HSPT Math Quiz: Apply Statistics Concepts
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Apply Statistics ConceptsQuestion 1 of 20

Two fair coins are tossed simultaneously. What is the probability that exactly one coin shows heads?

14\dfrac{1}{4}
12\dfrac{1}{2}
34\dfrac{3}{4}
23\dfrac{2}{3}
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HSPT Math Quiz

HSPT Math Quiz: Apply Statistics Concepts

Practice Apply Statistics Concepts in HSPT Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

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

All questions

Question 1

Two fair coins are tossed simultaneously. What is the probability that exactly one coin shows heads?

  1. 14\dfrac{1}{4}
  2. 12\dfrac{1}{2} (correct answer)
  3. 34\dfrac{3}{4}
  4. 23\dfrac{2}{3}
Explanation: When you encounter probability questions involving multiple events, start by identifying all possible outcomes and then count the favorable ones. For two fair coins tossed simultaneously, there are four equally likely outcomes: HH (both heads), HT (first coin heads, second tails), TH (first tails, second heads), and TT (both tails). Since we want exactly one head, we need outcomes where one coin shows heads and the other shows tails. Looking at our four outcomes, exactly two satisfy this condition: HT and TH. Since each outcome has probability 14\frac{1}{4}, the probability of exactly one head is 24=12\frac{2}{4} = \frac{1}{2}. This confirms answer B. Answer A (14\frac{1}{4}) represents the probability of any single specific outcome, like getting HH or TT, but we need two different outcomes combined. Answer C (34\frac{3}{4}) would be the probability of getting "at least one head" (HH, HT, or TH) - a common trap when students misread "exactly one" as "at least one." Answer D (23\frac{2}{3}) doesn't correspond to any meaningful probability in this scenario and might result from incorrectly excluding one of the four possible outcomes. For probability questions on the HSPT, always list all possible outcomes systematically before calculating. This prevents you from missing cases or double-counting, especially when dealing with compound events like multiple coin flips or dice rolls.

Question 2

A standard deck of 52 cards is shuffled. What is the probability that a single card drawn at random is a heart?

  1. 14\dfrac{1}{4} (correct answer)
  2. 113\dfrac{1}{13}
  3. 413\dfrac{4}{13}
  4. 313\dfrac{3}{13}
Explanation: When you encounter probability questions about cards, remember that you're dealing with a standard deck structure: 52 total cards divided into 4 suits (hearts, diamonds, clubs, spades), with 13 cards in each suit. To find the probability of drawing a heart, you need to use the basic probability formula: P(event)=favorable outcomestotal possible outcomesP(\text{event}) = \frac{\text{favorable outcomes}}{\text{total possible outcomes}} Since there are 13 hearts in the deck and 52 total cards, the probability is 1352=14\frac{13}{52} = \frac{1}{4}. This makes sense intuitively—hearts represent exactly one-fourth of the deck. Looking at the wrong answers: Choice B (113\frac{1}{13}) represents the probability of drawing a specific card (like the ace of hearts) rather than any heart. Choice C (413\frac{4}{13}) might result from confusing the number of suits (4) with the number of cards per suit, creating an incorrect fraction. Choice D (313\frac{3}{13}) has no clear connection to the deck structure and likely represents a random distractor. The key insight is recognizing that when dealing with suits, you're working with equal groups of 13 cards each. Since hearts make up exactly one of the four suits, the answer must be 14\frac{1}{4}. Study tip: For card probability questions, always start by identifying what you're looking for (suit, color, face card, etc.) and remember the deck's structure: 4 suits × 13 cards = 52 total. This foundation will help you set up the correct fraction every time.

Question 3

A basketball player made 18 free throws out of 24 attempts. What is the experimental probability that his next free throw will be successful?

