Ask a Mathematician!

[GhostWriter16]

Why do you think you'll need geometry?

I don't but I had no choice but to take that

 

[mayor]

So I've (well it's my co-worker) got a problem.

She needs to calculate the minimal size for a sample she's about to do.
The formula used for this is:

N'=(N*z^2*p(1-p))/(z^2*p(1-p)+(N-1)*F^2)

in this formula:
N'=amount of respondents needed
N=population
z=standard deviation
p=chance of a certain answer
F=marge of error

The known values are:

N'=33
N=35
z=1,96
p=50%
F=5%

Yes, I'm ashamed to say I'm not kidding. We know every single value that this formula has to offer. The answer was received trough a very nifty piece of software, unfortunately she has to write down how the problem is solved.

Now you say, that shouldn't be to hard. You have all the values. True, an again I'm ashamed to say, I can't seem to get to the right answer...

If someone could give me a hand how to write this down that would be much appreciated.

Spoiler :

The worst part is... I followed a class in this and had the highest score of said class and for some reason I can't seem to solve it

 

[Atticus]

Maybe you should post how you did it and what's the expected answer?

Wild guess of the error is that you entered probabilty as 50, while it should be 0.5. And I'd guess margin of error should be 0.05 then, but don't know much about statistics.

 

[CKS]

Look at the numerator first:
35 * (1.96)^2 * .5 * .5 = 33.614
(p = .5, so 1-p is also .5)

Now look at the denominator, which has two parts:
The first part: 1.96^2 * .5 * .5 = .9604
The second part: (35 - 1) * .05^2 = .085

So our answer is: 33.614 / (.9604 + .085) = 33.614/1.0454 = 32.15. Since we can't interview a sixth of a person, we must interview 33 people.

Hopefully there are no typos here, as the calculator I have with me doesn't show what I've typed, and I'm too lazy to do it myself.

 

[mayor]

Maybe you should post how you did it and what's the expected answer?

Wild guess of the error is that you entered probabilty as 50, while it should be 0.5. And I'd guess margin of error should be 0.05 then, but don't know much about statistics.

Hmm that might be prudent.
I did it the following way:

(35*1.96^2*0.5(1-0.5))/(1.96^2*0.5(1-0.5)+(35-1)*0.05^2)
=
(35*3.8416*0.5*0.5)/(3.8416*0.5*0.5+34*0.025)
=
33.614/1.8104
=
18,57

since the answer should be 33 (more or less) it is quite obvious I made a mistake... but for the life of me I can't figure out what

 

[mayor]

Look at the numerator first:
35 * (1.96)^2 * .5 * .5 = 33.614
(p = .5, so 1-p is also .5)

Now look at the denominator, which has two parts:
The first part: 1.96^2 * .5 * .5 = .9604
The second part: (35 - 1) * .05^2 = .085

So our answer is: 33.614 / (.9604 + .085) = 33.614/1.0454 = 32.15. Since we can't interview a sixth of a person, we must interview 33 people.

Hopefully there are no typos here, as the calculator I have with me doesn't show what I've typed, and I'm too lazy to do it myself.

GIANT PACEPALM

thanks I found my mistake

for some reason I had 0.05^2=0.025 instead 0.05^2=0.0025

 

[IdiotsOpposite]

Looks to me like I need to get up earlier... the problem's already solved by the time I get here!

 

[nc-1701]

Looks to me like I need to get up earlier... the problem's already solved by the time I get here!

Here's a puzzle for you...

You have 2 pieces of string of different, unspecified length, and some matches. Each piece of string takes exactly an hour to burn, but the burn rate is not constant. This means that it could take 59 minutes to burn the first 1⁄4, and 1 minute for the rest. The strings have different burn rates, and of course you don't know the rates anyway.

Using only the matches and the strings, measure 45 minutes.

This is one of my favorite puzzles that I have ever solved. The bonus version I came up with is to show that if given an infinite number of such strings you can measure any quantity of time greater than a certain amount to an arbitrary precision.

 

[CKS]

Here is my method.

Spoiler :

Use some matches and one string to make a pendulum. Burn the other string and count swings to determine the period of the pendulum. (This will be somewhat tedious.) Now use the pendulum as a clock to measure out 45 minutes.

 

[nc-1701]

Here is my method.

Spoiler :

Use some matches and one string to make a pendulum. Burn the other string and count swings to determine the period of the pendulum. (This will be somewhat tedious.) Now use the pendulum as a clock to measure out 45 minutes.

Spoiler :

But this would take an hour to calibrate, and you want to calculate 45 minutes now. Also this uses other assumptions like being on a planet that rotates etc. Which we don't make, there is a much more elegant mathematical solution.

