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help123please.
Group Title
Using the image at http://fotos.fotoflexer.com/307d9591812479fea36cb26258913578.jpg
Answer these questions:
Do you think the test was too hard for the students? Explain.
Would you expect the mean to be above or below the median? Explain.
 one year ago
 one year ago
help123please. Group Title
Using the image at http://fotos.fotoflexer.com/307d9591812479fea36cb26258913578.jpg Answer these questions: Do you think the test was too hard for the students? Explain. Would you expect the mean to be above or below the median? Explain.
 one year ago
 one year ago

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help123please. Group TitleBest ResponseYou've already chosen the best response.0
@Kitt020912
 one year ago

help123please. Group TitleBest ResponseYou've already chosen the best response.0
@timo86m
 one year ago

timo86m Group TitleBest ResponseYou've already chosen the best response.1
Do you think the test was too hard for the students? Explain. NO because it looks like some even got 102%
 one year ago

help123please. Group TitleBest ResponseYou've already chosen the best response.0
Hmm. Okay,thanks!
 one year ago

help123please. Group TitleBest ResponseYou've already chosen the best response.0
What about the second question?
 one year ago

timo86m Group TitleBest ResponseYou've already chosen the best response.1
got no idea.
 one year ago

help123please. Group TitleBest ResponseYou've already chosen the best response.0
Okay.
 one year ago

timo86m Group TitleBest ResponseYou've already chosen the best response.1
http://answers.yahoo.com/question/index?qid=20110523160727AA4i32v that is exactly what i was thinking it impossible to find mean.
 one year ago

help123please. Group TitleBest ResponseYou've already chosen the best response.0
Impossible?
 one year ago

timo86m Group TitleBest ResponseYou've already chosen the best response.1
Well i guess we can make an estimate (72+88+96+102)/4 Just an estimate and see if it higher than medium
 one year ago

help123please. Group TitleBest ResponseYou've already chosen the best response.0
Yes, but I have to explain.
 one year ago

help123please. Group TitleBest ResponseYou've already chosen the best response.0
O well. Thats alright   no worries.
 one year ago

timo86m Group TitleBest ResponseYou've already chosen the best response.1
explain that you assume that averaging the quartiles will give you an estimated mean :)
 one year ago
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