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December 21st, 2013, 05:14 PM  #1 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 602 Thanks: 82  Probability Distribution and Standard Deviation
I read something somebody else posted and it made me wonder something. If you guess on every multiple choice question without eliminating any choices first, your expected score will be 100/x where x is the number of choices for each question. Assume that all questions have exactly x choices and that there is no penalty for guessing. If you guessed like this on many tests: 1. Are your scores going to be normally distributed? 2. What is the standard deviation of this distribution in terms of x? 3. Can Questions 1 and 2 be answered in general or is it necessary to know the number of choices and/or number of questions? It's possible to calculate the probability of getting at least a certain value given a mean and standard deviation for normally distributed data, but even if this data is normally distributed the number of questions matters. For example, the probability of scoring at least a 20 (the mean) is only 0.2 if there are five choices and there is only one question. With two questions the probability rises to 0.36, with three questions it's 0.488, with four questions it's 0.5904, and with five questions it's 0.67232. With six questions the probability drops to 0.39328 (I think) because it is now necessary to get more than 1 question right to get at least a 20. 4. The number of choices is x and the number of questions is y. Is it a general rule that increasing the number of questions from y to y+1 will increase the probability of getting at least a score of 100/x if y is not a multiple of x and decrease the probability of getting at least a score of 100/x if y is a multiple x? 5. The number of choices is x and the number of questions is y. Let's say instead of wanting to get as high a score as possible you want to get a score of exactly 100/x. This is only possible y is a multiple of x. How many questions would you want there to be to maximize your chance of getting a score of exactly 100/x? I would guess that you would want x and y to be equal because if you flip a coin and want exactly half heads and half tails you would want to flip the coin 2 times. 
December 21st, 2013, 08:11 PM  #2 
Senior Member Joined: Aug 2012 Posts: 229 Thanks: 3  Re: Probability Distribution and Standard Deviation
Hey Evan3. For 1) You should think about whether the Central Limit Theorem (CLT) applies here or whether there is some other justification for the normal distribution assumption or approximation. If you think the CLT holds, this should help you with your other questions. 
December 22nd, 2013, 12:18 AM  #3  
Senior Member Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116  Re: Probability Distribution and Standard Deviation Quote:
It approximates to a Normal Distribution for a large number of questions. Quote:
Where n is the number of questions. Quote:
Quote:
Quote:
The most likely outcome is y/x (as it is a bellshaped curve). Also, as y increases, P(y/x) reduces, because there are more possibilities, so the distribution gets more spread out. So, your best chance of getting 100/x is to have 100 questions.  
December 23rd, 2013, 05:17 AM  #4 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 602 Thanks: 82  Re: Probability Distribution and Standard Deviation
Thank you. I'm surprised by 4 and 5. If there are 5 choices, the probability of getting at least 1/5th of the questions correct increases as the number of questions goes up from 1 to 5 because in all cases 1 correct answer is necessary to do that. As the number of choices goes up from 5 to 6, the number of correct answers necessary goes up from 1 to 2 (obviously you can't get 6/5 = 1.2 correct answers) and the minimum score necessary to get at least 1/5th of the questions right goes from 20 to 33 1/3. So how can the probability of getting at least 1/5th of the questions correct increase when the increase in number of questions cause the lowest possible score that is > or = to 1/5th of the questions to increase?

December 23rd, 2013, 08:35 AM  #5 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 602 Thanks: 82  Re: Probability Distribution and Standard Deviation
I decided to compare the probabilities to what you get from the normal distribution table given the standard deviation Pero posted for 100 questions with 5 choices each. Here are my results: p(16 to 24 successes from the normal distribution) = 0.6826 p(16 to 24 successes from the binomial formula) = 0.7381 That isn't extremely close. I'll do it for 1,000 questions and a smaller range of probabilities: p(190 to 210 successes from the normal distribution) = 0.5704 p(190 to 210 successes from the binomial formula) = 0.5935 I used a smaller range of probabilities because if I did p(160 to 240 success) it would have been almost 1 in both methods. 

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deviation, distribution, probability, standard 
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