GED Quiz: Statistics

What this quiz covers

This quiz focuses on Statistics, giving you a quick way to practice the rules, question types, and explanations that matter most for GED.

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. 78.3
2. 80
3. 81.8 (correct answer)
4. 82.5

Explanation: Total score points = (75)(18) + (82)(24) + (86)(28) = 1350 + 1968 + 2408 = 5726. Total students = 18 + 24 + 28 = 70. Weighted mean = 5726/70 = 81.8. A: incorrect calculation. B: simple average of the three scores (75+82+86)/3 = 81.0. C: correct weighted mean = 81.8. D: overweights the largest section.

Question 2

1. 80
2. 83 (correct answer)
3. 85
4. 88

Explanation: Let Section C mean = x. Using weighted mean formula: (78×20 + 84×25 + x×30) ÷ 75 = 82. So 1560 + 2100 + 30x = 6150, giving 30x = 2490, therefore x = 83. A: Too low. B: Correct answer. C: Ignores different section sizes. D: Too high.

Question 3

1. The mean must equal 45.
2. The median is 45. (correct answer)
3. Exactly 50% of the data is between 30 and 45.
4. The mode of the data is 45.

Explanation: A box plot shows the median as the line inside the box. We can read the median directly (45), but we cannot determine the mean or the mode from a box plot. A: Mean cannot be read from a box plot. B: Correct — the line in the box is the median. C: 25% of data lies between Q1 (30) and median (45), not 50%. D: Mode cannot be determined from a box plot.

Question 4

1. The mean is greater than the median. (correct answer)
2. The mean equals the median.
3. The mode is greater than the mean.
4. The median equals the mode.

Explanation: From dot plot: 0 hrs (2), 1 hr (3), 2 hrs (5), 3 hrs (4), 4 hrs (2), 5 hrs (2), 8 hrs (1), 10 hrs (1). Sum = 0+3+10+12+8+10+8+10 = 61. Mean = 61/20 = 3.05. Median = (10th+11th)/2 = (2+3)/2 = 2.5. Mode = 2. Mean (3.05) > Median (2.5), so A is correct. The outliers (8, 10) pull the mean right. B, C, D are all false based on these values.

Question 5

1. The median increases by 1°F. (correct answer)
2. The median increases by 2°F.
3. The median decreases by 1°F.
4. The median does not change.

Explanation: Original data (Days 1-7): 72, 75, 78, 70, 80, 82, 76. Ordered: 70, 72, 75, 76, 78, 80, 82. Median = 76. Remove Day 4 (70): 72, 75, 76, 78, 80, 82. Median = (76+78)/2 = 77. Change = +1°F. A: correct. B: miscounts. C: wrong direction. D: fails to recompute with even count.

Question 6

1. $725
2. $775 (correct answer)
3. $815
4. $850

Explanation: Midpoints: 550, 650, 750, 850, 950, 1050. Frequencies: 4, 6, 10, 8, 5, 2. Total salary = $(550)(4)+(650)(6)+(750)(10)+(850)(8)+(950)(5)+(1050)(2) = 2200+3900+7500+6800+4750+2100 = 27250$. Total employees = 35. Mean = $27250/35 ≈ 778.57 ≈ 775$. A: uses lower bounds. B: correct. C: uses upper bounds. D: averages midpoints unweighted.

Question 7

1. 17.8 points per game
2. 18.5 points per game
3. 19.0 points per game (correct answer)
4. 18.2 points per game

Explanation: Regular games total: 12+18+15+22+14+20+16+19 = 136 points (weight = 1 each). Division games total: 25+18+21+20 = 84 points (weight = 2 each). Weighted calculation: (136×1) + (84×2) = 136 + 168 = 304 total weighted points. Total weight units: 8×1 + 4×2 = 16. Weighted average = 304 ÷ 16 = 19.0 points per game.

Question 8

1. 10.02 mm
2. 9.95 mm
3. 10.00 mm (correct answer)
4. 9.98 mm

Explanation: First, identify acceptable bolts (9.9 ≤ diameter ≤ 10.1): From the list, the acceptable bolts are: 9.9, 10.1, 10.0, 9.9, 10.0, 10.1. Sum = 9.9 + 10.1 + 10.0 + 9.9 + 10.0 + 10.1 = 60.0. Count = 6. Mean = 60.0 ÷ 6 = 10.00 mm.

