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Continous Random Variable Questions P(X>0.2) different to P(X>=0.2)? watch

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    Is there any difference so I get that you integrate from 0.2 to being the lower limit, and the upper limit being infinity/highest value?
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    (Original post by Mihael_Keehl)
    Is there any difference so I get that you integrate from 0.2 to being the lower limit, and the upper limit being infinity/highest value?
    Nope!  > is the same as \geqslant for continuous distributions.
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    (Original post by aymanzayedmannan)
    Nope!  > is the same as \geqslant for continuous distributions.
    I thought so thank you.

    What about this question: Find P(-sigma)<X<sigma)

    Where sigma is the s.d.

    How do I solve this lmao

    I have worked out P(X<sigma) but I never know what to do next: It is page 52 Q6d of the edexcel book if you have it
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    (Original post by Mihael_Keehl)
    I thought so thank you.

    What about this question: Find P(-sigma)<X<sigma)

    Where sigma is the s.d.

    How do I solve this lmao

    I have worked out P(X<sigma) but I never know what to do next: It is page 52 Q6d of the edexcel book if you have it
    Which module?
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    (Original post by aymanzayedmannan)
    Which module?
    lmao sorry S2. I don't know if you are writing it though :P
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    (Original post by aymanzayedmannan)
    Nope!  &gt; is the same as \geqslant for continuous distributions.
    Is there any reason why this is btw, just feel like I am following it without really understanding it :/
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    (Original post by Mihael_Keehl)
    Is there any reason why this is btw, just feel like I am following it without really understanding it :/
    P(X \geq x) = P(X=x) + P(X &gt; x), but for continuous distributions: P(X=x) = 0 since, well, intuitively you're asking for one single possibility in a (continuous, hence massive/infinite/uncountably infinite) realm of possibilities.
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    (Original post by Mihael_Keehl)
    Is there any reason why this is btw, just feel like I am following it without really understanding it :/
    Think about it - time is a continuous variable and you can have 3.5 seconds but if you were selecting beads from a bowl you couldn't have 3.5 beads. Maybe look over S1 Chapter 8 and 9 if you're still a little confused.
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    (Original post by Zacken)
    P(X \geq x) = P(X=x) + P(X &gt; x), but for continuous distributions: P(X=x) = 0 since, well, intuitively you're asking for one single possibility in a (continuous, hence massive/infinite/uncountably infinite) realm of possibilities.
    (Original post by aymanzayedmannan)
    Think about it - time is a continuous variable and you can have 3.5 seconds but if you were selecting beads from a bowl you couldn't have 3.5 beads. Maybe look over S1 Chapter 8 and 9 if you're still a little confused.
    ofc ffs this is continuous and not discrete omg.

    Did you look at that question aymanzayedmannan
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    (Original post by Mihael_Keehl)
    I thought so thank you.

    What about this question: Find P(-sigma)<X<sigma)

    Where sigma is the s.d.

    How do I solve this lmao

    I have worked out P(X<sigma) but I never know what to do next: It is page 52 Q6d of the edexcel book if you have it
    You can do this by thinking about the area under the graph. Is X normally distributed? If so you can just look up the relevant values in the table of the cumulative normal function
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    (Original post by shamika)
    You can do this by thinking about the area under the graph. Is X normally distributed? If so you can just look up the relevant values in the table of the cumulative normal function
    Thanks but isnt it P(x<sigma) - P(x<sigma) = 0?
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    (Original post by Mihael_Keehl)
    Thanks but isnt it P(x<sigma) - P(x<sigma) = 0?
    Obviously. But how is that relevant here?
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    (Original post by Mihael_Keehl)
    Thanks but isnt it P(x<sigma) - P(x<sigma) = 0?
    Yep. But P(X<-sigma) isn't the same as P(X<sigma); note the minus sign!
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    (Original post by shamika)
    Yep. But P(X<-sigma) isn't the same as P(X<sigma); note the minus sign!
    Thanks yes, so how do you account for the negative sign?

    P(X< - sigma) = 1 - P(X<sigma)?
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    (Original post by Mihael_Keehl)
    Thanks yes, so how do you account for the negative sign?

    P(X< - sigma) = 1 - P(X<sigma)?
    Yes. so with some nice re-arranging, you should get: P(x < sigma) = 1/2. Which makes sense if you look at a normal distribution graph.
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    (Original post by Zacken)
    Yes. so with some nice re-arranging, you should get: P(x < sigma) = 1/2. Which makes sense if you look at a normal distribution graph.
    thnx
 
 
 
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