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Know if :

Population Data (N) : 1,000

Standard Deviation (σ) : 3,100,000

Average Revenue / Population Mean (μ) : 2,400,000

Sample data (n) : 50

Which data should be collected? Is it possible to find sample deviation, sample variance, sample mean?

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    $\begingroup$ Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments. $\endgroup$ – José Carlos Santos Oct 3 '18 at 10:28
  • $\begingroup$ I have found : ∑( xi- μ )2 = 9,610,000,000,000 ; Population variance (σ2) = 9,610,000,000 ; ∑1000 = 24,000,000,000 but I stuck here. I do not know what should I gonna do. $\endgroup$ – Adiputra Kho Oct 3 '18 at 10:30
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This is an extremely vaguely-worded problem, and you have been steadfast in not showing a meaningful attempt at solution. So we have no context to compensate for the vagueness. My first impulse was to vote to Close this as "unclear what you're asking." But there might be something to be learned here, so I didn't.

Therefore, this is an outline toward what I think may be required. I speculate that the idea is to get you to think about estimation in reverse: given parameters, what can be said about values of estimators. Others may have their own speculations.

Approximate probability bounds on sample mean. The sample size 50 is a small fraction of the population size 1000, so it doesn't seem worthwhile to consider the difference between sampling with and without replacement. By the CLT the sample mean $\bar X \stackrel{aprx}{\sim} \mathsf{Norm}(\mu, \sigma/\sqrt{n}).$ Thus, $Z = \frac{\bar X - \mu}{\sigma/\sqrt{n}}$ has $P(-1.96 < Z < 1.96) = 0.95.$ But remember that (unlike a usual sampling situation) $\mu$ and $\sigma$ are known.

So we can be 95% sure that the value of $\bar X$ is in the interval $\mu \pm 1.96\sigma/\sqrt{n}.$ That is, in $3.1m \pm 1.96\frac{2.4m}{\sqrt{50}},$ where $m = 1,000,000.$

Approximate probability bounds on sample SD. When the population mean $\mu$ is known, the sample variance is computed as $T^2 = \frac 1n\sum_i(X_i - \mu)^2.$ Also $\frac{nT^2}{\sigma^2}. \stackrel{aprx}{\sim} \mathsf{Chisq}(n).$ Let $L$ and $U$ be quantiles .025 and .975, respectively, of this chi-squared distribution.

qchisq(c(.025, .975), 50)
[1] 32.35736 71.42020

Then $P(\sigma^2L < nT^2 < \sigma^2U) = 0.96.$ Because $n$ and $\sigma^2$ are known, you can find an interval that is 95% sure to contain the estimate $T$ of the population SD.

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