- #1
Saladsamurai
- 3,020
- 7
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:
I am a little lost already at the forst 3 sentences. Do they really mean that all of the Xi will have the same normal distribution? Let's say that we have 10 000 ball-bearings with a population mean diameter of [itex]\mu[/itex] and pop variance [itex]\sigma^2[/itex]; 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 [itex]\mu \text{ and }\sigma^2[/itex]. 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:
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:
TEXTBOOK said:The first important sampling distribution to be considered is that of the mean [itex]\bar{X}[/itex]. Suppose that a random sample of n observations is taken from a normal population with mean [itex]\mu[/itex] and variance [itex]\sigma^2[/itex]. Each observation Xi, 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
[tex]\bar{X} = \frac{1}{n}(X_1 + X_2 + ...+X_n)[/tex]
has a normal distribution with mean
[tex]\mu_{\bar{X}}= \frac{1}{n}(\mu+\mu+...+\mu)[/tex]
I am a little lost already at the forst 3 sentences. Do they really mean that all of the Xi will have the same normal distribution? Let's say that we have 10 000 ball-bearings with a population mean diameter of [itex]\mu[/itex] and pop variance [itex]\sigma^2[/itex]; 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 [itex]\mu \text{ and }\sigma^2[/itex]. 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:
TEXTBOOK said:If X1, X2, ...Xn are independent random variables having normal distributions with means [itex]\mu_1,\mu2,...\mu_n[/itex] and variances [itex]\sigma_1^2,\sigma_2^2,...,\sigma_n^2
[/itex], then the random variable [itex] Y = a_1X_1+a_2X_2+...+a_nX_n[/itex] has a normal distribution with [itex]\mu_Y = a_1\mu_1+a_2\mu_2+...+a_n\mu_n[/itex] and variance [itex]\sigma_Y^2 = a_1^2\sigma_1^2+a_2^2\sigma_2^2+...+a_n^2\sigma_n^2
[/itex]