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the measured resistance of a sample 10 resistors from a box of 900 is as below. All values are in Ohms.

82.5, 82.45, 82.11, 81.56, 81.91, 82.31, 81.5, 82.08, 82.33, 82.7.

the two questions are : find the sample standard deviation and population standard deviation.

no distribution was specified so I take it, I should consider them uniformly distributed.

in this case the sample mean is the full sum divided by 10 which gives : 82.145

for the sample standard deviation we use the formula SD = $\sqrt{\frac{\sum(x-\overline{x})^{2}}{n-1}}$

here n = 10. and $\overline{x} = 82.145$

now what I'd like to know is what about the population standard deviation, is it possible to actually compute it just with the given data ? is there a formula ?

to me it feels like as if it's impossible because we'd need all values of all 900 resistors to get a correct answer but maybe I'm missing something.

Thanks for any clarification

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  • $\begingroup$ Not sure I follow. The distribution is just the values you see...that is, one case out of $10$ is $82.5$ and so on. For the other question, the difference between population and sample is the denominator. It should be $n$ if this is the entire population. As you surmise, there's no way to compute the standard deviation for the entire $900$ just using this sample. $\endgroup$ – lulu Jun 3 at 20:04
  • $\begingroup$ @lulu I just saw that formula online, thanks I'll try to understand the intuition behind the difference. $\endgroup$ – rapidracim Jun 3 at 20:06
  • $\begingroup$ Good luck. The idea, which I'm sure your sources will speak of, is that switching to $n-1$ removes bias. In practical situations (where $n$ tends to be very large), the distinction is nearly irrelevant. $\endgroup$ – lulu Jun 3 at 20:07
  • $\begingroup$ You can only estimate the population sd using the sample sd if the former is unknown. (And do you have any reason to believe the sample comes from a uniform distribution?) $\endgroup$ – StubbornAtom Jun 3 at 20:08
  • $\begingroup$ @StubbornAtom nope, it's just in the solved practice problems, the way the answers were computed always assumed that they came from a uniform distribution. $\endgroup$ – rapidracim Jun 3 at 20:12

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