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S1 mean/standard deviation

I've done all parts of this question but I'm not sure how to work out the mean in part .iv
I know the standard deviation will be the same 9.7.
@osayukiigbinoba
Every piece of data has been shifted down by 50 in the coding of the data. So the mean can be obtained by undoing shift. So 18.1+50=68.1
(edited 7 years ago)
Would you just add 50 to the mean given for part iv.
The way I was taught it was that for the mean you do everything on the tin, SD you just times or divide and then for PMCC you don't do anything to it

Megan :smile:
Original post by meganthegreat
Would you just add 50 to the mean given for part iv.
The way I was taught it was that for the mean you do everything on the tin, SD you just times or divide and then for PMCC you don't do anything to it

Megan :smile:


You are absolutely correct! :smile:
Reply 4
Original post by Cryptokyo
@osayukiigbinoba
Every piece of data has been shifted down by 50 in the coding of the data. So the mean can be obtained by undoing shift. So 18.1+50=68.1


Oh it's easy now you mention it, thank you!
Reply 5
Original post by meganthegreat
Would you just add 50 to the mean given for part iv.
The way I was taught it was that for the mean you do everything on the tin, SD you just times or divide and then for PMCC you don't do anything to it

Megan :smile:

Thank you!

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