anonymous
  • anonymous
I need help with "mean absolute deviation" Will give medal.
Mathematics
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anonymous
  • anonymous
I need help with "mean absolute deviation" Will give medal.
Mathematics
schrodinger
  • schrodinger
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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anonymous
  • anonymous
idk how to do it can you draw me a formula or something
Ibbutibbu.
  • Ibbutibbu.
The mean absolute deviation of a set of data is the average distance between each data value and the mean. 2. Find the distance between each data value and the mean. That is, find the absolute value of the difference between each data value and the mean.
anonymous
  • anonymous
Ok thankyou soooooooooo much I was doing a math assessment and then it said to find the mean absolute deviation and i didnt know what to do(didnt explain in my lesson)

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anonymous
  • anonymous
hi iGreen
iGreen
  • iGreen
^^^ Find the mean, and subtract all the values from the mean, then find the mean of that. For example. 2, 3, 3, 4, 6, 8, 9, 9 Find the mean. \(\sf \dfrac{2 + 3 + 3 + 4 + 6 + 8 + 9 + 9}{8} \rightarrow 5.5\) Now subtract all the values from it. \(\sf 5.5 - 2 = 3.5\) \(\sf 5.5 - 3 = 2.5\) \(\sf 5.5 - 3 = 2.5\) \(\sf 5.5 - 4 = 1.5\) \(\sf 6 - 5.5 = 0.5\) \(\sf 8 - 5.5 = 2.5\) \(\sf 9 - 5.5 = 3.5\) \(\sf 9 - 5.5 = 3.5\) Find the mean of these values. \(\sf \dfrac{3.5 + 2.5 + 2.5 + 1.5 + 0.5 + 2.5 + 3.5 + 3.5}{8} \rightarrow \color{#45D229}{\boxed{2.50}}\)
iGreen
  • iGreen

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