Interpret simple statistics and probability questions in context - TACHS Math

1. In a class of 30 students, 14 like chocolate ice cream, 18 like vanilla ice cream, and 7 like both flavors. If one student is selected at random, what is the probability that the student likes neither flavor?
   Answer: $\dfrac{1}{6}$ When you encounter a problem about overlapping groups, you're dealing with set theory and the inclusion-exclusion principle. The key insight is that students who like "both" flavors are being counted twice when you add the chocolate and vanilla lovers. Let's work systematically. First, find how many students like at least one flavor using the inclusion-exclusion principle: Students liking chocolate OR vanilla = (chocolate lovers) + (vanilla lovers) - (both flavors). That's 14 + 18 - 7 = 25 students who like at least one flavor. Since there are 30 total students, the number who like neither flavor is 30 - 25 = 5 students. Therefore, the probability is $\dfrac{5}{30} = \dfrac{1}{6}$, which is answer D. Now let's see where the wrong answers come from. Choice A ($\dfrac{11}{30}$) likely comes from incorrectly calculating 30 - 14 - 18 + 7, which misapplies the inclusion-exclusion principle. Choice B ($\dfrac{7}{30}$) mistakenly uses the number who like both flavors as the answer. Choice C ($\dfrac{25}{30}$) gives you the probability of liking at least one flavor instead of neither flavor - essentially the complement of what you want. Remember this pattern: in overlapping set problems, always subtract the overlap to avoid double-counting, then subtract the total who fit either category from the universal set to find "neither." Drawing a Venn diagram can help you visualize these relationships and catch calculation errors.
2. The table below shows the results of drawing colored marbles from a bag 200 times (with replacement). Refer to the table to answer the question. Based on this experiment, if the bag contains 40 marbles, what is the best estimate of how many are NOT green?
   Answer: 30 Green drawn: 50 out of 200 = $25\%$. So $75\%$ are not green: $0.75 \times 40 = 30$. A gives green count only (25% of 40). B uses red count ratio. C gives sum of red and blue proportions applied (70/200 × 40 = 14, plus 12 = 26, a mixed error).
3. The stem-and-leaf plot shows test scores for a class. Use the plot to answer the question. If a passing score is 70 or higher, what is the probability that a randomly chosen student passed AND scored above the class median?
   Answer: $\frac{1}{2}$ List the 16 scores in order: 52, 58, 60, 64, 65, 69, 70, 71, 72, 75, 78, 82, 84, 86, 88, 90. With 16 scores, the median is the average of the 8th and 9th: $\frac{71 + 72}{2} = 71.5$. The scores above 71.5 are 72, 75, 78, 82, 84, 86, 88 and 90. All of them are 70 or higher, so all 8 students passed. The probability is $\frac{8}{16} = \frac{1}{2}$. The value $\frac{5}{8}$ is the probability of passing alone (10 scores of 70 or more), ignoring the median. The value $\frac{9}{16}$ counts 71 as well, which comes from averaging the 7th and 8th scores to get a median of 70.5. The value $\frac{7}{16}$ counts only scores above 72, treating 72 as the median.
4. The double bar graph below compares the number of minutes Ana and Ben exercised each day from Monday to Friday. Refer to the graph to answer the question. On how many days did Ana's minutes differ from Ben's minutes by MORE than the mean of the daily differences?
   Answer: 2 Daily differences |Ana−Ben|: Mon |30−20|=10; Tue |40−35|=5; Wed |25−45|=20; Thu |50−30|=20; Fri |35−40|=5. Mean difference = $(10+5+20+20+5)/5 = 12$. Days with difference >12: Wed (20) and Thu (20) = 2 days. A uses median. C counts ≥10. D counts any nonzero difference.
5. Maria played five basketball games and scored 12, 15, 18, 21, and 24 points. After a sixth game, her overall mean rose to 20 points per game. How many points did she score in the sixth game?
