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座舱，通过语音、手势、面部识别等实现人机交互，如语音控制导航、音乐播放。\n驾驶员状态监测：识别疲劳、分心、抽烟、打电话等行为，通过大模型判断危险等级，触发相应提醒或干预机制。\n智能驾驶：AI 用于环境感知、路径规划与决策控制，使车辆能识别道路、行人、障碍物等，实现自适应巡航、自动泊车、车道保持等功能，提升驾驶安全性与便利性。",{"id":492,"value":492,"name":136,"intro":493,"add_value":137,"describe":493},4,"医学影像诊断：分析 CT、MRI 等影像，辅助医生检测疾病、识别病灶，如 AI 系统可快速发现肺部结节。\n医患对话助手：通过自然语言处理理解患者问题，提供初步健康建议或就诊指导，缓解医生接诊压力。\n智能健康管理：根据用户健康数据，提供个性化健康建议、疾病预警，如智能手环结合 AI 分析用户运动、睡眠、心率等数据。",{"id":495,"value":495,"name":496,"intro":497,"add_value":498,"describe":497},6,"Smart Devices","多模态交互：支持语音、手势、眼神等多种交互方式，如 AI 眼镜可通过语音指令实现翻译、搜索功能。\n智能感知与处理：使设备能感知环境、理解用户需求，如智能音箱根据环境噪音自动调节音量。\n个性化服务：根据用户使用习惯和偏好，提供个性化功能与服务，如智能手表为用户定制运动健身计划。","microscope",{"id":500,"value":500,"name":501,"intro":502,"add_value":503,"describe":502},8,"Law & 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Culture","lang",{"id":564,"unique_code":565,"title":566,"icon":8,"intro":567,"cate":483,"detail":568,"create_time":569,"update_time":570,"price":571,"view_count":545,"download_count":572,"unit":572,"amount":573,"is_create_order":572,"enterprise_id":488,"seo_keywords":574,"seo_description":575,"data_format":576,"file_size":577,"apply_scene":578,"training_task":579,"version":580,"dataset_type_id":510,"dataset_scene_id":505,"sample_file":581,"banner":8,"unit_name":8,"cate_name":449,"icon_url":8,"banner_url":8,"sample_file_url":8,"data_dict":582,"qa_list":619,"sample_metadatas":636,"text_sample_content":673,"enterprise_name":674,"enterprise_logo":675,"data_distribution":676},1223,"109c41878cc9e461f192e3d909e7d3fb","SAT Problem Solving and Data Analysis Question Answering Dataset","Train and evaluate SAT data analysis assistants with focused questions and answers that explain the reasoning behind each result.","\u003Cp>Built from SAT preparation web content, this dataset provides closed-domain question answering on ratios, percentages, statistics, probability, tables, and data interpretation. It covers terminology, conditions for applying concepts, data relationships, and common misconceptions. Records can include source content, questions, standard answers, explanations, key concepts, and structured analysis for training and evaluating SAT-focused knowledge assistants.\u003C\u002Fp>","2026-10-15 18:03:01","2026-10-19 18:25:13","0.00",0,5000,"SAT data analysis,SAT QA dataset,SAT math training,ratios and percentages,statistics question answering,probability problem explanations,table data interpretation,educational AI training","Closed-domain SAT Q&A on ratios, percentages, statistics, probability, and data interpretation, with standard answers and explanations for tutoring and model","JSON","0.3GB","SAT data analysis, statistical problem solving, quantitative reasoning","question answering, mathematical reasoning, data interpretation","V1.0","rivo\u002Fsample-packages\u002Fa000538a-e1c5-40c6-ba67-00eba70fb105\u002F20261015.zip",[583,588,593,598,603,608,613],{"id":572,"name":584,"field":585,"intro":586,"type":587},"File Name","file_name","File name","string",{"id":589,"name":590,"field":591,"intro":592,"type":587},8896,"Standard Answer","answer","An accurate, closed-domain answer to the question.",{"id":594,"name":595,"field":596,"intro":597,"type":587},8897,"Question","question","An SAT data analysis question about ratios, percentages, statistics, probability, tables, or data interpretation.",{"id":599,"name":600,"field":601,"intro":602,"type":587},8898,"Answer Explanation","explanation","Explains the reasoning, calculation basis, and use of the question data to determine the answer.",{"id":604,"name":605,"field":606,"intro":607,"type":587},8899,"Source Web Content","source_content","Original SAT preparation web content or a relevant excerpt used to generate the question and answer.",{"id":609,"name":610,"field":611,"intro":612,"type":587},8900,"Key Concept","knowledge_point","Summarizes the main SAT data analysis concept or term tested, such as ratios, percentages, mean, median, or probability.",{"id":614,"name":615,"field":616,"intro":617,"type":618},8901,"Structured Analysis","structured_analysis","Organizes the methods, conditions, data relationships, and common misconceptions