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December 15th, 2016, 01:31 PM   #1
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Chebychev, where mean is not in the middle of the estimated interval

One more question, can you maybe help me again?
Fifty-five percent of all students on a campus are in Humanities. If 150 students are selected at random, what is the probability that the proportion of Humanities students in this particular sample will be between 45% and 55%.
I did the following
150*0.55=82.5
150*0.45=67.5
P(67.5<X<82.5)
σ²=np(1-p)=37.125
Here comes where I´m not quite sure
15=kσ k=15/σ k=15/√σ² k=15/√37.25 k=0.1654
Thanks in advance!
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December 15th, 2016, 01:54 PM   #2
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15=kσ

Where did 15 come from? What is k supposed to be?
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December 15th, 2016, 02:01 PM   #3
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I posted the whole question.
That's all that was given.
15 is 82.5-67.5
Therefore the interval I want to calculate.
And if I understood it correctly
That distance (15) equals kσ
15=kσ k=15/σ k=15/√σ² k=15/√37.25 k=0.1654
As I already stated.
The Problem I'm having is that the mean is not in the middle of the interval.
If it would be in the middle it would result in 7.5=kq.
However I'm not sure whether I understood it correctly.

Last edited by skipjack; December 15th, 2016 at 04:18 PM.
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