ACT Math Quiz: Center Shape And Spread Of Data

What this quiz covers

This quiz focuses on Center Shape And Spread Of Data, giving you a quick way to practice the rules, question types, and explanations that matter most for ACT Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. No mode
2. 12
3. 9 and 12 (correct answer)
4. 9

Explanation: The mode is the value or values that occur most often in a data set, and a set can legitimately have more than one mode. Counting frequencies in $\{9,12,12,15,16,9\}$, the value 9 appears twice, 12 appears twice, and 15 and 16 each appear once, so two different values tie for the highest frequency and the answer is 9 and 12. Answering just 12 or just 9 comes from stopping at the first repeat found instead of tallying every value, and saying there is no mode comes from misremembering the rule as requiring a unique winner, when in fact no mode applies only if every value appears the same number of times. Always tally the count of each distinct value before deciding, and report every value tied for the top count, since a tie means the set is bimodal rather than modeless.

Question 2

1. 1.9 (correct answer)
2. 2
3. 2.1
4. 2.5

Explanation: This is a mean from frequency table question testing the weighted average calculation. Choice A (1.9) is correct — multiply each vehicle count by its frequency, sum the products, then divide by total families. Total vehicles = (1 × 15) + (2 × 25) + (3 × 10) = 15 + 50 + 30 = 95. Mean = 95/50 = 1.9. Choice B (2.0) comes from computing (1 + 2 + 3)/3 = 2 — averaging the vehicle numbers without weighting by frequency. Choice C (2.1) results from an arithmetic error in one product, perhaps computing (2 × 25) = 52 instead of 50: 15 + 52 + 30 = 97, 97/50 = 1.94 ≈ 2.1... or another minor error. Choice D (2.5) averages only the vehicle counts (1 + 2 + 3 + 4)/4 type error, or computes (15 + 25 + 10)/something incorrectly. Pro tip: For frequency tables, NEVER average the category values directly. You must weight each value by how many times it appears. Think of it as expanding the table: 15 families with 1 vehicle = fifteen 1s; 25 families with 2 = twenty-five 2s; etc. Then sum and divide by total families (50).

Question 3

1. {1, 1, 10, 10}
2. {2, 4, 6, 8}
3. {5, 5, 5, 5} (correct answer)
4. {5, 5, 6, 6}

Explanation: This is a standard deviation question testing conceptual understanding of spread. Choice C ({5, 5, 5, 5}) is correct — standard deviation measures how spread out the values are from the mean. A set with all identical values has zero spread, giving a standard deviation of exactly 0. No other set can have a smaller SD. Choice A ({1, 1, 10, 10}) has a mean of 5.5 with values far from the mean — large SD. Choice B ({2, 4, 6, 8}) has a mean of 5 with values spread 3 units apart on average — moderate SD ≈ 2.24. Choice D ({5, 5, 6, 6}) has a very small but nonzero SD ≈ 0.5. A student might choose D thinking it has the "smallest nonzero" deviation, but the question asks for smallest overall — and C achieves SD = 0. Pro tip: Standard deviation is zero when all values are identical. To find the set with the smallest SD without calculating, look for the set with the least variation. Identical values → SD = 0, always the minimum possible. If no set has identical values, choose the one where all values are closest to each other.

Question 4

1. Skewed right (correct answer)
2. Skewed left
3. Symmetric
4. Bimodal

Explanation: To determine skewness, examine the distribution of values and where the tail extends. The data [1, 2, 2, 3, 4, 4, 4, 5, 6] has most values clustered on the lower end with fewer values extending toward higher numbers. This creates a right-skewed (positively skewed) distribution where the tail extends to the right. Choice B (skewed left) would have the opposite pattern.

Question 5

1. 30 and 40
2. 35
3. 30
4. 40 (correct answer)

Explanation: The mode is the most frequent value in a data set. In {25,30,30,35,40,40,40,45,50}, count frequencies: 30 appears twice, 40 three times, others once. 40 appears most often. The mode is 40. Choice A suggests bimodal with 30 and 40, but 40 is clearly more frequent.