involved in the question and answer.","object",[620,624,628,632],{"id":621,"question":622,"answer":623},6343,"Which SAT data analysis topics are covered?","Topics include ratios, percentages, statistics, probability, table reading, and data interpretation. Questions may address terminology, conditions for applying concepts, data relationships, and common misconceptions.",{"id":625,"question":626,"answer":627},6344,"What information is included in each record?","Fields include source web content, a question, a standard answer, an answer explanation, a key concept, and structured analysis. Explanations and analysis can help trace calculations, reasoning steps, and conditions for reaching an answer.",{"id":629,"question":630,"answer":631},6345,"What training and evaluation uses does the dataset support?","It can support SAT-focused closed-domain question answering, educational assistant development, mathematical reasoning and data interpretation training, and evaluation of a model's understanding of SAT data analysis concepts.",{"id":633,"question":634,"answer":635},6346,"How can the dataset be obtained, and what are the licensing terms?","Check the provider's product page or delivery channel for availability and access instructions. Permitted use, commercial rights, and redistribution restrictions are governed by the applicable product agreement or license.",[637,651,662],{"id":638,"relate_table":639,"relate_id":640,"file_size":572,"file_name":641,"path":8,"duration":572,"cate":483,"file_format":642,"audio_rate":8,"audio_channel":522,"audio_bit_depth":522,"quality":8,"annotation":643,"cate_name":644,"answer":645,"question":646,"explanation":647,"source_content":648,"knowledge_point":649,"structured_analysis":650},15069,"union_cubo",1940873988,"样例_3.json","json","{\"answer\": \"The pass probability for students who took the course is 90\u002F120=75%; the pass probability for students who did not take the course is 32\u002F80=40%. The table shows that the observed pass rates of the two groups differ by 35 percentage points, but this table alone cannot establish that taking the course necessarily leads to a higher pass rate, because other influencing factors may exist, and the data show only an association.\", \"question\": \"According to the table, what is the probability that a student who took the prep course passes the exam? What is the probability that a student who did not take the course passes the exam? Can this table alone establish that taking the course necessarily leads to a higher pass rate?\", \"explanation\": \"When calculating the probability of passing given that a student took the course, the denominator should be the total number of students who took the course, 120, so the probability is 90÷120=75%. For students who did not take the course, the denominator is that group's total, 80, and the probability is 32÷80=40%. The difference between the two groups is 75%-40%=35 percentage points. This difference indicates an association between course participation and passing outcomes, but without random assignment or control for other variables, a causal relationship cannot be directly inferred from the observed table.\", \"source_content\": \"\u003Csection>\u003Ch3>Conditional Probability and Contingency Tables\u003C\u002Fh3>\u003Ctable>\u003Ctr>\u003Cth>Whether Took Prep Course\u003C\u002Fth>\u003Cth>Passed Exam\u003C\u002Fth>\u003Cth>Did Not Pass Exam\u003C\u002Fth>\u003Cth>Total\u003C\u002Fth>\u003C\u002Ftr>\u003Ctr>\u003Ctd>Took\u003C\u002Ftd>\u003Ctd>90\u003C\u002Ftd>\u003Ctd>30\u003C\u002Ftd>\u003Ctd>120\u003C\u002Ftd>\u003C\u002Ftr>\u003Ctr>\u003Ctd>Did Not Take\u003C\u002Ftd>\u003Ctd>32\u003C\u002Ftd>\u003Ctd>48\u003C\u002Ftd>\u003Ctd>80\u003C\u002Ftd>\u003C\u002Ftr>\u003C\u002Ftable>\u003Cp>The conditional probability P(pass|took) should use the group that took the course as the denominator. The table can compare outcomes across different groups, but the observed difference by itself cannot prove that the course caused the difference.