# Statistics and odds

1. Aug 16, 2008

### Diffy

I am wondering how one would go about solving a problem such as this.

Lets say all schools compete in a competition where students are awarded a score from 0 to 5. It has been shown that the chances of a student receiving a 5 are 10%, a 4, 20% and a 3 are 25%, and below a 3 are 45%

Each school may have a different number of competitors as this is an individual event. One particular school has 7 competitors. 3 get a 5, 3 get a 4 and 1 gets a 3.

This school wants to be able to say that only 1 in x did as well.

I am thinking that if I mulitplied .10 *.10*.10*.20*.20*.20*.25 = .000002 or 1 in 500,000, but that is way too high because that is just the odds of the school getting that exact score.

I am also thinking that since the probability of getting below a 3 are 45% then if I looked at (1 -.45)^7 this is the probability of a school with 7 kids all doing better than threes (which we have in this case) so that works out to 1 in roughly 65 schools.

But this now is too low as this particular school did better than say a school that had 7 kids getting just 3's.

So assume that if a school has just one kid that gets a 4, this is better than a school that gets all 3's. and if another school has just 1 5 this is better than a school that has all 4's.

Thanks

-dif

2. Aug 16, 2008

### HallsofIvy

The 7 competitors get an average score of
$$\frac{3*5+ 3*4+ 1*3}{7}= \frac{30}{7}= 4.2[/itex] 45% get below 3, 25% get 3 and 20% get a 4 so 90% get up to 4 and 10% get above 4. Looks to me like 1 in 10 do as well. 3. Aug 16, 2008 ### Diffy I think you can be a lot more accurate. If you look at all the combinations of a school with 7 kids the scores they can get, put them into 4 buckets: below 3, 3, 4, and 5. take all the ways to distribute the 7 children to the 4 buckets of scores and in each way calculate the probability. ie: 7 0 0 0 = .45^7 6 1 0 0 = .45^6 * .25^1 etc. Then you could add up all the percentages below your schools score to get the odds that your school would score 0 1 3 3. Wouldn't that also provide you with the odds I am looking for? 4. Aug 16, 2008 ### Diffy Anyone? 5. Aug 16, 2008 ### CRGreathouse Comparing this school to other schools with 7 competitors, only [tex]\sum_{n=3}^7{7\choose n}(1/10)^n(9/10)^{7-n}$$
such schools would be expected to have at least 3 scores of five.

If you feel like getting into combinatorics, you could find the chance that a school with 7 competitors gets at least 3 scores of five, at least 6 scores better than three, and all scores better than two. Actually that's probably not too hard.

6. Aug 16, 2008

### Diffy

From my numbers above wouldn't that just be:
.10 * .10 * .10 * (.10 + .20) * (.10 + .20) * (.10 + .20) * (.10 + .20 + .25)?

7. Aug 17, 2008

### CRGreathouse

No, that's the ordered case. You want the unordered case.

8. Aug 19, 2008

### HallsofIvy

I think this was confusing me. "1 in x" what? "1 in x" students or "1 in x" schools?

My original answer was based on it being "1 in x" students. But you start talking about the chances of 7 students getting the same or better school so you appear to be saying "1 in x" schools. But the fact that you assume that a "school" is 7 students contradicts

9. Aug 19, 2008

### CRGreathouse

I think the desire is to compare schools with schools. Lacking a good way to compare schools of different sizes, the OP restricted to the case of 7-student schools -- but I imagine a good way to compare heterogeneous schools would be appreciated.