Law of Large numbers(turning pen experiment)

#1
Hello everyone,

My class did an turning pen experiment in class,where we turn a pen 5 times eventually, each time there is a possibility of the pen pointing to 1,2,3 or 4 equally.We need to sum up the accumulated sum(e.g 1+3+1+2+4+1=12). My class has are 100 people.The results showed up and 21% got 12, 12% got 11, 9% got 10, 8% got 9, 11% got 13 etc.. It is like a bell shaped curve due to CLT after 5 turns of the pen.Initially with just one turn the shape is not a normal distribution.

I know the CLT is due to the increase in turning of pens(5 times), not the number of students in class(The number of students is just to show the distribution). I then compute the sample mean using (0.21X12+ 0.12x11+0.09x10......) from the percentage of people who got the respective results.It is roughly ard 12.5. I then compute the expected value by ((1+2+3+4)/4) X5=12.5 which is 12.5. Is it the correct way to do it?

Does this shows the Law of Large numbers? Im rlly confused if the number of students(100) or the number of turns of pens(5) cause the Law of Large numbers. Can someone clarify which is the deciding factor, the number of turns of pens or the number of students?

Thanks
 

Dragan

Super Moderator
#2
Hello everyone,

My class did an turning pen experiment in class,where we turn a pen 5 times eventually, each time there is a possibility of the pen pointing to 1,2,3 or 4 equally.We need to sum up the accumulated sum(e.g 1+3+1+2+4+1=12). My class has are 100 people.The results showed up and 21% got 12, 12% got 11, 9% got 10, 8% got 9, 11% got 13 etc.. It is like a bell shaped curve due to CLT after 5 turns of the pen.Initially with just one turn the shape is not a normal distribution.

I know the CLT is due to the increase in turning of pens(5 times), not the number of students in class(The number of students is just to show the distribution). I then compute the sample mean using (0.21X12+ 0.12x11+0.09x10......) from the percentage of people who got the respective results.It is roughly ard 12.5. I then compute the expected value by ((1+2+3+4)/4) X5=12.5 which is 12.5. Is it the correct way to do it?

Does this shows the Law of Large numbers? Im rlly confused if the number of students(100) or the number of turns of pens(5) cause the Law of Large numbers. Can someone clarify which is the deciding factor, the number of turns of pens or the number of students?

Thanks
Well, first, what you have are integer values of 1,2,3,4 with equal probability of being selected. As such, you have a probability mass function that is referred to as a Discrete Uniform Distribution.

You are correct regarding that it is the CLT that makes the distribution of the sum of the n=5 trials for the N=100 students more "normal-like." For example, if n=1 trial for each of the 100 students, then the distribution would be Discrete Uniform. A Discrete Uniform distribution has a population mean of (a + b)/2. In your case, (1 + 4)/2 = 2.5.

In terms of the (Strong) Law of Large Numbers, you need to repeat your sampling with N=100 new students. It is most likely best that you just compute the mean (i.e. Sum/5). So repeat the experiment, T times, and compute the mean on your T samples with N fixed at 100. So, you would have XBar1, XBar2,...,XBarT sample means that are going to vary from the population mean of 2.5 because of random sampling fluctuation.

Next, compute the "mean of means" i.e., (XBar1 + XBar2 + ... + XBart)/T. This overall mean will be approximately equal to the population parameter of 2.5.

Theoretically, you would need to push T to infinity and then the "mean of means" will converge to the population mean of 2.5. This is basically what the Strong Law of Large Numbers is about - without going into to more detail.
 
#3
The \(n\) in this example is 5, the number of pen turns per student.

Try experimenting with different \(n\)s in the comfort of your own home with the following R code:
Code:
n <- 5
x <- replicate(100,sum(sample(1:4,n,replace=T)))
mean(x)
hist(x)
 
#4
And just to clarify, the above simulation will demonstrate how the sum of \(n\) independent, identically distributed random variables will converge to normality with mean \(n\mu\) and standard deviation \(\sigma\sqrt{n}\) as \(n\) gets really large. The arithmetic mean of \(n\) random variables, on the other hand, will converge to normality with mean \(\mu\) and standard deviation \(\sigma/\sqrt{n}\). For that just replace "sum(sample..." in the code with "mean(sample..."
 
#5
Well, first, what you have are integer values of 1,2,3,4 with equal probability of being selected. As such, you have a probability mass function that is referred to as a Discrete Uniform Distribution.

In terms of the (Strong) Law of Large Numbers, you need to repeat your sampling with N=100 new students. It is most likely best that you just compute the mean (i.e. Sum/5). So repeat the experiment, T times, and compute the mean on your T samples with N fixed at 100. So, you would have XBar1, XBar2,...,XBarT sample means that are going to vary from the population mean of 2.5 because of random sampling fluctuation.

Next, compute the "mean of means" i.e., (XBar1 + XBar2 + ... + XBart)/T. This overall mean will be approximately equal to the population parameter of 2.5.

Theoretically, you would need to push T to infinity and then the "mean of means" will converge to the population mean of 2.5. This is basically what the Strong Law of Large Numbers is about - without going into to more detail.
When u say repeat the experiment T times, i assume its like turning the pen 1st, 2nd, 3rd times .. to 5 for the 100 people right? Since my experiment is cumulative,summing up the results of all tries, my "means of means" is going to be simply the (percentage of students x their results)on 5th try /5)?
Since cumulative would mean the result of the 5th try is going to be inclusive of my first 4 tries.