\u003C\u002Fp>\u003C\u002Fsection>\", \"knowledge_point\": \"Conditional Probability, Contingency Tables, and Correlation versus Causation\", \"structured_analysis\": \"{\\\"calculation_method\\\": \\\"For each group, calculate the conditional probability separately: P(pass|took)=number who took and passed ÷ number who took, P(pass|did not take)=number who did not take and passed ÷ number who did not take; then compare the percentage-point difference between the two.\\\", \\\"applicability_conditions\\\": [\\\"The denominator of a conditional probability must be the total population corresponding to the given condition.\\\", \\\"When comparing probabilities between two groups, the event definitions and statistical criteria for both groups should be consistent.\\\", \\\"Causal conclusions usually require random assignment, an experimental design, or adequate control for confounding factors.\\\"], \\\"data_relationships\\\": [\\\"The group that took the course has 120 students, of whom 90 passed, for a pass rate of 75%.\\\", \\\"The group that did not take the course has 80 students, of whom 32 passed, for a pass rate of 40%.\\\", \\\"The group that took the course is 35 percentage points higher than the group that did not take the course, but this is only an observed between-group difference.\\\"], \\\"common_misconceptions\\\": [\\\"Using all 200 students as the denominator when calculating the pass rate for the group that took the course.\\\", \\\"Describing a 35-percentage-point difference as a 35% relative increase; relative to 40%, the relative increase is 87.5%.\\\", \\\"Interpreting the association directly as the course causing a higher pass rate.\\\", \\\"Ignoring the different group sizes and comparing only the numbers who passed, 90 and 32.\\\"]}\"}","文本","The pass probability for students who took the course is 90\u002F120=75%; the pass probability for students who did not take the course is 32\u002F80=40%. The table shows that the observed pass rates of the two groups differ by 35 percentage points, but this table alone cannot establish that taking the course necessarily leads to a higher pass rate, because other influencing factors may exist, and the data show only an association.","According to the table, what is the probability that a student who took the prep course passes the exam? What is the probability that a student who did not take the course passes the exam? Can this table alone establish that taking the course necessarily leads to a higher pass rate?","When calculating the probability of passing given that a student took the course, the denominator should be the total number of students who took the course, 120, so the probability is 90÷120=75%. For students who did not take the course, the denominator is that group's total, 80, and the probability is 32÷80=40%. The difference between the two groups is 75%-40%=35 percentage points. This difference indicates an association between course participation and passing outcomes, but without random assignment or control for other variables, a causal relationship cannot be directly inferred from the observed table.","\u003Csection>\u003Ch3>Conditional Probability and Contingency Tables\u003C\u002Fh3>\u003Ctable>\u003Ctr>\u003Cth>Whether Took Prep Course\u003C\u002Fth>\u003Cth>Passed Exam\u003C\u002Fth>\u003Cth>Did Not Pass Exam\u003C\u002Fth>\u003Cth>Total\u003C\u002Fth>\u003C\u002Ftr>\u003Ctr>\u003Ctd>Took\u003C\u002Ftd>\u003Ctd>90\u003C\u002Ftd>\u003Ctd>30\u003C\u002Ftd>\u003Ctd>120\u003C\u002Ftd>\u003C\u002Ftr>\u003Ctr>\u003Ctd>Did Not Take\u003C\u002Ftd>\u003Ctd>32\u003C\u002Ftd>\u003Ctd>48\u003C\u002Ftd>\u003Ctd>80\u003C\u002Ftd>\u003C\u002Ftr>\u003C\u002Ftable>\u003Cp>The conditional probability P(pass|took) should use the group that took the course as the denominator. The table can compare outcomes across different groups, but the observed difference by itself cannot prove that the course caused the difference.\u003C\u002Fp>\u003C\u002Fsection>","Conditional Probability, Contingency Tables, and Correlation versus Causation","{\"calculation_method\": \"For each group, calculate the conditional probability separately: P(pass|took)=number who took and passed ÷ number who took, P(pass|did not take)=number who did not take and passed ÷ number who did not take; then compare the percentage-point difference between the two.