When you encounter problems involving categories and subcategories like this one, you're working with two-way tables that organize data by multiple characteristics. The key is to systematically break down each group and calculate step by step.

Start by finding how many students are in each study habit group. Since 45% study regularly, that's $$240 \times 0.45 = 108$$ regular studiers. The remaining students study sporadically: $$240 - 108 = 132$$ sporadic studiers.

Next, determine how many sporadic studiers score above average: $$132 \times 0.35 = 46.2$$, which rounds to 46 students. Since there are 132 total sporadic studiers and 46 score above average, the number scoring at or below average is $$132 - 46 = 86$$ students.

Looking at the wrong answers: Choice A (46 students) represents the number of sporadic studiers who score above average, not at or below average. Choice B (108 students) is the total number of regular studiers, which has nothing to do with sporadic studiers' performance. Choice C (132 students) is the total number of sporadic studiers, but the question asks specifically for those scoring at or below average.

The correct answer is D: 86 students.

Remember that two-way table problems require you to carefully track which subgroup you're calculating. Always identify the total for each category first, then use the given percentages to find the specific subgroups you need. Double-check that your final answer matches exactly what the question is asking for.

This is a conditional probability problem that requires you to work backwards from the given information. When you see questions asking "what percentage of Group A belongs to Category B," you're looking for a conditional probability.

Start by finding the actual numbers. Of 180 students, $$\frac{2}{3}$$ eat breakfast regularly: $$180 \times \frac{2}{3} = 120$$ regular breakfast eaters and $$180 - 120 = 60$$ irregular breakfast eaters.

Next, calculate how many students in each group perform well academically. Among regular breakfast eaters: $$120 \times 0.75 = 90$$ students perform well. Among irregular breakfast eaters: $$60 \times 0.40 = 24$$ students perform well.

The total number of students who perform well academically is $$90 + 24 = 114$$ students.

The question asks: of all students who perform well academically, what percentage are regular breakfast eaters? This means: $$\frac{90}{114} = 0.789 = 78.9%$$

Choice A (75.0%) represents the percentage of regular breakfast eaters who perform well, but that's not what the question asks. Choice B (66.7%) is simply the fraction of all students who eat breakfast regularly ($$\frac{2}{3}$$), ignoring academic performance entirely. Choice D (83.3%) might result from calculation errors, possibly confusing the conditional relationships.

Remember: conditional probability questions often flip the relationship. Pay careful attention to whether you're asked "percentage of A that are B" versus "percentage of B that are A" – these give very different answers.

Students without smartphones: 200 - 140 = 60. Students who complete homework on time but don't have smartphones: 110 - 85 = 25. Relative frequency = 25/60 = 0.417. Choice B uses total homework completers as denominator. Choice C uses smartphone owners who complete homework as numerator and total smartphone owners as denominator. Choice D incorrectly calculates students without smartphones who don't complete homework on time as numerator.

First, find the number of light users: 150 × (1 - 0.40) = 90 teens. Among light users, 30% have poor sleep quality, so 70% have good sleep quality. Therefore: 90 × 0.70 = 63 teens are light users with good sleep quality. Choice A incorrectly uses 70% of heavy users. Choice C gives total light users, not those with good sleep. Choice D gives light users with poor sleep quality (90 × 0.30).

To find the relative frequency of students who have a part-time job among those who play sports, we need the conditional probability P(part-time job | plays sports) = (number who do both)/(number who play sports) = 25/80 = 0.3125. Choice B uses the wrong denominator (total with jobs instead of total who play sports). Choice C uses the total surveyed as denominator. Choice D incorrectly uses students who play sports but don't have jobs.

This question tests constructing two-way tables for categorical data, calculating row/column relative frequencies (percentages), and identifying associations by comparing conditional rates. Two-way table: rows for one variable (season: summer/winter), columns for other (sport: yes/no), cells show counts (36 students: summer AND sport yes). Relative frequencies: row relatives show conditional—of summer students, 36/60=60% play sport vs 16/40=40% of winter students. Different rates (60% vs 40%) suggest association: favorite season relates to playing sports (summer fans more likely). Similar rates suggest independence. The correct comparison uses row relatives 36/60=60% and 16/40=40%, showing summer higher, as in choice B. A common error is using wrong denominators, like grand total in choice A (36/100=36%, 16/100=16%) or inverting fractions in choice C (60/36≈167%, 40/16=250%), leading to incorrect conclusions.

This question tests constructing two-way tables for categorical data, calculating row/column relative frequencies (percentages), and identifying associations by comparing conditional rates. Relative frequencies include row relatives, which show conditional probabilities, such as among students with curfew (30 total), 20/30 ≈ 67% have chores, versus 5/20 = 25% for those without curfew, with differing rates suggesting an association between curfew and chores, while similar rates would indicate independence. In this example with 50 students, the row relative frequency for chores among curfew students is 20/30 ≈ 67%, as calculated in choice B, using the row total as the denominator for the conditional percentage. This is the correct interpretation, as it focuses on the proportion within the curfew-yes row. A common error is using the wrong denominator, like the grand total (20/50 = 40% in A) or column total (20/25 = 80% in C), which computes a different relative frequency. To calculate row relatives: divide the cell count by its row total (e.g., 20/30 for curfew yes and chores yes); for association, compare these across rows. Mistakes include confusing row with column relatives or arithmetic errors, such as misdividing (e.g., treating 30/50 = 60% as the row relative in D).

This question tests interpreting associations in two-way tables by comparing conditional relative frequencies. Association is supported by differing rates, like 20/30≈67% chores among curfew-yes vs 5/20=25% among no-curfew. Data: curfew yes (20 chores,10 no), no (5,15), with unequal conditionals indicating relation. Choice A best supports this using row relatives. Errors: marginals (B:30 vs20), overall (C:25/50=50%), causation (D). Compare relatives across categories—if different, associated. Mistakes include using absolutes or inferring cause.

This question tests constructing two-way tables for categorical data, calculating row/column relative frequencies (percentages), and identifying associations by comparing conditional rates. Two-way table: rows for one variable (bring lunch: yes/no), columns for other (buy snack: yes/no), cells show counts (18: bring yes AND snack yes). Relative frequencies: row relatives show conditional—of bring lunch students, 18/24=75% buy snack vs 12/36≈33% of no-bring students. Different rates suggest association: bringing lunch relates to buying snacks (bringers more likely). Similar rates suggest independence. The correct set of row relatives is 18/24=75% and 12/36≈33%, as in choice A. A common error is using grand total like in choice B (18/60=30%, 12/60=20%) or inverting like in choice C (24/18=133%, 36/12=300%), or wrong cells like 6/24=25% in choice D (percent not buying).

This question tests constructing two-way tables for categorical data, calculating row/column relative frequencies (percentages), and identifying associations by comparing conditional rates. Relative frequencies include column relatives, which are conditional, such as among students with chores (25 total), 20/25 = 80% have curfew versus 10/25 = 40% without chores, with differences suggesting association, while similarity implies independence. For this 50-student survey, the column relative frequency of curfew among chores students is 20/25 = 80%, as in choice C, using the column total as the denominator. This correctly answers the question by conditioning on the chores-yes column. Errors often involve wrong denominators, like row total (20/30 ≈ 67% in A) or grand total (20/50 = 40% in B). To compute column relatives: divide cell by column total (e.g., 20/25); compare across columns for association. Common mistakes: mixing row/column (e.g., 25/50 = 50% in D) or not recognizing conditional nature.