[dougmwpsu]

i've been trying to figure out this staticics problem that i dreamed up once while getting graded tests handed back to me n class. i've asked friend who've taken stat classes and even e-mailed a stat professor, but havn't been able to get an answer on it. here's the problem as i commonly pose it.

> I’m in a class of N students and we are given a multiple choice test with Q
> questions and each question has C choices. No one studied for this test and so
> we all guessed on every single question. What will, on average, be the
> highest score on the test, scored by a single very lucky student?

i'm looking for a number greater than Q/C when N>1. if anyone can offer some insight on this, or show me how to go about solving it, i'd be very appriciative
Thanks!
DOug

 

[some1x]

I wrote this rather hastily.... so there might be mistakes.

Let X = the highest number of questions right
you need to find the E[X]=
sum: i=0 to Q i * p(x=i)


Finding p(x=i) is tricky. What you can do is use some combinatorics:

First, using the binomial theorem, the probability that an individual student answers n questions right is:
((Q n) * p^n * q^(Q-n) )

where (Q n) is Q choose n
p=1/C
q=(C-1)/C

Using the above equation, you can find the probability that an individual student scores less than or equal to n/Q (just sum from 0 to n).

Finally p(x=i) = p(everybody scores <= i) - p(everybody scores <= i-1)

note: p(everybody scores <= i) = (p(an individual student scores <= i))^N

edited to remove an error.

 

[morbo667]

Quote:

Originally Posted by some1x First, using the binomial theorem, the probability that an individual student answers n questions right is: ((Q n) * p^n * q^(Q-n) ) / (C^Q)

You dont need to divide by C^Q there. The probability ist just
(Q n) * p^n * q^(Q-n)
Or choose p=1 and q=C-1. Then yours is fine.

Rest seems fine to me but will give you some nasty formulas.

 

[Homeless]

Now I remember why I hated my stats class so much!