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Praxis Core · Mathematics (5733)

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Praxis Core Data Interpretation, Statistics, and Probability Summary

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★★★★★ (5 / 5)
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About 18 of 56 Mathematics (5733) questions (32%)

Reading tables and graphs, finding mean, median, mode, and range, and finding simple probabilities.

📖 See Data Interpretation, Statistics, and Probability concepts →

Data Interpretation, Statistics, and Probability: must-know points

  • The median is the middle value after the data are put in order.
  • An outlier changes the mean much more than the median.
  • Probability = favorable outcomes divided by all equally likely outcomes.

Data Interpretation, Statistics, and Probability: 10 sample questions

Practice questionReviewedMathematics (5733) › Data Interpretation, Statistics, and Probability★★★★★

Question 1. Find the mean of these numbers: 10, 15, 9, 22, 16

  1. ① 15
  2. ② 18
  3. ③ 13
  4. ④ 72
  5. ⑤ 14.4
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Answer: ⑤ 14.4

Key point: Mean = sum of the values ÷ number of values.

  1. Sum: 10 + 15 + 9 + 22 + 16 = 72.
  2. Mean: 72 ÷ 5 = 14.4.

Check: 14.4 × 5 = 72.

Wrong choices

  • "15": 15 is the median (the middle value when ordered), not the mean.
  • "18": This divides the total (72) by 4 instead of by 5 numbers.
  • "13": 13 is the range (22 − 9), not the mean.
  • "72": 72 is the total. It must be divided by how many numbers there are.
Practice questionReviewedMathematics (5733) › Data Interpretation, Statistics, and Probability★★★★★

Question 2. What is the median of 7, 3, 9, 4, 12, 8?

  1. ① 7
  2. ② 7.5
  3. ③ 8
  4. ④ 9
  5. ⑤ 6.5
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Answer: ② 7.5

Key point: Order the data; with an even count, average the two middle values.

  1. Order: 3, 4, 7, 8, 9, 12.
  2. The two middle values are 7 and 8.
  3. Median: (7 + 8) ÷ 2 = 7.5.

Check: three values (3, 4, 7) are below 7.5 and three (8, 9, 12) are above it.

Wrong choices

  • "6.5": This takes the middle two numbers without first putting the list in order (9 and 4).
  • "7": With six numbers there are two middle values (7 and 8); the median is their average.
  • "8": This picks only one of the two middle values.
  • "9": 9 is the range (12 − 3), not the median.
Practice questionReviewedMathematics (5733) › Data Interpretation, Statistics, and Probability★★★★★

Question 3. A bag holds 4 red, 6 blue, and 5 green marbles. One marble is picked at random. What is the probability that it is blue?

  1. ① 2/5
  2. ② 2/3
  3. ③ 3/5
  4. ④ 1/3
  5. ⑤ 1/6
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Answer: ① 2/5

Key point: Probability = favorable outcomes ÷ total outcomes.

  1. Total marbles: 4 + 6 + 5 = 15.
  2. Blue: 6.
  3. Probability: 6/15 = 2/5.

Check: 2/5 = 0.4, and 0.4 × 15 = 6 blue marbles.

Wrong choices

  • "2/3": This compares blue to non-blue (6 to 9). Probability compares blue to ALL marbles (6 out of 15).
  • "3/5": 3/5 is the probability of NOT picking blue.
  • "1/3": This assumes the three colors are equally likely. There are more blue marbles than red or green.
  • "1/6": This puts 1 over the number of blue marbles.
Practice questionReviewedMathematics (5733) › Data Interpretation, Statistics, and Probability★★★★★

Question 4. A coin is flipped and a six-sided die is rolled. What is the probability of getting heads AND a 6?

  1. ① 1/6
  2. ② 2/3
  3. ③ 7/12
  4. ④ 1/12
  5. ⑤ 1/8
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Answer: ④ 1/12

Key point: For independent events, P(A and B) = P(A) × P(B).

  1. P(heads) = 1/2. P(6) = 1/6.
  2. P(both) = 1/2 × 1/6 = 1/12.

Check: there are 2 × 6 = 12 equally likely pairs, and only one of them is (heads, 6).

Wrong choices

  • "2/3": This adds 1/2 + 1/6. For two independent events that BOTH must happen, multiply.
  • "7/12": 7/12 is the probability of heads OR a 6, not both.
  • "1/8": This puts 1 over 2 + 6. The total number of outcomes is 2 × 6 = 12.
  • "1/6": This gives only the chance of rolling a 6 and ignores the coin.
Practice questionReviewedMathematics (5733) › Data Interpretation, Statistics, and Probability★★★★★

Question 5. What is the mode of this data set: 4, 7, 7, 2, 9, 4, 7, 5?

  1. ① 3
  2. ② 5.625
  3. ③ 4
  4. ④ 6
  5. ⑤ 7
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Answer: ⑤ 7

Key point: The mode is the value that appears most often.

Count each value: 2 (once), 4 (twice), 5 (once), 7 (three times), 9 (once). The value 7 appears most often, so the mode is 7.

Check: no other value appears three or more times.

Wrong choices

  • "4": 4 appears twice, but 7 appears three times.
  • "6": 6 is the median ((5 + 7) ÷ 2), not the mode.
  • "3": 3 is HOW MANY times the mode appears, not the mode itself.
  • "5.625": 5.625 is the mean (45 ÷ 8), not the mode.
Practice questionReviewedMathematics (5733) › Data Interpretation, Statistics, and Probability★★★★★

Question 6. The low temperatures for five days were −3°F, 5°F, 12°F, −8°F, and 7°F. What is the range?