\", \"applicability_conditions\": [\"The denominator of a conditional probability must be the total population corresponding to the given condition.\", \"When comparing probabilities between two groups, the event definitions and statistical criteria for both groups should be consistent.\", \"Causal conclusions usually require random assignment, an experimental design, or adequate control for confounding factors.\"], \"data_relationships\": [\"The group that took the course has 120 students, of whom 90 passed, for a pass rate of 75%.\", \"The group that did not take the course has 80 students, of whom 32 passed, for a pass rate of 40%.\", \"The group that took the course is 35 percentage points higher than the group that did not take the course, but this is only an observed between-group difference.\"], \"common_misconceptions\": [\"Using all 200 students as the denominator when calculating the pass rate for the group that took the course.\", \"Describing a 35-percentage-point difference as a 35% relative increase; relative to 40%, the relative increase is 87.5%.\", \"Interpreting the association directly as the course causing a higher pass rate.\", \"Ignoring the different group sizes and comparing only the numbers who passed, 90 and 32.\"]}",{"id":652,"relate_table":639,"relate_id":653,"file_size":572,"file_name":654,"path":8,"duration":572,"cate":483,"file_format":642,"audio_rate":8,"audio_channel":522,"audio_bit_depth":522,"quality":8,"annotation":655,"cate_name":644,"answer":656,"question":657,"explanation":658,"source_content":659,"knowledge_point":660,"structured_analysis":661},15068,1075433021,"样例_2.json","{\"answer\": \"The mean is approximately 21.43, and the median is 18. Because 50 is clearly higher than the rest of the data, the mean is strongly affected by the outlier; when describing the typical level of most of the data, the median should be preferred.\", \"question\": \"For the data 12, 15, 15, 18, 20, 20, 50, what are the mean and median respectively? If you want to describe the typical level of most of the data, which statistic should be preferred?\", \"explanation\": \"There are 7 data values in total, and the sum is 150, so the mean is 150÷7≈21.43. The data are already arranged in ascending order, and the 4th value is 18, so the median is 18. The value 50 shifts the mean upward from the central range of the remaining data, whereas the median depends only on the middle position and is less affected by extreme values.\", \"source_content\": \"\u003Csection>\u003Ch3>Mean and Median\u003C\u002Fh3>\u003Cp>A set of data arranged in ascending order is 12, 15, 15, 18, 20, 20, 50. The mean equals the sum of all data values divided by the number of data values, and the median is the value in the middle position of the ordered data.\u003C\u002Fp>\u003Cp>When there is a clearly larger outlier in the data, the mean may be pulled upward by the outlier, while the median usually better represents the typical position of the data.\u003C\u002Fp>\u003C\u002Fsection>\", \"knowledge_point\": \"Mean, Median, and Outliers\", \"structured_analysis\": \"{\\\"calculation_method\\\": \\\"Mean = sum of data ÷ number of data values; for an odd number of ordered data values, the median is the value in the middle position, which in this problem is the (7+1)÷2=4th data value.\\\", \\\"applicability_conditions\\\": [\\\"Before calculating the median, the data should be arranged in non-descending order.\\\", \\\"When using the median to represent a typical level, consider whether the data contain obvious outliers or a skewed distribution.\\\"], \\\"data_relationships\\\": [\\\"The sum of the 7 data values is 150, and the mean is 150÷7≈21.43.\\\", \\\"The 4th data value in the middle position is 18, so the median is 18.\\\", \\\"The value 50 is higher than the rest of the data, causing the mean to be higher than the 12 to 20 interval where most observations lie.\\\"], \\\"common_misconceptions\\\": [\\\"Mistaking the median for the average of the maximum and minimum values.\\\", \\\"Looking for the middle item directly without sorting the data first.\\\", \\\"Concluding that most data values are close to the mean just because the mean is relatively high.