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    Please can someone help me with this. I've asked some friends and they worked it out and all got different answers. We only learnt s1 for a week and I missed a few of the lessons.
    I would really appreciate some guidance with this question and would appreciate any help. Thanks in advance.

    The lifetime of an electrical component may be modelled by the normal distribution X with mean 2500 hours and variance 900 hours ^2

    Find the value of P(X>a) = 0.8

    Please help

    All I've got so far is

    X~N(2500,900^2)

    P(X>a) = 0.8

    P(Z<(a-2500)/900) = 0.8
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    (Original post by Aty100)
    Please can someone help me with this. I've asked some friends and they worked it out and all got different answers. We only learnt s1 for a week and I missed a few of the lessons.
    I would really appreciate some guidance with this question and would appreciate any help. Thanks in advance.

    The lifetime of an electrical component may be modelled by the normal distribution X with mean 2500 hours and variance 900 hours ^2

    Find the value of P(X>a) = 0.8

    Please help

    All I've got so far is

    X~N(2500,900^2)

    P(X>a) = 0.8

    P(Z<(a-2500)/900) = 0.8
    I suspect that the variance is 900 rather than 900^2.

    Your last statement isn't correct. You should go from P(X>a) = 0.8 to P(Z> ....) = 0.8, and then apply what you know about normal distributions to solve that.
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    (Original post by SeanFM)
    I suspect that the variance is 900 rather than 900^2.

    Your last statement isn't correct. You should go from P(X>a) = 0.8 to P(Z> ....) = 0.8, and then apply what you know about normal distributions to solve that.
    Oh sorry about that it's just because the statement said variance is 900^2 so I got a little confused
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    (Original post by SeanFM)
    I suspect that the variance is 900 rather than 900^2.

    Your last statement isn't correct. You should go from P(X>a) = 0.8 to P(Z> ....) = 0.8, and then apply what you know about normal distributions to solve that.
    Would I do

    1-P(Z>a-2500/900) = 0.8

    So 1-0.8 =0.2

    So

    a-2500/900 = 0.2 ?

    Or do I just use 0.8 ??
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    (Original post by Aty100)
    Oh sorry about that it's just because the statement said variance is 900^2 so I got a little confused
    If it's 900^2 then I suppose the variance is 900^2.

    If it's 900 hours^2 then it's just the units that are squared.

    (Original post by Aty100)
    Would I do

    1-P(Z>a-2500/900) = 0.8

    So 1-0.8 =0.2

    So

    a-2500/900 = 0.2 ?
    Just a note about my previous post, when you go from P(X<a) to P(Z<z) or P(X>a) to P(Z>z) the same inequality is used, you don't flip it or anything.

    So you've correctly done that P(X>a) can be written as P(Z>a-2500/300), so they are both equal to 0.8.

    Remember that you want it of the form P(Z<z) where z is a positive number so that you can read it off of the tables.

    Once you've corrected the 0.8 bit to 0.2, you should have
    1-P(Z>a-2500/300) = 0.2.

    Then you can express the left hand side as something else, and then use a property of the normal distribution to find the z value that corresponds to what's there (you've already posted a question that does something like this before).

    Remember that P(X<a) = 1-P(X>a) and that for the normal distribution, P(Z<z) = P(Z>-z) for a fixed z (visualise with a graph).
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    (Original post by SeanFM)
    If it's 900^2 then I suppose the variance is 900^2.

    If it's 900 hours^2 then it's just the units that are squared.



    Just a note about my previous post, when you go from P(X<a) to P(Z<z) or P(X>a) to P(Z>z) the same inequality is used, you don't flip it or anything.

    So you've correctly done that P(X>a) can be written as P(Z>a-2500/300), so they are both equal to 0.8.

    Remember that you want it of the form P(Z<z) where z is a positive number so that you can read it off of the tables.

    Once you've corrected the 0.8 bit to 0.2, you should have
    1-P(Z>a-2500/300) = 0.2.

    Then you can express the left hand side as something else, and then use a property of the normal distribution to find the z value that corresponds to what's there (you've already posted a question that does something like this before).

    Remember that P(X<a) = 1-P(X>a) and that for the normal distribution, P(Z<z) = P(Z>-z) for a fixed z (visualise with a graph).
    Thanks ! Shouldn't the 300 be 30?
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    (Original post by Aty100)
    Thanks ! Shouldn't the 300 be 30?
    Yes, sorry
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    (Original post by SeanFM)
    Yes, sorry
    so

    a-2500 / 30 = -0.2

    a-2500 / 30 = -0.8416 ?

    a = 2474.75 ??
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    (Original post by Aty100)
    so

    a-2500 / 30 = -0.2

    a-2500 / 30 = -0.8416 ?

    a = 2474.75 ??
    That seems to be correct, though I'm not sure what you meant by that first line.
 
 
 
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