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ACT Math Help: Center Shape And Spread Of Data

Review real example questions for Center Shape And Spread Of Data in ACT Math.

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A store recorded the number of customers entering in 6 different hours: 9, 12, 12, 15, 16, 9. What is the mode of the data set {9,12,12,15,16,9}\{9,12,12,15,16,9\}?

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Question 1

A store recorded the number of customers entering in 6 different hours: 9, 12, 12, 15, 16, 9. What is the mode of the data set {9,12,12,15,16,9}\{9,12,12,15,16,9\}?

  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}\{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

A researcher surveyed 50 families and summarized her findings in the table. Of the families surveyed, what is the mean number of vehicles per family?

  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

Which of the following sets of data has the smallest standard deviation?

  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

Which best describes the shape of the distribution of the data [1, 2, 2, 3, 4, 4, 4, 5, 6]?

  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

The numbers of minutes a student studied for 9 days are: 25, 30, 30, 35, 40, 40, 40, 45, 50. What is the mode of the data set?

  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

A student has taken 4 tests and has an average score of 82. What score must the student get on the 5th test to raise the average to exactly 84?

  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

The table shows the test scores for two classes, Class A and Class B. Both classes have the same mean score. Which of the following statements about the standard deviations of the two classes is true?

  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

The numbers of goals scored by a team in 8 games are 2, 0, 3, 1, 2, 4, 1, 2. What is the median of {2,0,3,1,2,4,1,2}\{2,0,3,1,2,4,1,2\}?

  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

What is the median of the data set [19, 24, 22, 26, 20]?

  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

What is the mode of the data set [4, 4, 6, 9, 10, 4]?

  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.