   Answer: 30 When you see a problem asking how the mean changes after adding a new data point, you're working with the relationship between individual values and their average. The key insight is that if you know the new mean and how many total values you have, you can find the sum of all values. First, let's find Maria's total points from the first five games: $12 + 15 + 18 + 21 + 24 = 90$ points. After six games, her mean is 20 points per game. Since mean equals total divided by number of games, we can work backwards: $\text{Total points} = \text{Mean} \times \text{Number of games} = 20 \times 6 = 120$ points. If her total after six games is 120 points, and she had 90 points after five games, then her sixth game score must be $120 - 90 = 30$ points. Let's check why the other answers don't work. Choice (A) 24 points would give her a total of $90 + 24 = 114$ points, making her mean $114 ÷ 6 = 19$ points per game. Choice (B) 26 points would yield $90 + 26 = 116$ total points and a mean of $116 ÷ 6 = 19.33$ points per game. Choice (D) 32 points would result in $90 + 32 = 122$ total points and a mean of $122 ÷ 6 = 20.33$ points per game. Remember this strategy: when a new data point changes the mean, work backwards from the target mean to find the required total, then subtract what you already have.
6. The ages of seven cousins are $11,12,13,13,14,15,16$ years. If the youngest cousin is not included, what is the median age of the remaining six cousins?
   Answer: 13.5 When you see a question about finding the median after removing data points, you need to carefully identify the new dataset and apply the median formula correctly. Starting with the ages $11, 12, 13, 13, 14, 15, 16$, you first remove the youngest cousin (age 11). This leaves you with six ages: $12, 13, 13, 14, 15, 16$. To find the median of an even number of values, you take the average of the two middle values. With six numbers, the middle positions are the 3rd and 4th values when arranged in order. Here, those are both 13 and 14. The median is $\frac{13 + 14}{2} = \frac{27}{2} = 13.5$. Looking at the wrong answers: Choice A (13) would be correct if you mistakenly thought the median was just one of the middle values instead of their average. Choice C (14) represents the same error, taking the other middle value. Choice D (14.5) might result from incorrectly calculating the average of 14 and 15, which would happen if you miscounted the middle positions or forgot that arrays start counting from the first position. Remember that with an even number of data points, the median is always the average of the two middle values, which often results in a decimal answer even when all your original data points are whole numbers. Don't let decimal answers throw you off—they're often correct when working with even-sized datasets.
7. For the set of numbers $4,9,11,19,22$, what is the range?
   Answer: 18 When you encounter a question asking for the "range" of a data set, you're being asked to find the spread of the data — specifically, the difference between the largest and smallest values. To find the range of the set $4, 9, 11, 19, 22$, you need to identify the maximum and minimum values, then subtract. The largest number is 22 and the smallest is 4. Therefore, the range is $22 - 4 = 18$, which is answer choice C. Let's examine why the other options are incorrect. Choice A (11) is actually the median of this data set — the middle value when the numbers are arranged in order. This is a common confusion since both range and median are measures related to data sets. Choice B (17) doesn't correspond to any standard statistical measure for this set; it might result from incorrectly subtracting 22 - 5 or making another calculation error. Choice D (26) is the sum of the maximum and minimum values (22 + 4), which represents the trap of adding instead of subtracting. Remember that range is always about subtraction: maximum minus minimum. Don't confuse it with other statistical measures like mean, median, or mode. On the TACHS, statistics questions often include answer choices that are other statistical measures from the same data set, so make sure you're calculating exactly what the question asks for.
8. A standard deck contains 52 playing cards, 26 of which are red. If one card is drawn at random, what is the probability that the card is not red?
   Answer: $\dfrac{1}{2}$ When you encounter probability questions, remember that you're looking for the ratio of favorable outcomes to total possible outcomes. This question asks for the probability of drawing a card that is not red. Since there are 26 red cards in a standard 52-card deck, there must be 26 cards that are not red (these are the black cards). The probability of drawing a non-red card is therefore $\frac{26}{52} = \frac{1}{2}$. You can simplify this by dividing both numerator and denominator by 26. Looking at the wrong answers: Choice A ($\frac{1}{4}$) represents the probability of drawing cards from just one suit, like all hearts or all spades, since each suit contains 13 cards out of 52. Choice C ($\frac{3}{4}$) doesn't correspond to any meaningful probability in this context—it's too large since we know exactly half the deck is non-red. Choice D ($\frac{13}{52}$) is the unreduced form of $\frac{1}{4}$, representing the same one-suit probability as choice A. Notice that this problem tests the complement rule: if the probability of drawing a red card is $\frac{26}{52} = \frac{1}{2}$, then the probability of drawing a non-red card must also be $\frac{1}{2}$ since these probabilities must sum to 1. Strategy tip: For "not" probability questions, you can often work backwards from what you know. If half the deck is red, then half must be non-red. Always check that complementary probabilities add up to 1.
