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I came across this question: Pebbles on a beach have masses normally distributed with mean 70 and standard deviation 45. Samples made up of 25 pebbles are taken from the beach. 

(a) Which of these statements is correct?
     A    The sample means will have a Poisson distribution.
     B    The sample means will have a Binomial distribution.   
     C    The sample means will have a Normal distribution.
(b) What will be the mean and standard deviation of the sampling distribution of means?

It cannot be Poisson as mean is not equal to variance. Also, since the sample size <30, we cannot assume normality for the sampling distribution of means. I am not sure if its binomial. Any input is greatly appreciated (its not a homework question :) )

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    $\begingroup$ Sum of independent normal random variables is? $\endgroup$
    – Math Lover
    Nov 27, 2017 at 17:40
  • $\begingroup$ How about C.L.T? $\endgroup$
    – Joe
    Nov 27, 2017 at 17:40
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    $\begingroup$ The distribution of the masses of individual pebbles must be skewed. If the distribution were symmetric then the mass of individual pebbles at $-2\sigma$ would have negative masses which is impossible. $\endgroup$
    – MaxW
    Nov 27, 2017 at 17:56
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    $\begingroup$ Maybe confused standard deviation with variance? $\endgroup$
    – Gregory
    Nov 27, 2017 at 17:59

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I am answering this because between the statement of the Question and one of the Comments, there are misconceptions that should be clarified.

If the masses of individual pebbles are normally distributed with $\mu = 70$ and $\sigma^2 = 45,$ then batches of $n=25$ pebbles have masses $$T \sim \mathsf{Norm}(\mu_T=70n,\,\sigma_T^2 = 45n) = \mathsf{Norm}(\mu_T = 1750,\, \sigma_T^2 = 1125).$$

[Notice that I have used a variance of 45, not a standard deviation of 45, based on the Comments of @MaxW (+1) and @Gregory (+1). Then $70 - 3\sqrt{45} \approx 50$ (three SD below the mean), which is far above $0$ and there is negligible error in modeling pebble weights as normal.]

The statement "Also, since the sample size <30, we cannot assume normality for the sampling distribution of means." is not true.

If the pebbles have normally distributed masses, then the batches do also. This is a statement about normal distributions, and does not rely on the Central Limit Distribution (CLT). [Per the Comment of @MathLover (+1).]

[The CLT might be used if we were not sure individual pebbles have normally distributed masses. Then batches of 25 might have very nearly normally distributed masses--especially if the distribution for individual pebbles is not markedly skewed.]

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  • $\begingroup$ Oh yes, I missed the point that the population itself in normally distributed, so no need of CLT. Probably the question should have been variance instead SD. Thanks for pointing out :) $\endgroup$
    – Bitsy
    Nov 28, 2017 at 19:07

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