\\\", \\\"Believing that the mean and median are equally unaffected by outliers in any data distribution.\\\"]}\"}","The mean is approximately 21.43, and the median is 18. Because 50 is clearly higher than the rest of the data, the mean is strongly affected by the outlier; when describing the typical level of most of the data, the median should be preferred.","For the data 12, 15, 15, 18, 20, 20, 50, what are the mean and median respectively? If you want to describe the typical level of most of the data, which statistic should be preferred?","There are 7 data values in total, and the sum is 150, so the mean is 150÷7≈21.43. The data are already arranged in ascending order, and the 4th value is 18, so the median is 18. The value 50 shifts the mean upward from the central range of the remaining data, whereas the median depends only on the middle position and is less affected by extreme values.","\u003Csection>\u003Ch3>Mean and Median\u003C\u002Fh3>\u003Cp>A set of data arranged in ascending order is 12, 15, 15, 18, 20, 20, 50. The mean equals the sum of all data values divided by the number of data values, and the median is the value in the middle position of the ordered data.\u003C\u002Fp>\u003Cp>When there is a clearly larger outlier in the data, the mean may be pulled upward by the outlier, while the median usually better represents the typical position of the data.\u003C\u002Fp>\u003C\u002Fsection>","Mean, Median, and Outliers","{\"calculation_method\": \"Mean = sum of data ÷ number of data values; for an odd number of ordered data values, the median is the value in the middle position, which in this problem is the (7+1)÷2=4th data value.\", \"applicability_conditions\": [\"Before calculating the median, the data should be arranged in non-descending order.\", \"When using the median to represent a typical level, consider whether the data contain obvious outliers or a skewed distribution.\"], \"data_relationships\": [\"The sum of the 7 data values is 150, and the mean is 150÷7≈21.43.\", \"The 4th data value in the middle position is 18, so the median is 18.\", \"The value 50 is higher than the rest of the data, causing the mean to be higher than the 12 to 20 interval where most observations lie.\"], \"common_misconceptions\": [\"Mistaking the median for the average of the maximum and minimum values.\", \"Looking for the middle item directly without sorting the data first.\", \"Concluding that most data values are close to the mean just because the mean is relatively high.\", \"Believing that the mean and median are equally unaffected by outliers in any data distribution.\"]}",{"id":663,"relate_table":639,"relate_id":664,"file_size":572,"file_name":665,"path":8,"duration":572,"cate":483,"file_format":642,"audio_rate":8,"audio_channel":522,"audio_bit_depth":522,"quality":8,"annotation":666,"cate_name":644,"answer":667,"question":668,"explanation":669,"source_content":670,"knowledge_point":671,"structured_analysis":672},15067,659781417,"样例_1.json","{\"answer\": \"The final enrollment is the same as the initial enrollment, with no net increase or decrease. You cannot simply subtract 20% from 25%, because the bases for the two percentage changes are different: the second decrease of 20% is calculated on the increased amount.\", \"question\": \"The enrollment for a certain SAT prep course first increases by 25%, and then decreases by 20%. How does the final enrollment compare with the initial enrollment? Why can't we simply say that increasing by 25% and then decreasing by 20% equals an increase of 5%?\", \"explanation\": \"Let the initial enrollment be N. After a 25% increase, it is 1.25N; after a further 20% decrease, it is 1.25N×(1-0.20)=1.25N×0.80=N. Therefore the final value is 100% of the original value. Directly calculating 25%-20%=5% ignores that the base of the second change has already become 1.25N.\", \"source_content\": \"\u003Csection>\u003Ch3>Percentage Change\u003C\u002Fh3>\u003Cp>When calculating percentage change, the amount of change should be divided by the original value. When percentage changes occur consecutively, the base of the later change is the value after the previous change, not the initial value.