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    $\begingroup$ I knew there was some @... notation that would be involved. Thank @mikado. Could you ellaborate a little on the way it works? $\endgroup$ Commented Dec 8, 2019 at 16:38
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    $\begingroup$ It might be worth to add Through[{Mean, Apply@Subtract}[1175., 247.]] as the "inverse" approach (where a list is supplied and converted to a sequence of arguments) $\endgroup$ Commented Dec 9, 2019 at 10:46
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    $\begingroup$ @Lukas, Thank you! I took the liberty of editing the Mikado's answer to quote your comment (I also fixed a small typo in the code where you wrote [1175., 247.] for [{1175., 247.}]. Feel free to revert that edit if you desagree. $\endgroup$ Commented Dec 9, 2019 at 15:16