9. The line plot below shows the number of hours 15 students spent on homework last week. Use the line plot to answer the question. Which of the following statements is true?
   Answer: The mean is greater than the median, and the mode is 4. Data: 2,2,3,3,4,4,4,4,5,5,6,7,8,9,10. Mean = $76/15 \approx 5.07$. Median (8th value) = 4. Mode = 4. Mean > median and mode = 4. A wrong mode. B reverses mean/median. C says they're equal (false due to right skew).
10. The bar graph below shows the number of books read by members of a reading club in one month. Use the graph to answer the question. If one member is selected at random, what is the probability that the member read more than the mean number of books?
    Answer: $\frac{3}{5}$ Read the graph: 2 members read 1 book, 2 read 2, 3 read 3, 2 read 4 and 1 read 5. There are $2 + 2 + 3 + 2 + 1 = 10$ members. The total number of books is $1(2) + 2(2) + 3(3) + 4(2) + 5(1) = 2 + 4 + 9 + 8 + 5 = 28$, so the mean is $28 \div 10 = 2.8$ books. The members who read more than 2.8 books read 3, 4 or 5 books: $3 + 2 + 1 = 6$ members. The probability is $\frac{6}{10} = \frac{3}{5}$. The value $\frac{3}{10}$ counts only the members who read 4 or 5 books, as if the mean were 3 or more. The value $\frac{2}{5}$ is the probability of reading fewer books than the mean. The value $\frac{1}{2}$ assumes half of any group is above its mean.
11. Consider all consecutive integers from 7 through 15, inclusive. What is the median of this data set?
    Answer: 11 When you're finding the median of a data set, you're looking for the middle value that divides the data into two equal halves. This requires first identifying all values in the set, then arranging them in order. The consecutive integers from 7 through 15, inclusive, are: 7, 8, 9, 10, 11, 12, 13, 14, 15. That's 9 total numbers. For an odd number of data points, the median is simply the middle value. Since we have 9 numbers, the median is the 5th value when arranged in order. Counting from either end: 7, 8, 9, 10, 11, 12, 13, 14, 15. The middle value is 11, making B correct. Looking at the wrong answers: A) 10 is the 4th value, not the middle one - this might tempt you if you miscounted or forgot that "inclusive" means both 7 and 15 are included. C) 11.5 would be correct if we had an even number of data points, where you'd average the two middle values, but with 9 numbers we have a true middle value. D) 12 is the 6th value - you might choose this if you incorrectly thought the median was the value just above the middle position. Study tip: Always count your data points first. For odd numbers of values, the median is position $\frac{n+1}{2}$. For even numbers, you average the values at positions $\frac{n}{2}$ and $\frac{n}{2}+1$. Don't forget that "inclusive" means both endpoints count!
12. The box-and-whisker plot represents scores on a science test. Use the plot to answer the question. Which statement MUST be true?
    Answer: At least 25% of students scored 85 or higher. Q3 = 85, so by definition at least 25% of the data is ≥ 85. A confuses median (75) with mean—not determinable. B says 'exactly 50%'—the IQR contains approximately 50%, but with discrete data and ties it is not exactly 50%, and more importantly 'between 65 and 85' excludes endpoints. D range = max−min = 100−50 = 50, not 40; mode cannot be determined from a boxplot.
13. The circle graph below shows how students travel to Lincoln Middle School. Refer to the graph to answer the question. If the school has 480 students and 25% of bus riders are sixth-graders, how many sixth-graders ride the bus?
