A How Can We Analyze an Exam with Varying Multiple Choice Options?

AI Thread Summary
Analyzing a multiple-choice exam with varying options per question can be approached using a Binomial distribution if all questions have the same number of choices. However, when questions have different numbers of choices, a Multinomial distribution may be more appropriate. Grouping questions by their number of possible answers can facilitate this analysis. The discussion also humorously touches on a potential typo regarding the term "generalizing." Overall, the focus remains on finding a suitable statistical method for varied multiple-choice formats.
WWGD
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If we had a multiple choice exam with , say, 20 questions, with 4 choices for each question, we can analyze it as a Binomial(20, .25). What if instead , some of the questions offered 2,3, 4, etc., choices? Is there a " nice" way of analyzing the exam as a whole?
 
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Is it genitalizing or generalizing? :-p
 
Frabjous said:
Is it genitalizing or generalizing? :-p
I'm almost certain he meant "generalizing." I have changed the title to suit my assumption.
 
Mark44 said:
I'm almost certain he meant "generalizing." I have changed the title to suit my assumption.
It is now a less interesting thread. :cry:
 
WWGD said:
If we had a multiple choice exam with , say, 20 questions, with 4 choices for each question, we can analyze it as a Binomial(20, .25). What if instead , some of the questions offered 2,3, 4, etc., choices? Is there a " nice" way of analyzing the exam as a whole?
Multinomial distribution? You could group the questions by their number of possible answers.
 
Frabjous said:
Is it genitalizing or generalizing? :-p
Maybe PFs auto correct is a pervert.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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