\u003C\u002Fp>\u003Cp>For example, if a quantity first increases by 25% and then decreases by 20%, the final quantity is 1.25×0.80=1.00 times the original quantity.\u003C\u002Fp>\u003C\u002Fsection>\", \"knowledge_point\": \"Consecutive Percentage Changes and Change Base\", \"structured_analysis\": \"{\\\"calculation_method\\\": \\\"Convert each percentage change into a multiplication factor: an increase of r% multiplies by 1+r, a decrease of r% multiplies by 1-r, and then multiply in the order they occur.\\\", \\\"applicability_conditions\\\": [\\\"Each percentage change is calculated relative to the value before that change occurs.\\\", \\\"The original quantity can be represented as a positive number N so that the final value can be compared with the original value.\\\"], \\\"data_relationships\\\": [\\\"After the first change, the quantity is 1.25 times the original quantity.\\\", \\\"After the second change, the quantity is 0.80 times the quantity after the first change.\\\", \\\"The total change factor is 1.25×0.80=1, so the final quantity equals the original quantity.\\\"], \\\"common_misconceptions\\\": [\\\"Adding or subtracting consecutive percentage changes directly.\\\", \\\"Mistakenly believing that both changes use the initial quantity as the base.\\\", \\\"Confusing percentage-point differences with relative percentage changes.\\\"]}\"}","The final enrollment is the same as the initial enrollment, with no net increase or decrease. You cannot simply subtract 20% from 25%, because the bases for the two percentage changes are different: the second decrease of 20% is calculated on the increased amount.","The enrollment for a certain SAT prep course first increases by 25%, and then decreases by 20%. How does the final enrollment compare with the initial enrollment? Why can't we simply say that increasing by 25% and then decreasing by 20% equals an increase of 5%?","Let the initial enrollment be N. After a 25% increase, it is 1.25N; after a further 20% decrease, it is 1.25N×(1-0.20)=1.25N×0.80=N. Therefore the final value is 100% of the original value. Directly calculating 25%-20%=5% ignores that the base of the second change has already become 1.25N.","\u003Csection>\u003Ch3>Percentage Change\u003C\u002Fh3>\u003Cp>When calculating percentage change, the amount of change should be divided by the original value. When percentage changes occur consecutively, the base of the later change is the value after the previous change, not the initial value.\u003C\u002Fp>\u003Cp>For example, if a quantity first increases by 25% and then decreases by 20%, the final quantity is 1.25×0.80=1.00 times the original quantity.\u003C\u002Fp>\u003C\u002Fsection>","Consecutive Percentage Changes and Change Base","{\"calculation_method\": \"Convert each percentage change into a multiplication factor: an increase of r% multiplies by 1+r, a decrease of r% multiplies by 1-r, and then multiply in the order they occur.\", \"applicability_conditions\": [\"Each percentage change is calculated relative to the value before that change occurs.\", \"The original quantity can be represented as a positive number N so that the final value can be compared with the original value.\"], \"data_relationships\": [\"After the first change, the quantity is 1.25 times the original quantity.\", \"After the second change, the quantity is 0.80 times the quantity after the first change.\", \"The total change factor is 1.25×0.80=1, so the final quantity equals the original quantity.\"], \"common_misconceptions\": [\"Adding or subtracting consecutive percentage changes directly.\", \"Mistakenly believing that both changes use the initial quantity as the base.\", \"Confusing percentage-point differences with relative percentage changes.\"]}","{\"answer\":\"The pass probability for students who took the course is 90\\\u002F120=75%; the pass probability for students who did not take the course is 32\\\u002F80=40%. The table shows that the observed pass rates of the two groups differ by 35 percentage points, but this table alone cannot establish that taking the course necessarily leads to a higher pass rate, because other influencing factors may exist, and the data show only an association.