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December 15th, 2016, 01:31 PM  #1 
Newbie Joined: Dec 2016 From: Trinidad and Tobago Posts: 5 Thanks: 0  Chebychev, where mean is not in the middle of the estimated interval
One more question, can you maybe help me again? Fiftyfive 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(1p)=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! 
December 15th, 2016, 01:54 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,133 Thanks: 468 
15=kσ Where did 15 come from? What is k supposed to be? 
December 15th, 2016, 02:01 PM  #3 
Newbie Joined: Dec 2016 From: Trinidad and Tobago Posts: 5 Thanks: 0 
I posted the whole question. That's all that was given. 15 is 82.567.5 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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