  1. ① 2.6°F
  2. ② 4°F
  3. ③ 20°F
  4. ④ 15°F
  5. ⑤ 5°F
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Answer: ③ 20°F

Key point: Range = greatest value − least value.

  1. Greatest: 12. Least: −8.
  2. Range: 12 − (−8) = 12 + 8 = 20°F.

Check: on a thermometer, going from −8 up to 0 is 8 degrees, and from 0 up to 12 is 12 more; 8 + 12 = 20.

Wrong choices

  • "4°F": This subtracts 8 instead of −8. Subtracting a negative number means adding: 12 − (−8) = 20.
  • "15°F": This uses −3 as the lowest value. The lowest temperature is −8°F.
  • "5°F": 5°F is the median (the middle value), not the range.
  • "2.6°F": 2.6°F is the mean (13 ÷ 5), not the range.
Practice questionReviewedMathematics (5733) › Data Interpretation, Statistics, and Probability★★★★★

Question 7. A store's sales by month were: January $4,200; February $3,800; March $5,100; April $4,600; May $5,700. Between which two months in a row did sales increase the most?

  1. ① The two increases (February to March and April to May) were equal
  2. ② March to April
  3. ③ January to February
  4. ④ February to March
  5. ⑤ April to May
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Answer: ④ February to March

Key point: Find each change (later − earlier) and compare.

  1. Jan → Feb: 3,800 − 4,200 = −400.
  2. Feb → Mar: 5,100 − 3,800 = +1,300.
  3. Mar → Apr: 4,600 − 5,100 = −500.
  4. Apr → May: 5,700 − 4,600 = +1,100.

The largest increase is +1,300, from February to March.

Check: 3,800 + 1,300 = 5,100.

Wrong choices

  • "April to May": Sales rose $1,100 from April to May, which is less than the $1,300 rise from February to March.
  • "The two increases (February to March and April to May) were equal": The increases were $1,300 and $1,100, which are not equal.
  • "March to April": Sales went DOWN $500 from March to April.
  • "January to February": Sales went DOWN $400 from January to February.
Practice questionReviewedMathematics (5733) › Data Interpretation, Statistics, and Probability★★★★★

Question 8. A fair six-sided die (numbered 1 to 6) is rolled once. What is the probability of rolling a number greater than 4?

  1. ① 1/3
  2. ② 2/3
  3. ③ 5/6
  4. ④ 1/2
  5. ⑤ 1/6
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Answer: ① 1/3

Key point: "Greater than" does not include the number itself.

  1. Numbers greater than 4: 5 and 6, which is 2 outcomes.
  2. Total outcomes: 6.
  3. Probability: 2/6 = 1/3.

Check: the probability of 4 or less is 4/6, and 2/6 + 4/6 = 1.

Wrong choices

  • "1/2": This counts 4, 5, and 6. "Greater than 4" does not include 4.
  • "1/6": This counts only one number. Both 5 and 6 are greater than 4.
  • "2/3": 2/3 is the probability of rolling 4 or less.
  • "5/6": This counts every number except 4.
Practice questionReviewedMathematics (5733) › Data Interpretation, Statistics, and Probability★★★★★

Question 9. Class A has 20 students with a mean test score of 80. Class B has 30 students with a mean test score of 70. What is the mean score of all 50 students?

  1. ① 75
  2. ② 74
  3. ③ 76
  4. ④ 70
  5. ⑤ 3,700
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Answer: ② 74

Key point: Combined mean = total of all values ÷ total number of values.

  1. Class A total: 20 × 80 = 1,600.
  2. Class B total: 30 × 70 = 2,100.
  3. Combined mean: (1,600 + 2,100) ÷ 50 = 3,700 ÷ 50 = 74.

Check: 74 is closer to 70 than to 80, which makes sense because Class B is larger.

Wrong choices

  • "75": This averages the two class means. Class B has more students, so it counts more.
  • "76": This gives Class A the larger weight. Class B has 30 students, not 20.
  • "70": This uses only the larger class.
  • "3,700": 3,700 is the total of all scores. It must be divided by 50 students.
Practice questionReviewedMathematics (5733) › Data Interpretation, Statistics, and Probability★★★★★

Question 10. Jamal's first four test scores are 78, 85, 92, and 88. What score does he need on the fifth test to have a mean of 87 for all five tests?

  1. ① 88.25
  2. ② 85.75
  3. ③ 435
  4. ④ 87
  5. ⑤ 92
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Answer: ⑤ 92

Key point: Needed score = (target mean × count) − current total.

  1. Total needed for five tests: 87 × 5 = 435.
  2. Current total: 78 + 85 + 92 + 88 = 343.
  3. Needed: 435 − 343 = 92.

Check: (343 + 92) ÷ 5 = 435 ÷ 5 = 87.

Wrong choices

  • "87": Scoring 87 would not raise his current mean (85.75) all the way to 87.
  • "88.25": This adds the 1.25-point shortfall to 87 once. The fifth test must make up 1.25 for each of the four earlier tests too.
  • "85.75": 85.75 is his current mean, not the score he needs.
  • "435": 435 is the total needed for all five tests, not the fifth score.

Test yourself.