Sampling Distribution of the Means

[Saladsamurai]

Hey folks!

I am working through my text entitled Probability and Statistics for Engineers and Scientists by Walpole et al. I am getting a little stuck on their lackadaisical derivation of the standard deviation of the means. Perhaps I can get a little guidance here by talking it out with someone. Here is their intro to the matter:

The first important sampling distribution to be considered is that of the mean $\bar{X}$. Suppose that a random sample of n observations is taken from a normal population with mean $\mu$ and variance $\sigma^2$. Each observation X_i, i = 1,2...,n, of the random sample will then have the same normal distribution as the population being sampled. Hence, by the reproductive property of the normal distribution established in Theorem 7.11. we conclude that

$$\bar{X} = \frac{1}{n}(X_1 + X_2 + ...+X_n)$$

has a normal distribution with mean

$$\mu_{\bar{X}}= \frac{1}{n}(\mu+\mu+...+\mu)$$

I am a little lost already at the forst 3 sentences. Do they really mean that all of the X_i will have the same normal distribution? Let's say that we have 10 000 ball-bearings with a population mean diameter of $\mu$ and pop variance $\sigma^2$; I take a random sample of n = 40 ball-bearings. It seems they are saying that those 40 diameters will follow the same normal distribution as the population with $\mu \text{ and }\sigma^2$. But they offer no proof, and furthermore, I do not see why that should be self evident.

Am I misinterpreting their words?Here is Theorem 7.11 for reference:

If X1, X2, ...Xn are independent random variables having normal distributions with means $\mu_1,\mu2,...\mu_n$ and variances $\sigma_1^2,\sigma_2^2,...,\sigma_n^2$, then the random variable $Y = a_1X_1+a_2X_2+...+a_nX_n$ has a normal distribution with $\mu_Y = a_1\mu_1+a_2\mu_2+...+a_n\mu_n$ and variance $\sigma_Y^2 = a_1^2\sigma_1^2+a_2^2\sigma_2^2+...+a_n^2\sigma_n^2$

 

[Saladsamurai]

I know it is a little early for a "bump" but I am seeing a lot of views and no responses. I was just wondering if there was anything I could do to make my question a little clearer as I realize it is a little open-ended.

Thanks

 

[mathman]

I hope I can clarify a little bit. It is assumed that you are taking samples of something, where that something has a normal distribution. Therefore by definition the probability distribution of each sample is normal with the given mean and variance. Theorem 7.11 shows how you can get the mean and variance when you take a linear combination of the estimates.

 

[jim mcnamara]

I think he is asking: can I get a non-representative sample? Or what is the probability of getting a skewed sample of the population?

 

[Saladsamurai]

Hello folks Thanks for the replies. I think what I am really asking is this: Take the ball bearing example that I mentioned

Let's say that we have 10 000 ball-bearings with a population mean diameter of $\mu$ and pop variance $\sigma^2$; I take a random sample of n = 40 ball-bearings.

So we have these 40 observations x1,x2,...,xn. It is saying that these 40 observations follow a normal distribution with the same expectation value and variance as the population from which they came?

I am not sure why I am so unsure of this, but can someone just confirm or reject that this is in fact a true statement before I continue?

Thanks again!

~Casey

 

[mathman]

If these 40 are selected from the 10,000, then the statistics for each ball bearing is the same, normal with a mean diameter μ with a variance σ².

 

[statdad]

IF you take the sample from a single normally distributed population then the statement is correct. In fact, even without the normality assumption, as long as the population mean and standard deviation exist, the formulae for the mean and variance of the sample mean continue to hold.