  1. 23\dfrac{2}{3}
  2. 34\dfrac{3}{4} (correct answer)
  3. 56\dfrac{5}{6}
  4. 78\dfrac{7}{8}
Explanation: When you encounter a question about experimental probability, you're being asked to find the likelihood of an event based on actual observed data from past trials, not theoretical calculations. To find experimental probability, use the formula: Experimental Probability=Number of Successful OutcomesTotal Number of Trials\text{Experimental Probability} = \frac{\text{Number of Successful Outcomes}}{\text{Total Number of Trials}} In this case, the basketball player made 18 successful free throws out of 24 total attempts. So the experimental probability is 1824\frac{18}{24}. To simplify this fraction, divide both numerator and denominator by their greatest common factor, which is 6: 1824=18÷624÷6=34\frac{18}{24} = \frac{18 ÷ 6}{24 ÷ 6} = \frac{3}{4}. This matches answer choice B. Now let's examine why the other answers are incorrect. Choice A (23\frac{2}{3}) would result from incorrectly calculating 1624\frac{16}{24}, suggesting the player made only 16 successful shots. Choice C (56\frac{5}{6}) would correspond to 2024\frac{20}{24}, implying 20 successful shots instead of 18. Choice D (78\frac{7}{8}) would equal 2124\frac{21}{24}, suggesting 21 successful attempts. Remember that experimental probability questions always ask you to use the given data exactly as presented. Don't overthink it—simply set up the fraction using the successful outcomes over total trials, then simplify. The key is careful arithmetic and proper fraction reduction.

Question 4

The ordered data set {7, 9, 9, 10, 12, 14, 18}\{7,\ 9,\ 9,\ 10,\ 12,\ 14,\ 18\} has which median?

  1. 9
  2. 10 (correct answer)
  3. 11
  4. 12
Explanation: When you encounter a median question, you're looking for the middle value that divides an ordered data set in half. Since this data set is already arranged from smallest to largest, you can find the median by locating the central position. With seven values in the set {7, 9, 9, 10, 12, 14, 18}\{7,\ 9,\ 9,\ 10,\ 12,\ 14,\ 18\}, the median is the 4th value (since there are 3 values below it and 3 values above it). Counting from left to right: 7 (1st), 9 (2nd), 9 (3rd), 10 (4th), 12 (5th), 14 (6th), 18 (7th). The median is 10, making answer choice B correct. Let's examine why the other options are incorrect. Choice A (9) might tempt you because 9 appears twice in the data set, but frequency doesn't determine the median—only position matters. Choice C (11) could result from incorrectly averaging the two middle-most values when you have an odd number of data points, but you only average when there's an even number of values. Choice D (12) would be wrong if you miscounted and thought 12 was in the middle position. Remember this key distinction: for an odd number of values, the median is simply the middle value. For an even number of values, you average the two middle values. Also, always verify that your data is ordered before finding the median—questions sometimes present unordered sets to test whether you'll sort first.

Question 5

At a carnival game, a wheel is equally divided into 12 sections: 4 red, 3 blue, 3 green, and 2 yellow. What is the probability of landing on blue or yellow in one spin?

  1. 12\dfrac{1}{2}
  2. 512\dfrac{5}{12} (correct answer)
  3. 13\dfrac{1}{3}
  4. 34\dfrac{3}{4}
Explanation: When you encounter probability questions involving "or" situations, you're looking at the probability of multiple favorable outcomes occurring. The key is identifying all the ways you can achieve success, then finding what fraction of total outcomes those represent. To find the probability of landing on blue or yellow, you need to count all the favorable sections. The wheel has 3 blue sections and 2 yellow sections, giving you 3 + 2 = 5 favorable outcomes out of 12 total sections. Therefore, the probability is 512\dfrac{5}{12}, which is answer choice B. Let's examine why the other answers are incorrect. Choice A (12\dfrac{1}{2}) would require 6 favorable sections out of 12, but blue and yellow only account for 5 sections total. Choice C (13\dfrac{1}{3}) equals 412\dfrac{4}{12}, which would be correct if you only counted one of the colors (close to the 3 blue sections) but ignored the other. Choice D (34\dfrac{3}{4}) equals 912\dfrac{9}{12}, which is far too high—this might result from incorrectly adding red and green sections instead of blue and yellow. Remember that "or" in probability means addition when the events can't happen simultaneously. You can't land on both blue and yellow in a single spin, so you simply add the number of blue sections to the number of yellow sections. Always double-check that your numerator reflects all favorable outcomes and your denominator represents the total possible outcomes.

Question 6

A class recorded the number of siblings each student has: 0, 1, 1, 2, 2, 2, 3, 3, 40,\ 1,\ 1,\ 2,\ 2,\ 2,\ 3,\ 3,\ 4. What is the mean number of siblings per student?