 

[Leoreth]

Here is my method.

Spoiler :

Use some matches and one string to make a pendulum. Burn the other string and count swings to determine the period of the pendulum. (This will be somewhat tedious.) Now use the pendulum as a clock to measure out 45 minutes.

Channeling Richard Feynman?

 

[CKS]

Spoiler :

But this would take an hour to calibrate, and you want to calculate 45 minutes now. Also this uses other assumptions like being on a planet that rotates etc. Which we don't make, there is a much more elegant mathematical solution.

Spoiler :

Well, it does take an hour to calibrate, but the problem said "measure 45 minutes", not "measure 45 minutes starting now."

It does assume the existence of a gravitational field that does not vary in time, but it doesn't require anything else. It does not assume a rotating planet.

Whether a mathematical solution is more elegant than a physical solution is entirely a matter of preference.

 

[meromorph]

I had not seen that riddle before, nice one.

Spoiler :

Light both ends of string #1, and light one end of string #2. When string #1 is completely gone we know that one half hour has passed, then light the other end of string #2.

Note that this makes the physical assumption that we are in an environment with oxygen and so can burn the matches.

 

[IdiotsOpposite]

Here's a puzzle for you...



This is one of my favorite puzzles that I have ever solved. The bonus version I came up with is to show that if given an infinite number of such strings you can measure any quantity of time greater than a certain amount to an arbitrary precision.

Ooh! A puzzle! Fun. I'll try to avoid the spoilers, as I'm sure they all have the right answer! I'll get back to you on this one.

 

[Gamemaster77]

Have you ever thought about the possibility of going crazy ? In math class two years ago, we watched a video that showed us a bunch of famous mathematicians. Pretty much all of them went crazy. They started talking to numbers and stuff like that.

 

[azzaman333]

Have you ever thought about the possibility of going crazy ? In math class two years ago, we watched a video that showed us a bunch of famous mathematicians. Pretty much all of them went crazy. They started talking to numbers and stuff like that.

Insanity is an unfortunate consequence of being too good at thinking about maths.

 

[Kozmos]

Similar with chess, when I played extensively I started seeing in grids and began considering how to eat girls on a night out by approaching them in an L-shaped manner.

 

[Mise]

Solution 1:

Spoiler :

Assuming you have a lot of matches, burn as many matches as you can in one hour (measured by the length of the string). Matches should have a constant burn rate, so once you determine how long each match takes (60 minutes / number of matches burnt in an hour), you can light as many matches as necessary to get to 45 mins.

Solution 2:

Spoiler :

Light one end of one string, and both ends of the other string. It'll take half an hour to burn through the 2nd string. Once the 2nd string is burnt, immediately light the other end of the fisrt string. It will take 15 mins to burn. Total time = 45 mins.

Solution 3:

Spoiler :

Light one end of one string. At some point, you will have measured 45 minutes.

P.S. I don't see why there can't be several solutions to this puzzle. I mean, if you gave 10 people the same tools and the same problem, they'd come up with several different answers. Some may work better than others, some may be faster or easier, but they all get the job done. I don't see what's wrong with CKS's solution - it has the benefit of being able to measure out any length of time, without destroying as much of your equipment as <other solutions>. It's simple and it's reusable, which, to me, is much more valuable than <other solutions>. If you ask me, his is the best overall solution.

 

[CKS]

My "mathematical" solution is distinctly non-elegant:

Spoiler :

Light both ends of string one. It will take 30 minutes to burn. Then light string 2 at both ends and somewhere in the middle. It will be burning in 4 places. When one of the two sections of string 2 burns up, light the other section in the middle. Repeat until all the string is gone. This will have string 2 burning always at four places, so it will take 15 minutes to burn. It will be a major pain to actually do.

This method can be used to measure other amounts of time, too, but it isn't very convenient. (Keep the string burning in 6 places to get 10 minutes, for example.) Clearly, with lots of strings you can measure to any amount of time you like, just burning each string in the appropriate number of places. Your precision can be as good as you like, but your accuracy will clearly suffer if you can't light the strings infinitely quickly.

I'm a physicist, not a mathematician; we tend not to look so much for elegance, and our perspective is distinctly different from that of mathematicians. While I appreciate mathematics, doing it requires a different sort of pickiness that I don't have.

 

[Mise]

I'm a physicist, not a mathematician

Spoiler :

I suppose an engineer would make a working clock from the matches. A computer scientist would join the matches together to make a series of logic gates, eventually turning the contraption into a binary clock. And a 18th Century French physicist would define "1 minute" as the time it takes for a match to burn, and then burn 45 matches in succession.