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The question says
Quote:
 A random variable has a uniform distribution over the interval from 17 to 34.
And I'm supposed to look for the IQR, but I thought I was doing the IQR correctly, but apparently not. It says the answer is 8.5, but that's not what I get.

What I did was list numbers from 17 to 34, found the Median, then the Q1 and Q3, then subtracted Q3-Q1, but I got 10.
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 02-25-2011 #2 (permalink) Don't touch 8| Awards   Join Date: Jun 2008 Location: Why Posts: 887 iTrader: (0) IGN: Flarina/Lunata Class: SHEEP/ ???? Level: 142/5xx Guild: SOLOMODO/Wish Xfire: !!! D8 Well I listed all the numbers from 17 to 34 too ... and got First Set: 17, 18, 19, 20, 21, 22, 23, 24, 25 Second Set: 26, 27, 28, 29, 30, 31, 32, 33, 34 Then I subtracted those from one another ... and got 9 ... wait a second what ;_; ... however, when i added them all up and divided by 18 and divided that by 3, I got 8.5 Dx Is that the entire question? __________________ Mr. and Mrs. GGFTW contest!
 02-25-2011 #3 (permalink) [夢] Games   Join Date: Jun 2008 Posts: 1,497 Blog Entries: 5 iTrader: (0) IGN: Yume/Relief/夢 Guild: Drop/JW/Kirameki/Hores(FT) It's all the information is given. Shadow and Adura helped me out with it, and Shadow figured that just do 34-17 = 17/2 = 8.5, because you don't just count whole numbers. >_< I was doing whole numbers LOL. __________________ "Even a broken clock is right twice a day."
 03-04-2011 #4 (permalink) meow Games Awards   Join Date: Aug 2007 Location: Rodesia Posts: 2,900 iTrader: (0) Class: Dacy Guild: ForielUnion, OurStory (17+34)/2 = 25.5 Median (34+25.5)/2 = 29.75 Upper Quartile (17+25.5)/2 = 21.25 Lower Quartile 29.75 - 21.25 = 8.5 <-- IQR Although i don't really get the question. If there were odd number of whole numbers in that range counting the whole numbers would have worked. __________________ Fanart Anime/Manga Tees at http://www.facebook.com/metronomist Last edited by kag; 03-04-2011 at 09:36 PM.
 03-09-2011 #5 (permalink) [夢] Games   Join Date: Jun 2008 Posts: 1,497 Blog Entries: 5 iTrader: (0) IGN: Yume/Relief/夢 Guild: Drop/JW/Kirameki/Hores(FT) I swear, stats is going to be the death of me. Suppose A is an event with probability 0.12 and B is an event with probability 0.51. If A and B are mutually exclusive, determine the following probabilities: P(AC or BC) = If A and B are independent, determine the following probabilities: P( AC or BC) = meaning, NOT A, and NOT B So for the first one, I thought it was 0.37 and 0.4312 for the second one, but apparently they are marking it wrong. Formula should be P(AC or BC) = 1 - P(A or B) right? why is it telling me it's wronggggg. I got the P(A or B) correct for both, but why is P(NOT A and NOT B) wrong? edit: I accidentally did something to get the no response, ignore that. D; I can't get the P(Ac or Bc) questions though. :| __________________ "Even a broken clock is right twice a day." Last edited by Yume; 03-09-2011 at 11:43 AM.
 03-09-2011 #6 (permalink) [夢] Games   Join Date: Jun 2008 Posts: 1,497 Blog Entries: 5 iTrader: (0) IGN: Yume/Relief/夢 Guild: Drop/JW/Kirameki/Hores(FT) I was able to get the answers thanks to Shadow again. but my prof gave us the formula P(A^c or B^c) = 1-P(AorB) but it doesn't work... still don't understand. __________________ "Even a broken clock is right twice a day."
 03-11-2011 #7 (permalink) ggFTW Lurker   Join Date: Jul 2009 Posts: 18 iTrader: (0) Probably a typo. De Morgan's laws: (A or B)^c = A^c and B^c (A and B)^c = A^c or B^c So, P(A^c or B^c) is identical to P[(A and B)^c] = 1 - P(A and B).

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