Week-4: Confused about sampling distribution of sample means

[God_of_Calamity]

As shown in the first image we first assume that the distribution of heights is follows a normal distribution.
Then as shown in second image we say that the sampling distribution of sample means is also normally distributed with μ_sample = μ_population
and σ_sample = (σ_population / √n).
I want to ask is this because we assumed that the heights are normally distributed then as we learned in the lecture of Sum of Gaussians (Week-2) what we’re doing is:
1: E[(X1 + X2 + … + Xn) / n] = (μ1 + μ2 + … μn) / n = μ
(since μ1=μ2=…=μn=μ)
2: Var[(X1 + X2 + … + Xn) / n] = (1/n^2) * Var[(X1 + X2 + … + Xn)]
= (1/n^2) * [Var(X1) + Var(X2) + … + Var(Xn)], (X1, X2…Xn are iid)
= (1/n^2) * [n * Var(X)]
= Var(X) / n

Am i Correct or is there another reason behind the assumption that the mean and variance of the sampling distribution of the sample means has the same mean as population mean and deviation = population_dev / n

[Alireza_Saei]

Hi @God_of_Calamity

You are correct. Because we assumed that the heights follow a normal distribution, the Central Limit theorem applies. The CLT says that the sampling distribution of the sample mean will be normally distributed if the sample size is large enough. Now, the individual heights are normally distributed with mean \mu and variance \sigma^2 , which means that the sample mean will also have a mean \mu and a variance \sigma^2 / n . So, the std of the sample mean is \sigma / \sqrt{n} .

Hope it helps! Feel free to ask if you need further assistance.