Question 6

1. 86
2. 90
3. 92 (correct answer)
4. 94

Explanation: This is an averages question testing the sum method. Choice C (92) is correct — current total sum = 4 × 82 = 328. Required total sum for a new average of 84 over 5 tests = 5 × 84 = 420. Required 5th score = 420 − 328 = 92. Choice A (86) comes from simply adding 2 to the target average of 84 — an intuitive but incorrect shortcut that ignores how averages compound across multiple values. Choice B (90) is a guess midway between 84 and 94, with no calculation behind it. Choice D (94) may result from computing 82 + (5 × (84 − 82)) = 82 + 10 = 92... actually 94 could come from 82 + 2 × 6 = 94 — an incorrect scaling. Pro tip: The sum method never fails for average problems — (target average × new count) − (current average × current count) = the missing value. The missing score must be ABOVE the new average to pull the mean up.

Question 7

1. Class A has a greater standard deviation than Class B.
2. Class B has a greater standard deviation than Class A. (correct answer)
3. Class A and Class B have equal standard deviations.
4. Standard deviation cannot be determined from this data.

Explanation: The correct answer is B (Class B has a greater standard deviation). Both classes have a mean of 74 (verify: Class A: (70+72+74+76+78)/5 = 370/5 = 74 ✓; Class B: (60+68+74+80+88)/5 = 370/5 = 74 ✓). Standard deviation measures how spread out the values are from the mean. Class A's values (70–78) have a range of 8 and cluster tightly around 74. Class B's values (60–88) have a range of 28 and deviate much further from 74. Because Class B's data is more spread out, it has the greater standard deviation. No calculation is required — the question tests conceptual understanding of spread. A is the most common trap: students may confuse which class is more dispersed. C is wrong because the spreads are visibly different. D is always false for a complete data set.

Question 8

1. 2 (correct answer)
2. 1.5
3. 1
4. 2.5

Explanation: To find the median, first sort the data: 0, 1, 1, 2, 2, 2, 3, 4. With 8 values (even count), the median is the average of the 4th and 5th values. Median = (2 + 2)/2 = 2. Choice B incorrectly calculated (1 + 2)/2 = 1.5.

Question 9

1. 22 (correct answer)
2. 24
3. 23
4. 21

Explanation: The median is the middle value when data is sorted in order. Sort the data: [19, 20, 22, 24, 26]. With 5 values, the median is the 3rd value (middle position). The median is 22. Choice B gave 24, which is the 4th value in the sorted list.

Question 10

1. 4 (correct answer)
2. 6
3. 9
4. 10

Explanation: The mode is the value that appears most frequently in the data set. Looking at [4, 4, 6, 9, 10, 4], the value 4 appears three times. The values 6, 9, and 10 each appear only once. Therefore, the mode is 4.

Question 11

1. 20 (correct answer)
2. 15
3. 10
4. 5

Explanation: The range is calculated by subtracting the minimum value from the maximum value. In the data set [30, 25, 20, 15, 10], the maximum is 30 and the minimum is 10. Range = 30 - 10 = 20. Choice B gave 15, which would be an incorrect calculation.

Question 12

1. It will remain the same.
2. It will decrease by 0.5.
3. It will increase by 0.5.
4. It will decrease by 1. (correct answer)

Explanation: Median problems require understanding how the middle value changes when data points are removed. For the 9 values {2,3,3,5,7,8,8,8,10}, the median is the 5th value: 7. Removing one 10 leaves 8 values, so the median becomes the average of the 4th and 5th values: (5 + 7)/2 = 6. The median decreased from 7 to 6, a decrease of 1. Strategy: When the dataset size changes from odd to even (or vice versa), remember that the median calculation method changes too.

Question 13

1. 85
2. 87
3. 89 (correct answer)
4. 90

Explanation: This is a classic mean (average) problem that requires you to work backwards from the desired average. To find the needed score, first calculate what the total sum must be: 85 × 5 = 425 points. Then find the current sum of the four tests: 78 + 85 + 92 + 81 = 336 points. The difference gives you the required fifth score: 425 - 336 = 89. Strategy tip: With mean problems, always start by finding the total sum needed, then subtract what you already have.

