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Given a mean of $5.60$ and a standard deviation of $0.13$ is there a way of finding out how many individual values produced this mean and what those values were?

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    $\begingroup$ Impossible, no. $\endgroup$ Commented Nov 18, 2021 at 11:13
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    $\begingroup$ Start with the mean of $56.0$. That could come from $56/1, 560/10, 5600/100$ and many more. Knowing the SD too doesn't help determine between those possibilities, and vice versa. $\endgroup$
    – Nick Cox
    Commented Nov 18, 2021 at 11:58
  • $\begingroup$ @Nickcox how about if we know that there were 2 values. Can we find those values then? $\endgroup$ Commented Nov 18, 2021 at 14:21
  • $\begingroup$ @MichaelMunta are you asking now if it's possible to determine the sampled values given that you had a sample size of 2 with a mean of 5.6 and a standard deviation of 0.13? $\endgroup$
    – jros
    Commented Nov 18, 2021 at 14:54
  • $\begingroup$ @jros yes, that is what I'm asking. $\endgroup$ Commented Nov 18, 2021 at 15:05

1 Answer 1

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If you know there are two values, then you can figure out what they are.

$$ \bar x = \dfrac{1}{n}\sum_{i=1}^nx_i = \dfrac{1}{2}\big(x_1 + x_2\big)\\ s^2 = \dfrac{1}{n-1}\sum_{i = 1}^n\big(x_i - \bar x\big)^2 = (x_1 - \bar x)^2 + (x_2 - \bar x)^2 $$

Solve the system of equations. Wolfram Alpha gives two solutions, but the way of interpreting them is that the calculation does not know which value is $x_1$ (x) and which value is $x_2$ (y), and it is true that you do not know which value is which.

$$ x_1 = \dfrac{1}{2}\bigg( 2\bar x - \sqrt{2s^2} \bigg)\\ x_2 = \dfrac{1}{2}\bigg( 2\bar x + \sqrt{2s^2} \bigg)\\ $$

Let's do a quick check in software.

x1 <- 4
x2 <- 6
x <- sort(c(x1, x2))
xbar <- mean(c(x1, x2))
s2 <- var(c(x1, x2))
x1_solved <- (1/2) * (2*xbar - sqrt(2*s2))
x2_solved <- (1/2) * (2*xbar + sqrt(2*s2))
x_solved <- sort(c(x1_solved, x2_solved))
x == x_solved

This should reveal why it is problematic to figure out the original values from just the mean, standard deviation, and sample size, however. You wind up with two equations but more then two unknowns, so there are multiple solutions.

If you have two points where $\bar x = 5.6$ and $s = 0.13$, then you can apply my code to solve that the points are $5.50807611844575$ and $5.69192388155425$.

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  • $\begingroup$ In the case of numbers I provided, what would be the two values? $\endgroup$ Commented Nov 18, 2021 at 15:08
  • $\begingroup$ Your values have been edited into the answer, but it is important for you to understand how to use my code to calculate them. Are you able to get them on your own? // There's a wrinkle in this in that I assumed $0.13$ was the sample standard deviation. If that is, instead, the population standard deviation, do you see how to proceed? (You'll need to do some algebra, perhaps on Wolfram Alpha.) $\endgroup$
    – Dave
    Commented Nov 18, 2021 at 15:10
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    $\begingroup$ Let's make explicit that you are calculating the variance using sample size minus 1, hence just 1 here, so no division of the sum of squared deviations is needed. $\endgroup$
    – Nick Cox
    Commented Nov 18, 2021 at 15:23
  • $\begingroup$ @NickCox Good idea! // Michael, regarding not knowing which value is which, it also is true that we do not care which value is which. Do you see why? $\endgroup$
    – Dave
    Commented Nov 19, 2021 at 4:28

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