So I'm really quite awful at probability.
I have a homework problem (so i'm not asking for the answer, rather guidance on how to think about it)
A professor has a 20 question true/false reading quiz he's giving to his students. He wants to choose the passing grade such that there is less
than a .05 chance of a student who is just guessing can pass. What score should he set?
So I know it's binomial, the average score someone would get is a 50% with a standard deviation of sqrt(5).
Where do I go from there? Would I do 10 (number you get right on average) + 2 standard deviations for a 95% interval? Or am I missing something
Stats Question
Moderators: gmalivuk, Moderators General, Prelates

 Posts: 64
 Joined: Tue Nov 03, 2009 1:59 am UTC
 Contact:
Stats Question
EXPAND: (a+b)^n
(a+b)^n
(a+b)^n
(a+b)^n
GENERATION 191,184,382 : The first time you see this, copy it into your sig and divide the generation number by 2 if it's even, or multiply it by 3 then add 1 if it's odd. Social experiment.
(a+b)^n
(a+b)^n
(a+b)^n
GENERATION 191,184,382 : The first time you see this, copy it into your sig and divide the generation number by 2 if it's even, or multiply it by 3 then add 1 if it's odd. Social experiment.
 Xanthir
 My HERO!!!
 Posts: 5215
 Joined: Tue Feb 20, 2007 12:49 am UTC
 Location: The Googleplex
 Contact:
Re: Stats Question
Overall, yes, but note that you want the 1tail z score, because you want the 5% all at one end, not split between both ends. 2 stdevs is the 2tail score.
(defun fibs (n &optional (a 1) (b 1)) (take n (unfold '+ a b)))
Re: Stats Question
It looks like you are on the right track. Just be careful that for a 95% confidence interval, there is just a 2.5% chance of being at least 2 standard deviations above the mean. Do you see why?
Edit: Xanthir sees why
Edit: Xanthir sees why
 MartianInvader
 Posts: 772
 Joined: Sat Oct 27, 2007 5:51 pm UTC
Re: Stats Question
Another way to do this problem, rather than using the normal approximation, is to just calculate the probabilities directly. You can use the formula for the binomial distribution to determine the chance that a student randomly guessing gets all 20 questions right, then add that to the chance of getting 19/20 questions right, and continue until you get something bigger than 5%.
Let's have a fervent argument, mostly over semantics, where we all claim the burden of proof is on the other side!
Who is online
Users browsing this forum: No registered users and 8 guests