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02-25-2011   #1 (permalink)
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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)
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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?
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02-25-2011   #3 (permalink)
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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.
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03-04-2011   #4 (permalink)
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(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.
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Last edited by kag; 03-04-2011 at 10:36 PM.
 
03-09-2011   #5 (permalink)
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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. :|
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Last edited by Yume; 03-09-2011 at 12:43 PM.
 
03-09-2011   #6 (permalink)
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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.
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03-11-2011   #7 (permalink)
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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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