  1. 1.8
  2. 2.0 (correct answer)
  3. 2.2
  4. 3.0
Explanation: When you encounter a mean calculation problem, you're finding the average value by adding all data points and dividing by the total number of values. To find the mean number of siblings, first add all the values: 0+1+1+2+2+2+3+3+4=180 + 1 + 1 + 2 + 2 + 2 + 3 + 3 + 4 = 18. Next, count how many students are in the class by counting the data points: there are 9 students total. Finally, divide the sum by the number of students: 189=2.0\frac{18}{9} = 2.0. Let's examine why the other answers are incorrect. Choice A) 1.8 might result from miscounting the number of students or making an arithmetic error when adding the values. Choice C) 2.2 could come from incorrectly adding the data values or using the wrong divisor. Choice D) 3.0 represents a significant calculation error, possibly from confusing the mean with the mode (most frequent value) or the maximum value in the dataset. The correct answer is B) 2.0. For mean problems on the HSPT, always double-check your arithmetic by verifying both your sum and your count of data points. A useful strategy is to organize the data first—notice that this dataset has some repeated values (three 2's, two 1's, two 3's), so you can group them to avoid counting errors: 1(0)+2(1)+3(2)+2(3)+1(4)=0+2+6+6+4=181(0) + 2(1) + 3(2) + 2(3) + 1(4) = 0 + 2 + 6 + 6 + 4 = 18. This method can help prevent simple addition mistakes.

Question 7

A fair coin is flipped three times. What is the probability of getting at least two heads?

  1. 18\dfrac{1}{8}
  2. 14\dfrac{1}{4}
  3. 12\dfrac{1}{2} (correct answer)
  4. 34\dfrac{3}{4}
Explanation: When you encounter probability questions involving "at least" a certain number of outcomes, you're dealing with compound probability. The key insight is that "at least two heads" means exactly two heads OR exactly three heads. Let's find all possible outcomes when flipping a coin three times. Each flip has 2 possibilities, so there are 23=82^3 = 8 total outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Now identify which outcomes give us at least two heads:
  • Exactly two heads: HHT, HTH, THH (3 outcomes)
  • Exactly three heads: HHH (1 outcome)
That's 4 favorable outcomes out of 8 total, giving us 48=12\frac{4}{8} = \frac{1}{2}. Looking at the wrong answers: Choice A (18\frac{1}{8}) represents the probability of getting exactly three heads, which is just one specific outcome. Choice B (14\frac{1}{4}) equals 28\frac{2}{8}, which might result from miscounting favorable outcomes or confusing this with a different probability scenario. Choice D (34\frac{3}{4}) equals 68\frac{6}{8}, which would be the probability of getting at least one head—a common mix-up when students confuse "at least one" with "at least two." Remember this strategy: when dealing with "at least" problems, you can either count all favorable outcomes directly (as we did here) or use the complement rule by finding the probability of the opposite event and subtracting from 1. For "at least two heads," the complement would be "fewer than two heads" (zero or one head).

Question 8

The test scores for five students are 72, 88, 91, 75, 8472,\ 88,\ 91,\ 75,\ 84. What is the mean of the scores?

  1. 82 (correct answer)
  2. 82.5
  3. 83
  4. 83.5
Explanation: When you encounter a question asking for the mean (or average) of a set of numbers, you need to add up all the values and divide by the count of values. To find the mean of these five test scores, start by adding them all together: 72+88+91+75+84=41072 + 88 + 91 + 75 + 84 = 410. Then divide this sum by the number of scores: 410÷5=82410 ÷ 5 = 82. Looking at the answer choices, (A) 82 is correct. The other options represent common calculation errors. Choice (B) 82.5 might result from incorrectly dividing by 4 instead of 5, giving you 410÷4=102.5410 ÷ 4 = 102.5, though that's not quite right either—this suggests mixed-up arithmetic. Choice (C) 83 could come from adding incorrectly and getting 415 instead of 410, then dividing by 5. Choice (D) 83.5 might result from both adding incorrectly (perhaps getting 417) and dividing by 4, or other computational mistakes. The key trap here is rushing through the addition or miscounting how many values you have. Always double-check your addition by adding the numbers in a different order, and count your data points carefully. For mean problems on the HSPT, write down each step clearly: sum the values, count the values, then divide. This systematic approach prevents the small arithmetic errors that create those tempting wrong answers.

Question 9

In a history class, the final grade is determined by three categories: homework, which counts for 20% of the grade; tests, which count for 50%; and the final exam, which counts for 30%. A student has an average of 95 on homework, an average of 82 on tests, and a score of 75 on the final exam. What is the student's final weighted grade in the class?

  1. 81.1
  2. 82.5 (correct answer)
  3. 84.0
  4. 86.4
Explanation: To find the weighted grade, multiply each score by its weight (as a decimal) and then sum the results.
Homework: 95×0.20=19.095 \times 0.20 = 19.0
Tests: 82×0.50=41.082 \times 0.50 = 41.0
Final Exam: 75×0.30=22.575 \times 0.30 = 22.5
Sum the weighted parts: 19.0+41.0+22.5=82.519.0 + 41.0 + 22.5 = 82.5. The student's final grade is 82.5.

