**What Is Sampling Distribution?**

If explained without any mathematical formulas, sampling distribution is the survey for all possible answers for the given query or the selection of all possible sample outcomes from the given huge number of population.

Let us get an idea about sampling distribution with a simple example, for example instead of asking 1000 people what kind of food they like, you can select limited number of people and conduct the survey repeatedly and find the results.

**Everything Everywhere is the Formulae**

- In order to find the kind of distribution first of all you should find the standard deviation of the sample.
- Sampling distribution can be classified into two types.

- Sampling distribution of mean
- Sampling distribution of Proportion

In sampling distribution of mean, standard deviation can be calculated using the below formula,

**Illustration**

**Case 1:**

Let us consider the group of population of four members, which means the size of the sample is N = 4.

Consider the age of each people as X, which is the random variable, Here X = 18, 20, 22, 24 ( Age ).

The standard deviation can be calculated from the formula,

In the formula above

**μ** is the mean

Standard Deviation is **σ**

**Mean = 18 +20+22+24 / 4 = 21**

**Where X**_{i }**= sum of all X**

By substituting all the values in the above formula we get Standard Deviation ( σ ) = 2.236, this would be uniformly distributed.

**Case 2:**

After replacing let us consider all the possibilities of two kinds of samples from the above,

(18,18), (20,18), (22, 18), (24,18)

(18,20), (20,20 ),(22,20),( 24,20)

(18,22 ),(20,22 ),(22,22),( 24,22)

(18,24 ),(20,24),( 22,24 ),(24,24)

n =2 but this will not be uniformly distributed, and the values will be n=2 , N = 16, mean = 21 and the standard deviation ( **σ** ) = 1.58

After getting the standard deviation you need to plot the graph, If you get the bell shaped curve it is said to be normally distributed else skewed.

The above figure is the example of normal distribution.

When the sample is normally distributed then the sampling distribution of their means is also normally distributed according to **central limit theorm. **That is even the sample size increases from the given population the sample is normally distributed.

Conclusion is that if the population is normal with μ and standard deviation ( σ ) then mean and standard deviation after replacement will also be normal.

** Diagrammatic representation of case 1 and case 2**

z value of sampling distribution of mean can be calculated by the formula,

z = (x – μ) / (σ / √n)

**Sampling Distribution of Proportion**

Let us consider proportion of people passed the examination is 87%, out of 200.

In such cases the expected number of students passed is calculated by the formulae,

np = 200 x 0.87 =174

Failed = n (1-p ) = 200 ( 1- 0.87 ) = 26

Both are also normally distributed since gre8ater than or equal to 10.

*μp*^*= p*

standard deviation (σ* p*^ ) = Square root of ( p (1-P ) / n )

*In this case standard deviation of **p*^ = 0.0237

P can be transferred into z score by the following formula,

Z Score for sample proportion: z = (p̄ – p) / SE.

**Importance of Sampling Distribution in the Practical World**

Selecting the sample and carrying out the process to predict the results is easier than the entire population.

It is time saving and more practical than the entire population. They provide simplified way in the statistical analysis.

**Reference :**

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