    Answer: 60 Bus riders: $50\%$ of 480 = 240. Sixth-graders riding bus: $25\%$ of 240 = 60. A computes $25\%$ of $25\%$ of 480. B gives $10\%$ of 480. D uses 30% instead of 25%.
14. Two fair coins are tossed simultaneously. What is the probability that at least one coin shows heads?
    Answer: $\dfrac{3}{4}$ When you encounter probability questions involving "at least one" of something, you have two main approaches: calculate directly or use the complement rule. The complement approach is often easier. Let's first identify all possible outcomes when tossing two coins. Each coin can be either heads (H) or tails (T), giving us four equally likely outcomes: HH, HT, TH, and TT. Since we want "at least one heads," this means one head OR two heads (any outcome except both tails). Using the complement rule: P(at least one heads) = 1 - P(no heads) = 1 - P(both tails). There's only one way to get both tails (TT) out of four possible outcomes, so P(both tails) = $\frac{1}{4}$. Therefore, P(at least one heads) = 1 - $\frac{1}{4}$ = $\frac{3}{4}$, which is answer D. You can verify this by counting directly: HH, HT, and TH all contain at least one heads. That's 3 favorable outcomes out of 4 total, confirming $\frac{3}{4}$. Answer A ($\frac{1}{4}$) represents the probability of getting both tails—the opposite of what we want. Answer B ($\frac{1}{3}$) might come from incorrectly thinking there are only three possible outcomes. Answer C ($\frac{1}{2}$) could result from mistakenly calculating the probability for just one specific coin rather than considering both coins together. Remember: for "at least one" probability questions, the complement rule (1 minus the probability of none) is usually your fastest path to the answer.
15. A jar contains 4 red marbles, 5 blue marbles, and 7 green marbles. If one marble is selected at random, what is the probability that the marble is blue?
    Answer: $\dfrac{5}{16}$ When you encounter probability questions, remember that probability equals the number of favorable outcomes divided by the total number of possible outcomes. First, let's find the total number of marbles in the jar: 4 red + 5 blue + 7 green = 16 marbles total. Since we want the probability of selecting a blue marble, the favorable outcomes are the 5 blue marbles. Therefore, the probability is $\dfrac{5}{16}$, which is answer choice C. Now let's examine why the other answers are incorrect. Choice A gives $\dfrac{5}{14}$, which suggests someone miscounted the total marbles as 14 instead of 16 - perhaps they forgot to include one of the colored marble groups. Choice B shows $\dfrac{5}{15}$, indicating a total count of 15 marbles, which is also incorrect but closer to the actual total. Choice D gives $\dfrac{1}{3}$, which might result from thinking there are three colors so each has a $\dfrac{1}{3}$ probability - this ignores the fact that each color group contains a different number of marbles. The key strategy for probability problems is to always verify your total count carefully. Write down each group and add them up methodically: 4 + 5 + 7 = 16. Then identify exactly what outcome you're looking for (blue marbles = 5) and set up your fraction. Double-check your arithmetic, as counting errors are common traps on standardized tests.
16. The Venn diagram below shows the number of students in a class who play basketball (B), soccer (S), or both. Refer to the Venn diagram to answer the question. If a student who plays at least one sport is chosen at random, what is the probability they play ONLY soccer?
    Answer: $\frac{10}{25}$ Only soccer = 10. Students who play at least one sport = $8 + 7 + 10 = 25$ (excluding the 5 who play neither). $P = \frac{10}{25} = \frac{2}{5}$. A uses 30 (total class) as denominator. C uses incorrect numerator. D uses total soccer players (7+10=17, but shows 15) over total class.
17. The word "STATISTICS" is written on individual letter tiles, one letter per tile. If a tile is selected at random, what is the probability that it shows a vowel?
