{"id":"cb_question_bank:1204a73a-9c89-4781-9160-1bad831edbb7","source_id":"cb_question_bank","source_item_id":"1204a73a-9c89-4781-9160-1bad831edbb7","external_id":null,"ibn":"11167-DC","question_id":"4c774b00","program":"SAT","module":"math","domain_code":"Q","domain":"Problem-Solving and Data Analysis","skill_code":"Q.C.","skill":"One-variable data: Distributions and measures of center and spread","difficulty":"M","score_band":4,"answer_type":"mcq","answer_source":"key","layout_warning":null,"form_id":null,"module_number":null,"position":null,"stem":"The table above shows the distribution of ages of the 20 students enrolled in a college class. Which of the following gives the correct order of the mean, median, and mode of the ages?","stimulus":"Ages of 20 Students Enrolled in a College Class Age Frequency 18 6 19 5 20 4 21 2 22 1 23 1 30 1","stimulus_html":"
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\n \n
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Ages of 20 Students Enrolled in a College Class 

\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n
AgeFrequency
186
195
204
212
221
231
301
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\n","stem_html":"

The table above shows the distribution of ages of the 20 students enrolled in a college class. Which of the following gives the correct order of the mean, median, and mode of the ages?

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Choice A is correct. The mode is the data value with the highest frequency. So for the data shown, the mode is 18. The median is the middle data value when the data values are sorted from least to greatest. Since there are 20 ages ordered, the median is the average of the two middle values, the 10th and 11th, which for these data are both 19. Therefore, the median is 19. The mean is the sum of the data values divided by the number of the data values. So for these data, the mean is \"the.

Since the mode is 18, the median is 19, and the mean is 20, \"mode.

Choices B and D are incorrect because the mean is greater than the median. Choice C is incorrect because the median is greater than the mode.

Alternate approach: After determining the mode, 18, and the median, 19, it remains to determine whether the mean is less than 19 or more than 19. Because the mean is a balancing point, there is as much deviation below the mean as above the mean. It is possible to compare the data to 19 to determine the balance of deviation above and below the mean. There is a total deviation of only 6 below 19 (the 6 values of 18); however, the data value 30 alone deviates by 11 above 19. Thus the mean must be greater than 19.

\n","correct_answer":["A"],"has_media":true,"has_mathml":false,"has_table":true,"source_created_at":"2023-08-02T20:25:59.638000+00:00","source_updated_at":"2023-08-02T20:25:59.638000+00:00","options":[{"label":"A","content_html":"

mode < median < mean

\n","ord":0},{"label":"B","content_html":"

mode < mean < median

\n","ord":1},{"label":"C","content_html":"

median < mode < mean

\n","ord":2},{"label":"D","content_html":"

mean < mode < median

\n","ord":3}],"attribution":{"source_id":"cb_question_bank","source_name":"College Board digital SAT question bank (community dump)","rights_holder":"College Board","canonical_url":"https://satsuitequestionbank.collegeboard.org/","retrieved_from":"https://raw.githubusercontent.com/mdn522/sat-question-bank/main/data/cb-digital-questions.json","attribution":"SAT practice questions © College Board, from the official SAT Suite Question Bank. Retrieved via the community dump at github.com/mdn522/sat-question-bank.","disclaimer":"SAT® and PSAT/NMSQT® are trademarks registered by the College Board, which is not affiliated with, does not sponsor, and does not endorse this project.","license":"College Board copyright; source repository carries no licence. Local use only.","redistributable":false,"item_count":2017}}