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Mathematics 21 Online
OpenStudy (grumpystumpy):

Help needed, question below. Will Fan & Medal

OpenStudy (grumpystumpy):

Which best describes the spread of a set of data that has an interquartile range of 18 and a mean absolute deviation of 10? A. The average distance of all data values from the mean is 18 and the middle 50% of data values has a range of 10. B. The average distance of all data values from the mean is 14 and the middle 50% of data values has a range of 8. C. The average distance of all data values from the mean is 14 and the middle 50% of data values has a range of 28. D. The average distance of all data values from the mean is 10 and the middle 50% of data values has a range of 18.

OpenStudy (grumpystumpy):

@Ms-Brains

OpenStudy (grumpystumpy):

@Miss.Rose

Vocaloid (vocaloid):

hint: mean absolute deviation = average distance from mean interquartile range =range of the middle 50% of data

OpenStudy (grumpystumpy):

So, would it be C?

Vocaloid (vocaloid):

no. match up the numbers carefully.

OpenStudy (grumpystumpy):

I'm stumped.

Vocaloid (vocaloid):

think about my hint carefully. read the numbers in the problem. read the definitions.

OpenStudy (grumpystumpy):

Uh, A? Sorry for annoying you...

Vocaloid (vocaloid):

the mean absolute deviation is 10 the mean absolute deviation is also the average distance from the mean therefore, what is the average distance from the mean?

OpenStudy (grumpystumpy):

18

OpenStudy (grumpystumpy):

You said 10

OpenStudy (grumpystumpy):

D

OpenStudy (grumpystumpy):

(ok)

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