    Answer: $\dfrac{3}{10}$ When you see a probability question involving letters, you need to identify the total number of outcomes and the number of favorable outcomes, then calculate the ratio. Let's examine the word "STATISTICS" letter by letter: S-T-A-T-I-S-T-I-C-S. Counting each tile, we have 10 total letters. Now identify the vowels (A, E, I, O, U): we find A (position 3), I (position 5), and I (position 8). That's 3 vowels out of 10 total letters. The probability of selecting a vowel is $\frac{\text{number of vowels}}{\text{total number of letters}} = \frac{3}{10}$, which is answer choice B. Let's examine why the other answers are incorrect. Choice A ($\frac{1}{5}$) equals $\frac{2}{10}$, suggesting someone miscounted and found only 2 vowels instead of 3. Choice C ($\frac{2}{9}$) reflects two errors: miscounting vowels as 2 and miscounting total letters as 9 (perhaps forgetting that repeated letters still count as separate tiles). Choice D ($\frac{1}{2}$) is far too large and suggests a fundamental misunderstanding of the setup. Remember that in probability problems involving words, every letter gets counted separately, even if it's repeated. Always write out or carefully count each position, identify your favorable outcomes systematically, and double-check your counts before calculating. The most common error on these problems is miscounting due to repeated letters.
18. The scatter plot below shows the relationship between hours studied and test scores for 12 students. Refer to the scatter plot to answer the question. If a student studied 6 hours, which is the best prediction of the test score based on the trend line?
    Answer: 84 Trend line passes through approximately (0, 60) and (10, 100), giving slope = 4 and equation $y = 4x + 60$. At $x = 6$: $y = 24 + 60 = 84$. A uses slope 2. B reads one actual data point near x=6. D extrapolates using wrong slope.
19. A fair number cube with faces numbered 1 through 6 is rolled once. What is the probability that the number shown is either a prime number or divisible by 4?
    Answer: $\dfrac{2}{3}$ When you encounter probability questions involving "or" conditions, you need to identify all favorable outcomes and avoid double-counting any overlaps. First, let's identify the outcomes on a standard die (1, 2, 3, 4, 5, 6) that satisfy each condition: - Prime numbers: 2, 3, and 5 (remember, 1 is not considered prime) - Numbers divisible by 4: only 4 Since there's no overlap between these sets (no number appears in both categories), you can simply add the favorable outcomes: {2, 3, 4, 5}. That's 4 favorable outcomes out of 6 possible outcomes, giving us $\frac{4}{6} = \frac{2}{3}$. Looking at the wrong answers: Choice A ($\frac{1}{2}$) represents only 3 out of 6 outcomes, which you'd get if you missed one of the prime numbers or forgot about 4 being divisible by 4. Choice C ($\frac{3}{4}$) suggests 4.5 out of 6 outcomes, which isn't possible with whole numbers. Choice D ($\frac{5}{6}$) represents 5 favorable outcomes, which you might get if you incorrectly included 1 as a prime number. The correct answer is B: $\frac{2}{3}$. Strategy tip: For "or" probability problems, list all favorable outcomes systematically, check for any overlaps (use addition principle), and remember that 1 is not prime. Always verify your count matches one of the answer choices when simplified.
20. The daily high temperatures (in °F) for one week were $68,70,70,72,68,70,74$. What is the mode of these temperatures?
    Answer: 70 When you encounter questions about measures of central tendency, you need to distinguish between mean (average), median (middle value), and mode (most frequent value). This question asks specifically for the mode. To find the mode, count how often each temperature appears in the data set: $68, 70, 70, 72, 68, 70, 74$. Let's tally each value: - 68 appears 2 times - 70 appears 3 times - 72 appears 1 time - 74 appears 1 time Since 70 appears most frequently (three times), the mode is 70. Looking at the wrong answers: (A) 68 is tempting because it also appears multiple times, but it only shows up twice, making it less frequent than 70. (C) 71 doesn't appear in the data set at all—this might catch students who incorrectly calculate the mean (which is 70.3) and round it. (D) 74 is the highest temperature but appears only once, so it cannot be the mode. Remember that the mode is simply the value that appears most often in a data set. When tackling these problems, always count frequencies systematically rather than trying to spot patterns visually. Also note that a data set can have no mode (if all values appear equally), one mode (like this problem), or multiple modes (if several values tie for most frequent).

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