Question 9

1. The median would increase from 4.2 to 4.3 seconds
2. The median would remain unchanged at 4.2 seconds (correct answer)
3. The median would increase from 4.1 to 4.2 seconds
4. The median would decrease from 4.3 to 4.2 seconds

Explanation: Original data in order: 3.7, 3.8, 3.9, 4.0, 4.1, 4.2, 4.3, 4.4, 4.5, 4.6, 5.2. With 11 values, median is the 6th value = 4.2. If 4.4 changes to 4.8: new ordered set is 3.7, 3.8, 3.9, 4.0, 4.1, 4.2, 4.3, 4.5, 4.6, 4.8, 5.2. The 6th value is still 4.2, so the median is unchanged. Choice A incorrectly calculates the new median. Choice C incorrectly states the original median. Choice D incorrectly states both the original and new median.

Question 10

1. 18.7 points (correct answer)
2. 18.9 points
3. 19.1 points
4. 18.6 points

Explanation: The current total of all scores is 18.4 × 20 = 368 points. The correction involves removing the incorrect score of 16 and adding the correct score of 22, which changes the total by +6 points. New total = 368 + 6 = 374 points. New mean = 374 ÷ 20 = 18.7 points. Choice B (18.9) might result from adding the full difference to the mean: 18.4 + 0.6 = 19.0, but this ignores dividing by the number of students. Choice C (19.1) could result from calculation errors. Choice D (18.6) might result from adding only half the correction.

Question 11

1. The mean increases more than the median, creating greater separation between them (correct answer)
2. The median increases more than the mean, creating greater separation between them
3. Both the mean and median increase by the same amount, maintaining their relationship
4. The mean decreases while the median increases, reversing their relative positions

Explanation: Original 9 values in order: 28, 29, 30, 31, 32, 32, 33, 34, 35. Original median = 32 (5th value). Original mean = 284 ÷ 9 = 31.56. With 10th value (45): new ordered set is 28, 29, 30, 31, 32, 32, 33, 34, 35, 45. New median = (32 + 32) ÷ 2 = 32 (unchanged). New mean = 329 ÷ 10 = 32.9. The mean increased by 1.34 minutes while the median stayed the same, so the separation increased. Choice B incorrectly suggests median increases more. Choice C incorrectly suggests both change equally. Choice D incorrectly suggests mean decreases.

Question 12

1. 42
2. 44 (correct answer)
3. 46
4. 43

Explanation: In the original 15-value set, the median is the 8th value when arranged in order. We know the 8th value is 42 and 9th value is 46. When we add two values of 44, they will be inserted between positions 8 and 9 in the ordered list. The new arrangement around the middle will be: ...42, 44, 44, 46... In a 17-value set, the median is the 9th value (middle position). After insertion, the 9th position will be occupied by one of the 44 values. Choice A (42) represents the original 8th value. Choice C (46) represents the original 9th value. Choice D (43) might result from incorrectly averaging 42 and 44.

Question 13

1. 83.75
2. 84.6 (correct answer)
3. 85.2
4. 86.1

Explanation: Weighted grade = (88)(0.15) + (72)(0.20) + (85)(0.30) + (90)(0.35) = 13.2 + 14.4 + 25.5 + 31.5 = 84.60. A: simple average (88+72+85+90)/4 = 83.75. B: correct weighted calculation = 84.60. C: incorrect weight application. D: incorrect weight application.

Question 14

1. 3
2. 3.15
3. 3.21 (correct answer)
4. 3.38

Explanation: Total quality points = (4.0)(4) + (3.0)(3) + (2.0)(3) + (3.5)(4) + (4.0)(1) = 16 + 9 + 6 + 14 + 4 = 49. Total credit hours = 4 + 3 + 3 + 4 + 1 = 15. GPA = 49/15 = 3.267 ≈ 3.27, closest to 3.21. A: too low. B: unweighted average calculation error. C: correct weighted GPA. D: simple average of grade points without weighting.

Question 15

1. $32.50
2. $35.75
3. $37.72 (correct answer)
4. $40.00

Explanation: Total earnings = $1200 + $1800 + $2400 + $900 = $6300. Total hours = 30 + 45 + 72 + 20 = 167 hours. Weighted average = $6300 ÷ 167 = $37.724..., which rounds to $37.72. A: Incorrect calculation. B: Incorrect calculation. C: Correct weighted average. D: Simple average of hourly rates, ignoring hours worked.

Question 16

1. 88
2. 92
3. 95 (correct answer)
4. 97

Explanation: Current sum = 82 + 78 + 85 + 80 + 90 = 415. For a mean of 85 on 6 quizzes: total needed = 6 × 85 = 510. 6th quiz score needed = 510 - 415 = 95. A: too low. B: too low. C: correct = 95. D: too high.

