n! / (r!(n−r)!) Example: Choose 3 of 5 books 5C3 = 5!/(3!×2!) = 120/12 = 10 Think: "How many groups?" Teams, committees, card hands, pizza toppings selection Fundamental counting principle: If task 1 has m ways and task 2 has n ways, the total is m × n.
Example 1: A restaurant offers 4 entrees and 3 desserts. Total meal combinations = 4 × 3 = 12.

Example 2: How many 3-letter codes can be made from the letters A through E if no letter is repeated?
5 × 4 × 3 = 60 (this is 5P3).

Example 3: A committee of 3 is to be chosen from 8 people. How many different committees are possible?
8C3 = 8!/(3! × 5!) = (8 × 7 × 6)/(3 × 2 × 1) = 336/6 = 56.

Data Interpretation

The ACT often presents data in tables, bar graphs, pie charts, or scatter plots. Read labels carefully and pay attention to units. For scatter plots, identify the trend (positive, negative, or no correlation).

Weighted Averages: When groups have different sizes, use weighted average = (sum of all values)/(total count), not the average of averages.

Example: Class A has 20 students with a mean score of 80. Class B has 30 students with a mean score of 90. What is the overall mean?
Total = 20(80) + 30(90) = 1600 + 2700 = 4300. Overall mean = 4300/50 = 86.
(Note: it is NOT simply (80+90)/2 = 85 because the classes have different sizes.)

Common Mistakes

Top 5 Statistics and Probability Traps 1. Averaging the averages instead of using weighted average (different group sizes!). 2. Forgetting to order data before finding the median. The middle of unordered data is meaningless. 3. Confusing permutations with combinations. Ask: "Does the order of selection matter?" 4. Adding probabilities for "and" problems instead of multiplying (independent events). 5. Not accounting for "without replacement" — probability changes after each draw.

Quick Reference — Formula Sheet

Statistics and Probability Quick Reference Central Tendency Mean = sum / count Median = middle value (ordered) Mode = most frequent Key Relationship Total = Mean × Count Probability P(A) = favorable / total P(not A) = 1 − P(A) P(A and B) = P(A) × P(B) P(A or B) = P(A)+P(B)−P(A∩B) At Least One Counting nPr = n!/(n−r)! nCr = n!/(r!(n−r)!) Fundamental: m × n ways Spread Range = max − min SD: adding constant → no change SD: multiply by k → SD × |k| Weighted Average = total of all values / total count ≠ average of the averages! Closer to the larger group's mean

ACT Strategies for Statistics and Probability

  • Total = Mean × Count: This relationship solves most "missing value" and "new average" problems instantly.
  • Outlier awareness: The median is resistant to outliers; the mean is not. If asked how an outlier affects measures, the mean shifts more.
  • Sample space first: Always identify the total number of outcomes before calculating probability.
  • "At least one" shortcut: P(at least one) = 1 − P(none). This is almost always faster than computing directly.
  • Order matters? Ask this question first: If yes → permutations. If no → combinations.
  • Seesaw method: A weighted average is always closer to the group with more members.
  • Read chart labels carefully: Many ACT data questions are missed due to misreading axes, units, or scale.
  • With vs. without replacement: Check if items are returned after selection — this changes the probability for subsequent draws.

Practice Walkthrough

ACT-Style Problem: The mean of five numbers is 12. Four of the numbers are 8, 10, 15, and 18. What is the fifth number?

A. 7    B. 9    C. 11    D. 12

Step 1 — Find the total: Mean × Count = 12 × 5 = 60.
Step 2 — Sum the known values: 8 + 10 + 15 + 18 = 51.
Step 3 — Find the missing value: 60 − 51 = 9.
Elimination: Choice A (7) would give a total of 58, mean = 11.6. Choice C (11) gives total 62, mean = 12.4. Choice D (12) gives total 63, mean = 12.6. Only B (9) gives a total of 60 and a mean of exactly 12.
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