Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a couple of maths question I want to ask, I am a computing student and I really want help with the answers to the following:

I am trying to state that if a test has more questions, then the relevance of getting questions answered correctly through guessing decreases. So if there is one question which has yes/no option to choose from, the chances of the student getting 100% in the exam is 50 per cent.

But what is the percentage if there are 50 questions each having yes/no options to gain the following full marks:

  • 100% (Full Marks)
  • 70% (Grade A)
  • 60% (Grade B)
  • 50% (Grade C)

If you can show formula on how this can be worked out then that would be great but if not it does not matter, the relevant thing is the correct percentags for each scenario.

Thanks

share|improve this question
1  
Simply giving the correct answer might be acceptable in a corporate environment, but this is an academic site. You are more likely to get a description of how to solve a problem than to get just an answer. –  robjohn Mar 26 '13 at 22:51
    
If no formula is given, how do you know the percentages are correct? And if a formula is given, how do you know it is correct? It should make sense, right? –  oks Mar 26 '13 at 23:11
add comment

3 Answers

You can use binomial distribution that gives you the probability of $k$ successes out of $n$ trials when the success probability is $p$.

enter image description here

In your case, $n=50$ and $k$ varies based on the percentage of correct answers. For example, $100\%$ means $k=50$, i.e. $50$ correct answers with probability $p=0.5$(since two choices yes or no are made with equal probability). Can you solve now for rest of the grades?

share|improve this answer
    
Only issues is what is f, p and X suppose to be? If you can show me step by step what the actual calculation should be to work out 70%, then I should be able to work out rest of grades using this example. –  user1394925 Mar 26 '13 at 21:08
    
$X$ is the random variable, i.e., number of questions that were correctly answered. I have already indicated what is $p$ above. If you don't know how to calculate the binomial probabilities, you should not be dealing with question or course anyway. Please read up! –  jay-sun Mar 26 '13 at 21:40
    
Im doing masters on database and web programming, I don't do maths, this is something I just want to include, I don't have to include it but I do need answers for each percentage. Let me ask you this, do you expect a person like myself who has not done maths for 7 years to just understand this formula by looking at it? :) –  user1394925 Mar 26 '13 at 21:45
    
Also how can I calculate this? I don't have a scientific calculator, is there an online feature which may help do this? –  user1394925 Mar 26 '13 at 21:47
    
@user1394925 vassarstats.net/binomialX.html. This should help you. For the 70% answer, you should find $k$ by multiplying 70% with $n=50$ to get $k=35$ and $p=0.5$. The answer for this case is 0.001999138255 –  jay-sun Mar 26 '13 at 22:44
add comment

$50$ random guesses is a binomial distribution.

Mean $25$, Standard Deviation(SD) = $\sqrt{Npq}$ about $3.5$. This approaches a normal distribution with same mean and sigma.

Since each answer is $2$% mean = $50$, SD = $7$

$68.2$% between $43-57$% $95.4$% between $36-64$% $99.7$% between $29-71$%

see wikipedia normal and binomial

share|improve this answer
    
It really helps to format using MathJax (see FAQ). Regards –  Amzoti Mar 26 '13 at 21:18
add comment

The answers are:

The probability of getting a 50% is 11.23%

The probability of getting a 60% is 4.19%

The probability of getting a 70% is .2%

The probability of getting a 80% is 0%

share|improve this answer
    
Care to explain? –  robjohn Mar 26 '13 at 22:49
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.