schmidtdancer
Consider a uniform distribution created by a random number generator. The distribution looks like a square with a length of 1 and a height of 1. The random number generator creates any number between 0 and 1. Find the following probabilities:
a) P(0 <= X <= 0.4)
b) P(0.4 <= X <= 1)
c) P(X > 0.6)
d) P(X <= 0.6)
e) P(0.23 <= X <= 0.76)
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schmidtdancer
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@ParthKohli @hba ??
schmidtdancer
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(a) I know a is 0.4
mathmate
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|dw:1358086275602:dw|
Here's how you get it.
schmidtdancer
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ok so a would then be 0.4?
mathmate
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Yes, as you said.
Remember the area under a probability distribution always add up to 1.0.
schmidtdancer
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Ok! how do i find b?
mathmate
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Probability(a<X<b)=area between the vertical lines a and b.
schmidtdancer
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is it .6?
mathmate
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Exactly!
schmidtdancer
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c?
mathmate
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Use your knowledge!
schmidtdancer
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um....4?
schmidtdancer
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.4?
mathmate
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Yep! Good again!
schmidtdancer
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Ok! would d be .4 too then
schmidtdancer
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P(X <= 0.6)
mathmate
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Can you repeat for d? Draw the diagram and check.
schmidtdancer
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x is less then or eqaual to .6.....
mathmate
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The area is....
schmidtdancer
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hmm
schmidtdancer
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im not sure how to find this probability?
mathmate
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The figure is similar to the one I drew, but the vertical line is at 0.6.
schmidtdancer
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.7?
mathmate
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X varies from 0 to 1, so
X<0.6 means from 0 to 0.6
mathmate
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|dw:1358086899713:dw|
schmidtdancer
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.6?
schmidtdancer
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thats 60 % of the diagram
mathmate
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Right again!
schmidtdancer
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ok so d is .6? :)
schmidtdancer
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and e is .53?
mathmate
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You do realize that we are dealing with a square of 1x1.
So 0<X< 0.6 means simply 60% of the square.
Sometimes, and most of the time, the distribution is not a straight line (uniform), the calculation may not be as easy. Such as this:
|dw:1358087084888:dw|
You're doing great with this particular case.
ALSO NOTICE that we did not care between <= (less than or equal) and < less than.
It is because P(x=0.6) is zero, because it is almost impossible for X to be exactly 0.6.
Finally, for (e), you got it right again, congratulations!