- Ryosuke Ishii is a researcher from Tokyo, Japan who is studying the Central Limit Theorem through online courses including MITx and HarvardX. - The Central Limit Theorem states that the distribution of sample means approaches a normal distribution as the sample size increases, even if the population is not normally distributed. - Ishii provides an example where samples are taken from a population following a normal distribution N(27.6, 28.3) and calculates the mean and standard deviation of the sample means, finding them to be consistent with the theoretical values predicted by the Central Limit Theorem.

1. Central Limit Theorem( )
for 14.310x students.
Ryosuke ISHII
(ryouen)

2. About author
• Ryosuke ISHII (call me ryo / ryouen)
• From Tokyo, Japan
• Graduated from The University of Tokyo.
• Current: Researcher, Grad School of
System Design and Management, Keio Univ.
• Enjoining MITx 14.310x and learn from a lot.
• Also MITx 14.100x (Microeconomics) and
HarvardX PHP525.x (Statistics) on edX.

3. According to CLT,
When the population is following 𝜇(population mean) and 𝜎2(population variance),
we took some sample and the sample size = 𝑛,
This 𝑛 means how many items in the group. It is different from “the number of samples”
If we took many samples repeatedly, we can calculate each of sample’s mean (this is
sample mean ഥ𝑥𝑖) and the sample mean is also a random variable. And the sample mean
follows: ҧ𝑥 ~ N(𝜇,
𝜎2
𝑛
)
↑
𝜎
𝜇
𝑠 𝑥 =
𝜎
𝑛

4. Sample size is different from the number of samples.
If we compare 10 males and 15 females
The sample size of the male group is 10.
The sample size of the female group is 15.
The number of samples (or the number of groups) is 2.
The number of samples and the sample size can potentially be
confusing. Sample size is the number of items within a group. Number
of samples is the number of groups.”
*Metin Çakanyıldırım,
Computing the Standard Deviation of Sample Means

5. (if you wish, you can simulate with the R code below)
x <- rnorm(3300, mean=27.6,sd=sqrt(28.3))
n=10 #sample size
N=1000 #the number of trials
set.seed(1)
ys <- vector("numeric",N)
ysmean <- vector("numeric",N)
ysvar <- vector("numeric",N)
yssd <- vector("numeric",N)
yalldata <- vector("numeric",0)
for(i in 1:N){
ys <- sample(x, n)
ysmean[i] = mean(ys,na.rm = TRUE)
ysvar[i]= var(ys,na.rm = TRUE)
yssd[i] = sd(ys,na.rm = TRUE)
yalldata = c(yalldata,ys)
}

6. In order to understand deeper, this time assume that we know the TRUE
population parameter N(𝜇, 𝜎2
).
TRUE Parameter
mean 𝜇 = 27.6
variation 𝜎2 = 28.3
SD 𝜎 = 5.31
(This number is only for example)
↑
𝜎
𝜇
Set up

7. From a population following N 𝜇, 𝜎2
𝑛 = 10
Let us try sampling the first time!
And we set the sample size n=10
𝑥1 34 31 25 28 26 NA 25 20 27 25
②

8. 𝜇
𝜎
We repeat it 6 times. It means we have 6 groups of samples and the sample
size of each group is 10

9. These 6 samples are different because each of sampling is an random
sampling.
But the result is not perfectly random because it is taken from a population
distribution.
So, we can say ”data is a representation of random variable gain from
sampling.”*
𝑥2 = 25.4𝑥1 = 26.8 𝑥4 = 27.6𝑥3 = 27.5 𝑥6 = 26.9𝑥5 = 27.6
And also, we can calculate each of samples’ mean.
You can see the sample mean is also a random variable.

10. How to calculate the sample mean? Yes, we must know.
𝑥1 34 31 25 28 26 NA 25 20 27 25 𝑥1 =26.8
𝑥2 20 NA 22 25 NA 24 21 29 39 23 𝑥2 =25.4
𝑥3 19 16 24 29 42 27 41 21 34 22 𝑥3 =27.5
𝑥4 24 35 24 25 28 20 26 38 28 28 𝑥4 =27.6
𝑥5 27 26 28 31 23 24 NA 34 30 26 𝑥5 =27.7
𝑥6 25 26 24 28 29 NA 28 26 21 35 𝑥6 =26.9
How do you think if we take more sample?
For example, we take 200 samples, and calc sample mean.

