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Mathematics 7 Online
OpenStudy (anonymous):

If the outliers are not included in the data set below, what is the mean of the data set? 42, 43, 46, 48, 57, 60, 96, 59, 38, 68, 29 47 48 49 52

OpenStudy (anonymous):

to find the value of the mean, add up all the numbers you have in your set of numbers, then divide that sum by the amount of numbers you added up, make sense?

OpenStudy (anonymous):

Yes but what are the outliers?

OpenStudy (anonymous):

I believe we first have to find the mean, then find the standard deviation, and then do mean + or - 1.5*(std devation) and if a number is above or below this, then it is an outlier

OpenStudy (cloverracer):

The mean of the data set would be 53.8

undeadknight26 (undeadknight26):

hmm are you sure clove?

OpenStudy (anonymous):

53.3

OpenStudy (anonymous):

How do I find standard deviation?

OpenStudy (anonymous):

Victoria, have you learned about standard deviations yet? or is this question just suppose to teach you to look at the sequence and pick out the outliers?

undeadknight26 (undeadknight26):

i keep getting 51.22...

OpenStudy (anonymous):

51.2 is correct

OpenStudy (anonymous):

I have learned it I just don't remember it

OpenStudy (anonymous):

So now I do 1.5(51.2) = 76.8 and -1.5(51.2) = -76.8? And so anything below -76.8 or above 76.8 is an outlier?

OpenStudy (anonymous):

I'm hesitant on how to teach you.. because the solution is 51.22... but there isn't an option for that. what grade/ class level is this?

OpenStudy (anonymous):

I am in the 8th grade but this is Algebra 1

OpenStudy (anonymous):

And the question wants the mean without the outliers not the standard deviation.

OpenStudy (anonymous):

find outliers by this method: if a number from the original set is above or below the original mean by a value of 1.5*(std dev) then that number is an 'outlier' so, we need to figure out what the standard deviation is. pull out your notes or your book and look for that equation ^_^

OpenStudy (anonymous):

I just submitted my answer and I got it right the mean without the outlier (which was 96) was 49

OpenStudy (anonymous):

if that is good enough of an explantion for you, then it's good enough for me ^_^

OpenStudy (anonymous):

Ok thanks I have to go :)

OpenStudy (anonymous):

^_^

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