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Mathematics 10 Online
OpenStudy (emani3):

What is the standard deviation of the following data? If necessary, round your answer to two decimal places. 82, 44, 67, 52, 120?

OpenStudy (phi):

can you first find the average of those numbers ?

OpenStudy (emani3):

The average is 73

OpenStudy (emani3):

@phi ^

OpenStudy (quennlee):

it might help you: https://www.mathsisfun.com/data/standard-deviation-formulas.html

OpenStudy (quennlee):

P.S. - the steps are already there.. you will just read that and apply to your given data. :)

OpenStudy (phi):

the next step is to do number - 73 for each number on your list (you will get both positive and negative numbers, that is ok) next, square each of those differences (you get all positive numbers) find the average of the numbers finally , find the square root of that average that will be the standard deviation.

OpenStudy (emani3):

The numbers were 81,841,36,441,2209 & the a average= 4699 The average squared= 22080601 is that correct ? @phi

OpenStudy (phi):

what did you get for 82-73 44-73 67-73 52-73 120-73 ?

OpenStudy (phi):

ok your numbers look good except for the last step. you don't square the average, you take the square root.

OpenStudy (acespeedfighter):

phi are you a mod?

OpenStudy (phi):

The numbers were 81,841,36,441,2209 that part is ok. but I got a different average

OpenStudy (emani3):

68.54 @phi is that correct now ?

OpenStudy (phi):

getting closer. what did you get for the average of 81,841,36,441,2209

OpenStudy (emani3):

721.6 for the average @phi

OpenStudy (phi):

yes, that looks good. now take the square root to get the "standard deviation"

OpenStudy (emani3):

26.86 @phi ?

OpenStudy (phi):

yes

OpenStudy (emani3):

Thank you so much @phi

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