> ## Documentation Index
> Fetch the complete documentation index at: https://v1-learn.neoartd.my.id/llms.txt
> Use this file to discover all available pages before exploring further.
# The Central Limit Theorem
## Distribution of Sample Means
**Distribution of Sample means:** A
sampling distribution of sample
means is a distribution obtained
by using the means computed
from random samples of a specific
size taken from a population.
> **Distribusi rata-rata sampel:** Distribusi sampling dari rata-rata sampel adalah distribusi yang diperoleh dengan menggunakan rata-rata yang dihitung dari sampel acak dari ukuran tertentu yang diambil dari populasi.
**Sampling error** is the difference
between the sample measure and
the corresponding population
measure due to the fact that the
sample is not a perfect
representation of the population.
> **Kesalahan sampling** adalah perbedaan antara ukuran sampel dan ukuran populasi yang sesuai karena sampel tidak mewakili populasi dengan sempurna.
### Properties of the Distribution of Sample Means
The mean of the sample means will be
the same as the population mean.
> Rata-rata rata-rata sampel akan sama dengan rata-rata populasi.
The standard deviation of the sample
means will be smaller than the standard
deviation of the population, and it will
be equal to the population standard
deviation divided by the square root of
the sample size.
> Simpangan baku rata-rata sampel akan lebih kecil daripada simpangan baku populasi, dan akan sama dengan simpangan baku populasi dibagi dengan akar kuadrat dari ukuran sampel.
### Properties of the Distribution of Sample Means - Example
Suppose a professor gave an 8-point
quiz to a small class of four students.
The results of the quiz were 2, 6, 4, and
8\. Assume the four students constitute
the population.
The mean of the population is $\mu = \dfrac{2+6+4+8}{4} = 5$.
The standard deviation of the population is
$~~~~~\sigma = \sqrt{\dfrac{(2-5)^2 + (6-5)^2 + (4-5)^2 + (8-5)^2}{4}} = 2.236$.
The graph of the distribution of the scores is uniform.
Next we will consider all samples of size 2 taken with replacement.
| Sample | Mean | Sample | Mean |
| :----: | :--: | :----: | :--: |
| 2,2 | 2 | 6,2 | 4 |
| 2,4 | 3 | 6,4 | 5 |
| 2,6 | 4 | 6,6 | 6 |
| 2,8 | 5 | 6,8 | 7 |
| 4,2 | 3 | 8,2 | 5 |
| 4,4 | 4 | 8,4 | 6 |
| 4,6 | 5 | 8,6 | 7 |
| 4,8 | 6 | 8,8 | 8 |
Frequency Distribution of the Sample Means
| X-bar (mean) | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| :----------: | - | - | - | - | - | - | - |
| Frequency | 1 | 2 | 3 | 4 | 3 | 2 | 1 |
Graph of the Sample Means
### Mean and Standard Deviation of the Sample Means
Mean of Sample Means
$~~~~~\mu_{\bar{X}} = \dfrac{\sum X}{n} = \dfrac{2+3+\cdots+8}{16} = \dfrac{80}{16} = 5$.
which is the same as the population mean. Thus $\mu\_{\bar{X}} = \mu$.
The standard deviation of the sample means is
$~~~~~\sigma_{\bar{X}} = \sqrt{\dfrac{(2-5)^2 + (3-5)^2 + \cdots + (8-5)^2}{16}} = \dfrac{2.236}{\sqrt{2}} = 1.581$.
This is the same as $\dfrac{\sigma}{\sqrt{2}}$.
The Standard Error of the Mean
The standard deviation of the sample
means is called the standard error of
the mean. Hence $\sigma_{\bar{X}} = \dfrac{\sigma}{\sqrt{n}}$.
## The Central Limit Theorem
As the sample size **n** increases, the
shape of the distribution of the sample
means taken from a population with
mean **$\mu$** and standard deviation of **$\sigma$** will
approach a normal distribution. As
previously shown, this distribution will
have a mean **$\mu$** and standard deviation $\dfrac{\sigma}{\sqrt{n}}$.
> Saat ukuran sampel **n** meningkat, bentuk distribusi rata-rata sampel yang diambil dari populasi dengan rata-rata **$\mu$** dan simpangan baku **$\sigma$** akan mendekati distribusi normal. Seperti yang telah ditunjukkan sebelumnya, distribusi ini akan memiliki rata-rata **$\mu$** dan simpangan baku $\dfrac{\sigma}{\sqrt{n}}$.
The central limit theorem can be used
to answer questions about sample means
in the same mannerthat the normal distribution
can be used to answer questions about
individual values. The only difference is that
a new formula must be used for the z- values.
It is $z = \dfrac{\bar{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}$.
> Teorema batas pusat dapat digunakan untuk menjawab pertanyaan tentang rata-rata sampel dengan cara yang sama seperti distribusi normal dapat digunakan untuk menjawab pertanyaan tentang nilai individu. Satu-satunya perbedaan adalah bahwa rumus baru harus digunakan untuk nilai z. Ini adalah $z = \dfrac{\bar{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}$.
### The Central Limit Theorem - Example
#### Question 1
A.C. Neilsen reported that children between
the ages of 2 and 5 watch an average of 25
hours of TV per week. Assume the variable is
normally distributed and the standard
deviation is 3 hours. If 20 children between
the ages of 2 and 5 are randomly selected
find the probability that the mean of the
number of hours they watch TV is greater
than 26.3 hours.
Solution :
* The standard deviation of the sample
means is $\dfrac{\sigma}{\sqrt{n}} = \dfrac{3}{\sqrt{20}} = 0.671$.
* The z-value is $z = \dfrac{\bar{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}} = \dfrac{26.3-25}{0.671} = 1.94$.
* Thus $P(z > 1.94) = 0.5 - 0.4732 = 0.0268$.
That is, the probability of obtaining a
sample mean greater than 26.3 is
0.0262 = 2.62%.
#### Question 2
The average age of a vehicle registered
in the United States is 8 years, or 96
months. Assume the standard
deviation is 16 months. If a random
sample of 36 cars is selected
find the
probability that the mean of their age is
between 90 and 100 months.
Solution :
* The standard deviation of the sample
means is $\dfrac{\sigma}{\sqrt{n}} = \dfrac{16}{\sqrt{36}} = 2.6667$.
* The two z-values are :
* $~$
* $~~~~~z_1 = \dfrac{90-96}{2.6667} = -2.25 $. and
* $~~$
* $~~~~~z_2 = \dfrac{100-96}{2.6667} = 1.5 $.
* $~~$
* Thus $P(-2.25 < z < 1.5) = 0.4878 + 0.4332 = 0.921$ or 92.1%.
