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I was wondering what is/are the fundamental difference(s) between a probability density function for a mean and sampling distribution of a mean?

Can we say that sampling distribution of sample means is a probability density function of the mean, as a random variable?

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    $\begingroup$ The sampling distribution of a mean often has no probability density at all. For instance, the mean of a sample drawn from a discrete distribution will itself have a discrete distribution. $\endgroup$
    – whuber
    Sep 9 '18 at 22:38
  • $\begingroup$ @whuber, Dear whuber, thanks. My question was are both of the curves in ETBruce's answer describing 2 continuous random variables? Are both these curves 2 probability density functions? $\endgroup$
    – rnorouzian
    Sep 9 '18 at 22:56
  • $\begingroup$ @whuber: respectfully, I disagree with the qualifier "often" in your comment above. What you've discribed is a highly unusual circumstance of little practical use in any applied setting I can think of. For most practical purposes, it is useful to think of the mean of a sample as always having a continuous sampling distribution except for some extremely unusual circumstances such as the one you mention, which could be relegated to the fine print without much loss of generality. $\endgroup$ Aug 5 '20 at 14:37
  • $\begingroup$ @Color I am amazed at your implicit belief that discrete distributions are not often used! Continuous approximations are fine--until they are not, which is always the case with small samples. Despite being in a world of big data, plenty of people still deal with small samples of discrete quantities. $\endgroup$
    – whuber
    Aug 5 '20 at 14:57
  • $\begingroup$ @whuber: but in order for the sampling distribution of the mean to have no density you'd need not just any discrete distribution but a highly contrived one with values say {2,4,6} otherwise the sampling distribution of the mean would take you to a density 99.9% of the time. $\endgroup$ Aug 5 '20 at 15:15
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If a normal population is distributed $\mathsf{Norm}(\mu = 100, \sigma = 16)$ and $\bar X$ is the mean of a random sample of size $n=4$ from that distribution then $\mu_{\bar X} = E(\bar X) = \mu = 100,$ $\sigma^2_{\bar X} = Var(\bar X) = \sigma^2/n$ and $\sigma_{\bar X} = SD(\bar X) = \sigma/\sqrt{n} = 16/2 = 8.$

Thus, the sampling distribution of the sample mean is $\bar X \sim \mathsf{Norm}(100, 8).$ In general, the distribution of the sample mean $\bar X$ is less variable than the population from which the sample was taken.

enter image description here

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  • $\begingroup$ Dear Bruce, thanks. My question was are both of these curves describing 2 continuous random variables? Are both these curves 2 probability density functions? $\endgroup$
    – rnorouzian
    Sep 9 '18 at 22:54
  • $\begingroup$ Yes twice. Both curves describe normal distributions (which are continuous); both are density functions. $\endgroup$
    – BruceET
    Sep 10 '18 at 3:47

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