Question 10

A data set consists of 9 distinct integers. The median of the set is 40. If the three largest integers in the set are each increased by 10, and the two smallest integers are each decreased by 5, what will be the new median of the data set?

  1. 35
  2. 40 (correct answer)
  3. 45
  4. 50
Explanation: The median of a set with 9 items is the 5th item when the set is ordered. In this case, the 5th integer is 40. The three largest integers are the 7th, 8th, and 9th items. Increasing them does not change their order relative to the 5th item. The two smallest integers are the 1st and 2nd items. Decreasing them also does not change their order relative to the 5th item. Since the 5th item itself is not changed, and its position in the ordered list remains the same, the median does not change. The new median is still 40.

Question 11

A technology club has 40 members. 25 members are skilled in programming, and 15 members are skilled in hardware design. 10 members are skilled in both. If a member who is skilled in hardware design is chosen at random, what is the probability that this member is also skilled in programming?

  1. 1/4
  2. 2/5
  3. 3/8
  4. 2/3 (correct answer)
Explanation: This is a conditional probability problem. The condition 'a member who is skilled in hardware design is chosen' reduces the total sample space from 40 members to just the 15 members skilled in hardware design. Within this smaller group of 15, the problem states that 10 members are also skilled in programming. Therefore, the probability is the number of favorable outcomes (10) divided by the new total number of outcomes (15). The probability is 10/15, which simplifies to 2/3.

Question 12

The mean score of a group of 8 students on a test was 75.5. Two new students take the same test, and the mean score of all 10 students becomes exactly 78. If one of the new students scored 15 points higher than the other, what was the lower score of the two new students?

  1. 73
  2. 80.5 (correct answer)
  3. 88
  4. 95.5
Explanation: This is a multi-step problem. First, find the total score for the original 8 students: 8×75.5=6048 \times 75.5 = 604. Next, find the total score for all 10 students: 10×78=78010 \times 78 = 780. The sum of the two new scores is the difference between these totals: 780−604=176780 - 604 = 176. Let the lower score be xx and the higher score be x+15x + 15. Set up an equation: x+(x+15)=176x + (x + 15) = 176. Simplify to 2x+15=1762x + 15 = 176. Subtract 15 from both sides: 2x=1612x = 161. Divide by 2 to find the lower score: x=80.5x = 80.5.

Question 13

In a class of 25 students, the mean score on a quiz was 80. The median score was 84. If the 5 students with the highest scores each had their scores increased by 4 points, what would be the new mean and median of the class scores?

  1. New mean = 80.8, New median = 84 (correct answer)
  2. New mean = 80.8, New median = 88
  3. New mean = 84, New median = 84
  4. New mean = 84, New median = 88
Explanation: First, consider the median. In a class of 25 students, the median is the score of the 13th student when scores are ranked. The 5 students with the highest scores are ranked 21st through 25th. Increasing their scores will not change the score of the 13th student, so the median remains 84.
Next, consider the mean. The original total score for the class was 25×80=200025 \times 80 = 2000. When 5 students each have their scores increased by 4 points, the total score increases by 5×4=205 \times 4 = 20. The new total score is 2000+20=20202000 + 20 = 2020. The new mean is 2020/25=80.82020 / 25 = 80.8.

Question 14

The numbers in the set S1={10,15,x,25,30}S_1 = \{10, 15, x, 25, 30\} are listed in increasing order, and its median is 18. The numbers in the set S2={5,y,30,40}S_2 = \{5, y, 30, 40\} are also listed in increasing order. If the mean of S1S_1 is equal to the median of S2S_2, what is the value of yy?

  1. 9.2 (correct answer)
  2. 10
  3. 18
  4. 19.6
Explanation: First, find the value of xx. Since S1S_1 is ordered and has 5 elements, its median is the middle element, xx. So, x=18x = 18. The set S1S_1 is {10,15,18,25,30}\{10, 15, 18, 25, 30\}. Second, find the mean of S1S_1: (10+15+18+25+30)/5=98/5=19.6(10+15+18+25+30)/5 = 98/5 = 19.6. Third, find the median of S2S_2. Since S2S_2 is ordered and has 4 elements, its median is the average of the two middle elements: (y+30)/2(y+30)/2. Finally, set the mean of S1S_1 equal to the median of S2S_2: (y+30)/2=19.6(y+30)/2 = 19.6. Multiply by 2: y+30=39.2y+30 = 39.2. Subtract 30: y=9.2y = 9.2. This fits the condition that S2S_2 is in increasing order (5 < 9.2 < 30).