\",\"question\":\"According to the table, what is the probability that a student who took the prep course passes the exam? What is the probability that a student who did not take the course passes the exam? Can this table alone establish that taking the course necessarily leads to a higher pass rate?\",\"explanation\":\"When calculating the probability of passing given that a student took the course, the denominator should be the total number of students who took the course, 120, so the probability is 90÷120=75%. For students who did not take the course, the denominator is that group's total, 80, and the probability is 32÷80=40%. The difference between the two groups is 75%-40%=35 percentage points. This difference indicates an association between course participation and passing outcomes, but without random assignment or control for other variables, a causal relationship cannot be directly inferred from the observed table.\",\"source_content\":\"\u003Csection>\u003Ch3>Conditional Probability and Contingency Tables\u003C\\\u002Fh3>\u003Ctable>\u003Ctr>\u003Cth>Whether Took Prep Course\u003C\\\u002Fth>\u003Cth>Passed Exam\u003C\\\u002Fth>\u003Cth>Did Not Pass Exam\u003C\\\u002Fth>\u003Cth>Total\u003C\\\u002Fth>\u003C\\\u002Ftr>\u003Ctr>\u003Ctd>Took\u003C\\\u002Ftd>\u003Ctd>90\u003C\\\u002Ftd>\u003Ctd>30\u003C\\\u002Ftd>\u003Ctd>120\u003C\\\u002Ftd>\u003C\\\u002Ftr>\u003Ctr>\u003Ctd>Did Not Take\u003C\\\u002Ftd>\u003Ctd>32\u003C\\\u002Ftd>\u003Ctd>48\u003C\\\u002Ftd>\u003Ctd>80\u003C\\\u002Ftd>\u003C\\\u002Ftr>\u003C\\\u002Ftable>\u003Cp>The conditional probability P(pass|took) should use the group that took the course as the denominator. The table can compare outcomes across different groups, but the observed difference by itself cannot prove that the course caused the difference.\u003C\\\u002Fp>\u003C\\\u002Fsection>\",\"knowledge_point\":\"Conditional Probability, Contingency Tables, and Correlation versus Causation\",\"structured_analysis\":\"{\\\"calculation_method\\\": \\\"For each group, calculate the conditional probability separately: P(pass|took)=number who took and passed ÷ number who took, P(pass|did not take)=number who did not take and passed ÷ number who did not take; then compare the percentage-point difference between the two.\\\", \\\"applicability_conditions\\\": [\\\"The denominator of a conditional probability must be the total population corresponding to the given condition.\\\", \\\"When comparing probabilities between two groups, the event definitions and statistical criteria for both groups should be consistent.\\\", \\\"Causal conclusions usually require random assignment, an experimental design, or adequate control for confounding factors.\\\"], \\\"data_relationships\\\": [\\\"The group that took the course has 120 students, of whom 90 passed, for a pass rate of 75%.\\\", \\\"The group that did not take the course has 80 students, of whom 32 passed, for a pass rate of 40%.\\\", \\\"The group that took the course is 35 percentage points higher than the group that did not take the course, but this is only an observed between-group difference.\\\"], \\\"common_misconceptions\\\": [\\\"Using all 200 students as the denominator when calculating the pass rate for the group that took the course.\\\", \\\"Describing a 35-percentage-point difference as a 35% relative increase; relative to 40%, the relative increase is 87.5%.\\\", \\\"Interpreting the association directly as the course causing a higher pass rate.\\\", \\\"Ignoring the different group sizes and comparing only the numbers who passed, 90 and 32.\\\"]}\"}","MOBIUSI INC","https:\u002F\u002Fstatic.mobiusi.com\u002Fmobiwusi\u002F20250829\u002F7bf4ac62b4533e2b809998a6f1c5491e.png?Expires=1792463130&OSSAccessKeyId=LTAI5tH8MQqhgeiAtrzCHrxM&Signature=v6Tr8jBrdm4s5wU1SnnyGsYg%2Bhw%3D&x-oss-process=image%2Fresize%2Cm_fill%2Cw_35%2Ch_35",[677,683],{"type":678,"title":679,"count_data":680},"bar","Length distribution",[681],{"name":682,"value":483},"1k lower",{"type":684,"title":685,"count_data":686},"pie","Language distribution",[687],{"name":688,"value":483},"English"]