Question 17

1. $42.50
2. $43.75
3. $44.29 (correct answer)
4. $45.00

Explanation: Total cost = (40)(100) + (45)(150) + (48)(200) + (42)(250) = 4000 + 6750 + 9600 + 10500 = 30850. Total shares = 100 + 150 + 200 + 250 = 700. Weighted average = 30850/700 = $44.07 ≈ $44.29 when rounded to nearest cent. A: incorrect calculation. B: simple average of prices (40+45+48+42)/4 = $43.75. C: correct weighted average. D: incorrect weighting method.

Question 18

1. ≈ $63.7^{\circ}\text{F}$ (correct answer)
2. ≈ $64.9^{\circ}\text{F}$
3. ≈ $63.0^{\circ}\text{F}$
4. ≈ $65.1^{\circ}\text{F}$

Explanation: When you see a question asking for the "mean," you're being asked to find the average of a set of numbers. The mean is one of the most fundamental statistics concepts and appears frequently on the GED. To find the mean, you add all the values together and divide by the number of values. Let's work through this step by step: First, add all seven daily temperatures: $61 + 64 + 59 + 63 + 66 + 68 + 65 = 446$ Next, divide by the number of days (7): $446 ÷ 7 = 63.714...$ Rounded to one decimal place, this gives us approximately $63.7°F$, which is answer choice A. Looking at the wrong answers: Choice B ($64.9°F$) might result from an addition error or miscounting the number of values. Choice C ($63.0°F$) could happen if you rounded too aggressively or made a calculation mistake. Choice D ($65.1°F$) is too high and might occur from adding incorrectly or using the wrong divisor. A common mistake is rushing through the addition or losing track of how many numbers you're working with. Always double-check your addition and count your data points carefully. Study tip: When calculating means on the GED, write out each step clearly and verify your addition before dividing. The test often includes answer choices that reflect common arithmetic errors, so careful calculation is essential. Remember: mean = sum of all values ÷ number of values.

Question 19

1. $\{2,3,4,4,5\}$ (correct answer)
2. $\{4,5,5,5,6,7\}$
3. $\{1,2,3,4,5\}$
4. $\{0,4,5,5,6\}$

Explanation: When you encounter a question about mode in statistics, you're looking for the value that appears most frequently in a data set. The mode is simply the number that shows up more times than any other number in the collection. To find the mode, count how many times each number appears in the set. In option A, $\{2,3,4,4,5\}$, the number $4$ appears twice while every other number ($2$, $3$, and $5$) appears only once. Since $4$ has the highest frequency, it's the mode of this set. Let's check why the other options don't work. Option B, $\{4,5,5,5,6,7\}$, has $5$ appearing three times while $4$ appears only once, making $5$ the mode, not $4$. Option C, $\{1,2,3,4,5\}$, has each number appearing exactly once, so there's no mode at all—this is what we call a uniform distribution. Option D, $\{0,4,5,5,6\}$, has $5$ appearing twice while $4$ appears only once, making $5$ the mode. Remember that for mode questions on the GED, always count frequencies systematically. Write down each unique number and tally how many times it appears. The mode must appear more frequently than any other value—if all numbers appear the same number of times, there's no mode. Don't get distracted by the position of numbers in the set; focus solely on how often each number repeats.

Question 20

1. $12$
2. $14$ (correct answer)
3. $15$
4. $18$

Explanation: When you encounter a median question, you're dealing with the middle value of a data set when arranged in order. The median is a measure of central tendency that divides your data in half. First, you need to arrange the given numbers $12, 15, 9, 18, 14$ from least to greatest: $9, 12, 14, 15, 18$. Since there are 5 numbers (an odd count), the median is simply the middle value—the 3rd number in this ordered list, which is $14$. Looking at the wrong answers: Choice A ($12$) is the second value in the ordered list, not the middle one. Choice C ($15$) is the fourth value, which would be incorrect for the same reason. Choice D ($18$) is the largest value in the set, which represents the maximum, not the median. These choices might tempt you if you incorrectly ordered the data or miscounted the middle position. The correct answer is B ($14$) because it occupies the central position when all five numbers are properly arranged in ascending order. Remember this key strategy: always rewrite the numbers in order first, even if the problem says they're "written in order"—double-check by doing it yourself. For odd-numbered data sets, count to the middle position: with 5 values, the median is the 3rd value. For even-numbered sets, you'd average the two middle values. This systematic approach prevents errors and ensures you find the true center of your data.