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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: roseeblosssom on November 05, 2013, 01:24:56 pm

Title: 2012 exam 1 question 8
Post by: roseeblosssom on November 05, 2013, 01:24:56 pm

I've been having trouble up standing what question 8a in the 2012 exam 1 is asking me! Probability has never exactly been strong for me. If anyone could help that'd be great! XD

Title: Re: 2012 exam 1 question 8
Post by: clueless123 on November 05, 2013, 01:30:08 pm

Since it is a normal distribution, you know that the curve is symmetrical.
Pr ( X<106 ) = Pr( X > 94)  *because the deviation from the mean (100 ) is 6 units*
Therefore
Pr(94<x<100)
= Pr(100<x<106)
= Pr(x<106) - Pr(X<100)
= q - 0.5

Title: Re: 2012 exam 1 question 8
Post by: Flor on November 05, 2013, 02:53:28 pm

I think the trick to this question is just recognising they're the same distance from the mean.
Also with these types of questions, it's always good to draw it out to see what you're dealing with.
Anyways, I know that someone else has already answered this question, but I thought I'd just show it to you visually http://oi43.tinypic.com/2v93xxx.jpg

Edit: There is a mistake in the working sorry. 100+6 = 106 haha, not 94 :) Also, for some reason I've used the letter p in the diagram. It should be Q :)

Title: Re: 2012 exam 1 question 8
Post by: roseeblosssom on November 05, 2013, 03:45:07 pm

Thank you both, I understand now. Both the explanation and the diagram help me understand. I don't know why but in probability, distributions have been my enemy all year!

Thanks heaps! XD

Title: Re: 2012 exam 1 question 8
Post by: Flor on November 05, 2013, 04:01:39 pm

Quote from: roseeblosssom on November 05, 2013, 03:45:07 pm

Thank you both, I understand now. Both the explanation and the diagram help me understand. I don't know why but in probability, distributions have been my enemy all year!

Thanks heaps! XD

No problem! All the best tomorrow :)