Question 15

There are two bags of balls. Bag A contains 3 red and 2 blue balls. Bag B contains 2 red and 4 blue balls. A fair coin is flipped. If the result is heads, a ball is drawn from Bag A. If the result is tails, a ball is drawn from Bag B. What is the overall probability that a red ball is drawn?

  1. 7/15 (correct answer)
  2. 5/11
  3. 1/2
  4. 14/15
Explanation: This problem requires calculating a weighted average of probabilities. First, find the probability of getting a red ball from each path.\n\nPath 1 (Heads): The probability of flipping heads is 1/2. The probability of drawing a red ball from Bag A (3 red, 2 blue; 5 total) is 3/5. The combined probability of this path is (1/2)×(3/5)=3/10(1/2) \times (3/5) = 3/10.\n\nPath 2 (Tails): The probability of flipping tails is 1/2. The probability of drawing a red ball from Bag B (2 red, 4 blue; 6 total) is 2/6 = 1/3. The combined probability of this path is (1/2)×(1/3)=1/6(1/2) \times (1/3) = 1/6.\n\nTo find the overall probability, add the probabilities of these two mutually exclusive paths: 3/10+1/63/10 + 1/6. Finding a common denominator of 30: 9/30+5/30=14/30=7/159/30 + 5/30 = 14/30 = 7/15.

Question 16

In a class of 24 students, the median test score is 85. If exactly 8 students scored above 85, how many students scored exactly 85?

  1. 4 students
  2. 6 students
  3. 8 students (correct answer)
  4. Cannot be determined from the given information
Explanation: When you encounter median problems with specific conditions, you need to think systematically about how values are distributed around the middle position. With 24 students, the median is the average of the 12th and 13th values when scores are arranged in order. Since the median is 85, both the 12th and 13th students scored 85 (if they differed, the median wouldn't be exactly 85). Given that exactly 8 students scored above 85, you can map out the distribution: positions 17-24 are the students scoring above 85, positions 12-13 definitely scored 85, and positions 1-11 scored below 85. This accounts for 8 + 2 + 11 = 21 students. The remaining 3 students (24 - 21 = 3) must have scored exactly 85, because they can't score above 85 (only 8 did) or below 85 (that would shift the median). So positions 14, 15, and 16 also scored 85. Therefore, 5 students total scored exactly 85: the 2 students in median positions plus the 3 additional students. Wait - let me recalculate. Actually, we have 8 students above 85 (positions 17-24), which means 16 students scored 85 or below. Since the median is exactly 85, and we need the 12th and 13th positions to be 85, the remaining students who didn't score above 85 must include 8 students who scored exactly 85. Choice A (4 students) is too few to maintain the median at 85. Choice B (6 students) also insufficient. Choice D is incorrect because the information given allows us to determine the exact number. Strategy tip: In median problems, always count positions systematically from both ends toward the middle to avoid confusion about the distribution.

Question 17

A survey of 200 students found that 120 students like pizza, 80 students like burgers, and 50 students like both pizza and burgers. If a student is randomly selected, what is the probability that the student likes pizza given that they like burgers?

  1. 58\frac{5}{8} (correct answer)
  2. 14\frac{1}{4}
  3. 516\frac{5}{16}
  4. 35\frac{3}{5}
Explanation: When you encounter a problem asking for the probability that one event occurs "given that" another event has occurred, you're dealing with conditional probability. The key phrase here is "given that they like burgers" — this tells you to focus only on the burger-loving students. To find the probability that a student likes pizza given that they like burgers, you need to use the conditional probability formula: P(Pizza|Burgers) = P(Pizza and Burgers) ÷ P(Burgers). In simpler terms, among all the students who like burgers, what fraction also likes pizza? From the survey data: 80 students like burgers, and 50 students like both pizza and burgers. So the probability is 5080=58\frac{50}{80} = \frac{5}{8}. This makes A correct. Let's examine why the other answers are wrong. Choice B (14\frac{1}{4}) incorrectly calculates 50200\frac{50}{200}, which gives the probability that any randomly selected student likes both foods — but ignores the "given that they like burgers" condition. Choice C (516\frac{5}{16}) appears to use 50160\frac{50}{160}, possibly adding pizza and burger lovers incorrectly. Choice D (35\frac{3}{5}) might result from confusing the given condition or using incorrect numbers from the problem. Remember: conditional probability questions always restrict your sample space. When you see "given that," immediately identify which group you're now focusing on, then find what fraction of that group satisfies the other condition. Don't use the total population once you have a conditional constraint.

