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Mathematics 13 Online
OpenStudy (anonymous):

Compute the standard deviation for the set of data. 2, 4, 6, 8, 10 @kelliegirl33

OpenStudy (anonymous):

A. 2 sqr root of 2 B.8 c.2 sqr root of 10 D.64

OpenStudy (amistre64):

its a fairly short list .... the mean is obviously 6; sum the squares of the differences and divide by 5; then sqrt it

OpenStudy (anonymous):

A

OpenStudy (anonymous):

would it be 30 divided by 5 wait, idk

OpenStudy (amistre64):

i get a when i use a pencil ....

OpenStudy (amistre64):

4,2,0,2,4 2(16+4) = 40/5 = 8 sqrt(8) = 2.2

OpenStudy (amistre64):

30 is the sum of the set, /5 gives the mean of 6 the differences from the mean are: 4,2,0,2,4 square, add, divide and sqrt

OpenStudy (anonymous):

Compute the standard deviation for the set of data. 13.3, 14.3, 15.3, 16.3, 17.3

OpenStudy (amistre64):

the process is the same ...

OpenStudy (amistre64):

15.3 is obviously the mean 1,2,0,2,1 1,4,0,4,1 = 10 10/5 = 2 and sqrt

OpenStudy (anonymous):

I dont have a calc on me

OpenStudy (amistre64):

well, 2,1,0,1,2 but thats immaterial

OpenStudy (amistre64):

neither do i ....

OpenStudy (anonymous):

lol

OpenStudy (amistre64):

sqrt 2 is about 1.414 if memory serves

OpenStudy (anonymous):

wow, awesome..... On a box-and-whiskers plot, what is the far left line called? a. minimum value c. median b. maximum value d. interquartile range

OpenStudy (anonymous):

Last one for now, please help :_)

OpenStudy (amistre64):

tell me which one youd pick first; this is pretty much conceptual and theres no way to really guide towards the answer

OpenStudy (anonymous):

D

OpenStudy (anonymous):

@amistre64 I would put D.

OpenStudy (amistre64):

D isnot a good choice since the IQR range is in the middle ... |<-----|____|_______|-------------->| *** this denotes the leftmost line it either has to be the biggest or smallest value

OpenStudy (anonymous):

Okay, well minimum value

OpenStudy (amistre64):

minimum is what i would go with as well

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