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If one uses the following: $$(5-x)/5 = y $$ For 5 values (n=5)

And I have the mean of y. To get the mean of x I understand that I just move it around so I have: $$\bar x = 5 - (5 \times \bar y)$$

If I also have the SD of mean y; do I just get the mean of x by multiplying it by 5?

Example: n = 5, mean y = 0.8, SD y = 0.09

so will mean x = 1, and SD x = 0.45?

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  • $\begingroup$ I'm a little confused by your notation @SPro $\endgroup$
    – Taylor
    Commented Sep 19, 2017 at 12:23
  • $\begingroup$ does this help? (edited above) $\endgroup$
    – SPro
    Commented Sep 19, 2017 at 12:33
  • $\begingroup$ @Taylor (first time using this sorry) $\endgroup$
    – SPro
    Commented Sep 19, 2017 at 12:41
  • $\begingroup$ It's all good. Still a little confused though--should the first equation involve $\bar{x}$ or $x_i$ for $i=1,2,3,4,5$? $\endgroup$
    – Taylor
    Commented Sep 19, 2017 at 13:16
  • $\begingroup$ The first equation is xi correct $\endgroup$
    – SPro
    Commented Sep 19, 2017 at 13:20

1 Answer 1

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Here are two rules that you might find useful. If you have a random variable $W$, and a constant $c$, then

  1. $\operatorname{Var}(cW) = c^2 \operatorname{Var}(W)$
  2. $\operatorname{Var}(W + c) = \operatorname{Var}(W)$

If $(5-X_i)/5 = Y_i$ for $i=1,\ldots,5$, then summing both sides over $i$ gives you what you have written: $$ \bar{X} = 5 - 5 \bar{Y}. $$ So take the variance of both sides of that. You should get, using the two rules above, that $$ \operatorname{Var}(\bar{X}) = 25 \operatorname{Var}(\bar{Y}). $$ And if you take the square root of both sides of this, you should get $$ \operatorname{SD}(\bar{X}) = 5 \operatorname{SD}(\bar{Y}). $$

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