Question 14

1. $29$
2. $30$ (correct answer)
3. $31$
4. $33$

Explanation: The correct answer is B (30). The mean of 5 integers is 10, so their sum is 5 × 10 = 50. The median (3rd value) is 8. To maximize the largest integer, minimize all other values. The two values below the median must be distinct positive integers less than 8 — use the smallest possible: 1 and 2. The value immediately above the median must be a distinct integer greater than 8 — use 9. Maximum value = 50 − 1 − 2 − 8 − 9 = 30. A (29) uses d = 10 instead of d = 9 (doesn't minimize the 4th value): 50 − 1 − 2 − 8 − 10 = 29. C (31) uses d = 8 (equal to the median), forgetting that all integers must be distinct: 50 − 1 − 2 − 8 − 8 = 31. D (33) makes an error in setting up the minimization constraints. Strategy: always minimize every value you can control to maximize the one you want.

Question 15

1. 32
2. 38
3. 40
4. 42 (correct answer)

Explanation: This is a mean and averages question testing the sum method. Choice D (42) is correct — original total sum = 5 × 30 = 150. New required sum = 6 × 32 = 192. Age of 6th person = 192 − 150 = 42. Choice A (32) confuses the new mean with the 6th person's age. The new mean IS 32, but the 6th person must be older than 32 to pull the average up — a person who is exactly 32 would leave the mean unchanged. Choice B (38) results from an arithmetic error in the sum calculation. Choice C (40) likely comes from computing 30 + (5 × 2) = 40 — an incorrect shortcut that doesn't account for how sums work when a new member is added. Pro tip: Whenever a mean changes after adding a new member, use total sums: new person's value = (new mean × new count) − (old mean × old count). The new person must always be above the new mean if the average increased.

Question 16

1. 80
2. 85 (correct answer)
3. 88
4. 90

Explanation: The correct answer is B (85). To find the required 4th test score, first calculate the target sum: 4 × 85 = 340. Then find the current sum of the first 3 scores: 85 + 92 + 78 = 255. The needed score is 340 − 255 = 85. A (80) is a trap for students who notice the current average is already 85 and guess a nearby number. C (88) results from an arithmetic error in computing the sum or target. D (90) overestimates how much the low score of 78 pulls the average down. Pro tip: always compute the target sum first, then subtract the known values — this method works for any weighted average problem on the ACT.

Question 17

1. 0
2. 1 (correct answer)
3. 1.5
4. 2

Explanation: The correct answer is B (1). With 50 values, the median is the average of the 25th and 26th values. Map cumulative positions: 0 pets covers positions 1–10 (10 people); 1 pet covers positions 11–30 (20 people). Both the 25th and 26th values fall in the '1 pet' group, so the median is 1. A (0) confuses the median with the mode of the smallest group. C (1.5) averages the two middle category labels (1 and 2) rather than finding the 25th and 26th data points. D (2) selects the middle category by count rather than by cumulative position. Pro tip: always build a cumulative frequency table to locate the median position.

Question 18

1. $10$
2. $12$
3. $20$ (correct answer)
4. $24$

Explanation: The correct answer is C (20). The sum of 5 integers with mean 12 is 5 × 12 = 60. After removing the largest, the 4 remaining integers have mean 10, so their sum is 4 × 10 = 40. The removed value = 60 − 40 = 20. A (10) reports the new mean of the remaining 4 integers. B (12) reports the original mean, confusing the mean with the removed value. D (24) likely comes from an arithmetic error in computing one of the sums. This is a clean mean-difference problem: always compute both sums and subtract to find the removed element.

Question 19

1. 25 (correct answer)
2. 20
3. 15
4. 30

Explanation: The range is calculated by subtracting the minimum value from the maximum value. In the data set [95, 85, 90, 100, 110], the maximum is 110 and the minimum is 85. Range = 110 - 85 = 25. Choice B gave 20, which would be an incorrect calculation.

Question 20

1. Bimodal
2. Skewed left
3. Skewed right
4. Symmetric (correct answer)

Explanation: To determine the shape, examine how values are distributed around the center. The data [5, 10, 10, 15, 15, 20, 25] shows values distributed relatively evenly around the center (15), with similar frequencies on both sides. This creates a roughly symmetric distribution where the left and right sides mirror each other. Choice C (skewed right) would show clustering on the left with a tail extending right.