11. We can plot a histogram of𝑥1~𝑥200
There are 200 averages (of samples) and each of the average is random
variable.
Next, we would like to
calculate the distribution’s (this histogram’s)
-mean of sample means ( ҧ𝑥)
-variation of sample means (𝑉𝑥)
-standard deviation of sample means (𝑠 𝑥)

12. We can calculate it by definition. (I used R to calculate)
mean of sample means ( ҧ𝑥)
ҧ𝑥 =
1
𝑛
෍
𝑖=1
𝑛
ഥ𝑥𝑖 =
ഥ𝑥1 + ഥ𝑥2 + ⋯ + 𝑥199 + 𝑥200
200
= 27.541
variation of sample means
𝑉𝑥 =
1
𝑛 − 1
෍
𝑖=𝑖
𝑛
ഥ𝑥𝑖 − ҧ𝑥 2
=
𝑥1 − ҧ𝑥 2
＋ 𝑥2 − ҧ𝑥 2
+ ⋯ 𝑥200 − ҧ𝑥 2
200 − 1
= 2.595608
standard deviation of sample means
𝑠 𝑥 = 𝑉𝑥 = 2.595608 = 1.611089

13. We can plot a Normal distribution using the result of the calculation on a
histogram we draw before.
↑
Mean ҧ𝑥 = 27.5
𝑁 ҧ𝑥, 𝑉𝑥 =
𝑁(27.5,2.6)
SD:
𝑠 𝑥 = 1.6

14. Let’s compare these distributions: population and sample means
↑
Mean ҧ𝑥 = 27.5
𝑁 ҧ𝑥, 𝑉𝑥 =
𝑁(27.5,2.6) 𝑆𝐷
𝑠 𝑥 = 1.6
↑
𝜎 = 5.3
Population mean 𝜇 = 27.6
𝑁 𝜇, 𝜎2 =
𝑁(27.6,28.3)
Remember,
First of all, we have a population distribution showing left.
We took randomly pick up samples 200 times and the number of items within
the each trial are n=10.
And we calculated each samples’ mean and the distribution of the 200 sample
means is showing right.

15. To compare, we can integrate these graphs.
What do you realize?

16. We know now…
The population mean is nearly samples’ mean.
The samples’ variation is smaller than population’s.

17. ↑
𝜎
𝜇
Central Limit Theorem : CLT
From a distribution
that have 𝝁 𝒂𝒏𝒅 𝝈 𝟐
(it must NOT be
following normal)
We repeatedly try to
take a many samples and
the sample size is n.
The distribution of “means
of samples” are distributed
and it follows
𝑁 𝜇,
𝜎2
𝑛
↑
𝜇 = ҧ𝑥
𝑠 𝑥 =
𝜎
𝑛
Also, we call
𝜎
𝑛
as Standard Error of the mean ഥ𝑥𝑖 SE

18. Numerically examine it!
The goal is to show 𝜇 = 𝑥 and 𝑠 𝑥 =
𝜎
𝑛
↑
ҧ𝑥 = 27.5
𝑁 ҧ𝑥, 𝑉𝑥 =
𝑁(27.5,2.6) SE＝𝑠 𝑥 = 1.61
↑
𝜎 = 5.3
𝜇 = 27.6
𝑁 𝜇, 𝜎2 =
𝑁(27.6,28.3)
𝜇 = 27.6 ≅ ҧ𝑥 = 27.5
𝜎
𝑛
= 𝑆𝐸 =
5.3
10
=
5.3
3.16277
= 1.68 ≅ 𝑠 𝑥(𝑆𝐸) = 1.61
Almost
Same!
True value we already
know
Theoretically calculate using true value Derived from R trial

19. n=2
n=5
n=10
𝑥1 34 31 25 28 26
𝑥2 20 NA 22 25 NA
⋮ 19 16 24 29 42
𝑥1000 24 35 24 25 28
𝑥1 34 31
𝑥2 20 NA
⋮ 27 26
𝑥1000 25 26
n is here
If we change sample size n (and fix the number of trial)