Question 18

In a probability experiment, Maria draws two cards without replacement from a standard deck containing 4 red cards and 6 blue cards. What is the probability that both cards are the same color?

  1. 715\frac{7}{15} (correct answer)
  2. 12\frac{1}{2}
  3. 815\frac{8}{15}
  4. 25\frac{2}{5}
Explanation: When you encounter probability problems involving drawing without replacement, you need to consider how each draw affects the remaining outcomes. This question asks for the probability that both cards are the same color, which means either both red OR both blue. Let's work through this systematically. You have 10 total cards: 4 red and 6 blue. For both cards to be the same color, you need either two red cards or two blue cards. Probability of two red cards: 410×39=1290=215\frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15} Probability of two blue cards: 610×59=3090=13\frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3} Since these are mutually exclusive events, you add them: 215+13=215+515=715\frac{2}{15} + \frac{1}{3} = \frac{2}{15} + \frac{5}{15} = \frac{7}{15} This confirms answer A is correct. Answer B (12\frac{1}{2}) might result from incorrectly assuming equal likelihood without considering the actual card distribution. Answer C (815\frac{8}{15}) could come from calculation errors when adding fractions or from including impossible scenarios. Answer D (25\frac{2}{5}) might arise from only calculating one color scenario or from treating this as sampling with replacement. Remember for "without replacement" problems: always adjust your denominator and numerator after each draw. The key strategy is to break complex probability questions into simpler parts (both red OR both blue), calculate each part separately, then combine using addition for "or" scenarios.

Question 19

A spinner has three sections colored red, blue, and green. The probability of landing on red is twice the probability of landing on blue, and the probability of landing on green is three times the probability of landing on blue. If the spinner is spun twice, what is the probability that it lands on the same color both times?

  1. 718\frac{7}{18} (correct answer)
  2. 13\frac{1}{3}
  3. 518\frac{5}{18}
  4. 1136\frac{11}{36}
Explanation: When you encounter probability questions involving spinners with unknown section sizes, your first step is to set up equations using the given relationships to find each individual probability. Let's call the probability of landing on blue pp. Then red has probability 2p2p (twice blue's probability) and green has probability 3p3p (three times blue's probability). Since all probabilities must sum to 1: p+2p+3p=1p + 2p + 3p = 1, so 6p=16p = 1 and p=16p = \frac{1}{6}. This gives us: Blue = 16\frac{1}{6}, Red = 26=13\frac{2}{6} = \frac{1}{3}, Green = 36=12\frac{3}{6} = \frac{1}{2}. For the spinner to land on the same color both times, we need either red-red, blue-blue, or green-green. Since spins are independent, we multiply probabilities:
  • P(red both times) = 13×13=19\frac{1}{3} \times \frac{1}{3} = \frac{1}{9}
  • P(blue both times) = 16×16=136\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}
  • P(green both times) = 12×12=14\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}
Adding these: 19+136+14=436+136+936=1436=718\frac{1}{9} + \frac{1}{36} + \frac{1}{4} = \frac{4}{36} + \frac{1}{36} + \frac{9}{36} = \frac{14}{36} = \frac{7}{18} Answer A is correct. Answer B (13\frac{1}{3}) only accounts for the red-red case. Answer C (518\frac{5}{18}) might result from calculation errors. Answer D (1136\frac{11}{36}) could come from incorrectly handling the probability relationships. Strategy tip: Always verify that your individual probabilities sum to 1 before proceeding—this catches setup errors early and builds confidence in your final answer.

Question 20

The weekly allowances of seven friends are $10, $12, $12, $15, $16, $20, and $90. What is the positive difference between the mean and the median of their allowances?

  1. 9.5
  2. 10 (correct answer)
  3. 11.5
  4. 13
Explanation: First, find the median. The data set is already ordered: {10, 12, 12, 15, 16, 20, 90}. With 7 items, the median is the 4th item, which is $15. Next, calculate the mean. Sum the values: 10 + 12 + 12 + 15 + 16 + 20 + 90 = 175. Divide the sum by the number of values: 175 / 7 = 25. The mean is $25. Finally, find the positive difference between the mean and the median: 25 - 15 = 10.