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    please can someone guide me through how to do this? I have no idea
    Any help will be much apprecaited. Thanks in advance to anyone does help !

    Question:
    the lifetime of an electrical component may be modelled by a normal distribution, X, which means 2500 hours and variance 900 hours 2. Find the value of P(X>a)=0.8

    Ive started off kinda :/
    Mean= 2500
    Variance = 900^2

    X~N(2500 , 900^2)
    P(X > a) = 0.8
    ??
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    (Original post by Aty100)
    please can someone guide me through how to do this? I have no idea
    Any help will be much apprecaited. Thanks in advance to anyone does help !

    Question:
    the lifetime of an electrical component may be modelled by a normal distribution, X, which means 2500 hours and variance 900 hours 2. Find the value of P(X>a)=0.8

    Ive started off kinda :/
    Mean= 2500
    Variance = 900^2

    X~N(2500 , 900^2)
    P(X > a) = 0.8
    ??
    Variance is 900 not 9002
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    (Original post by Aty100)
    x
    As per TeeEm, the normal distribution is set out like: X ~ N(mean, variance) or X ~ N(mean, standard deviation^2)

    So what you have is X ~ N(2500, 900) - second, convert this into a standardised z-value thingy.
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    (Original post by TeeEm)
    Variance is 900 not 9002
    Oh yah sorry 😁
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    (Original post by TeeEm)
    Variance is 900 not 9002
    If you don't mind would you be able to assist me on what do next?
    I'm assuming that I should look at the table of normal distribution functions and look for the value of 0.8 ??
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    (Original post by Aty100)
    If you don't mind would you be able to assist me on what do next?
    I'm assuming that I should look at the table of normal distribution functions and look for the value of 0.8 ??
    page 23 onwards
    http://madasmaths.com/archive/maths_...lculations.pdf
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    Hey thanks

    I alway get confused on this bit:

    You see in the attachment how does the 0.95 become -1.645??
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    (Original post by Aty100)
    Hey thanks

    I alway get confused on this bit:

    You see in the attachment how does the 0.95 become -1.645??
    You do 1-0.95 which is 0.05. The z value for 0.05 is 1.6449. Because what you're looking for is on the left hand side, the value is negative.
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    (Original post by Aty100)
    Hey thanks

    I alway get confused on this bit:

    You see in the attachment how does the 0.95 become -1.645??
    look at the normal tables backwards

    you are inverting probability of 0.95

    (the minus is because the "cut" is on the left of the mean)
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    (Original post by TeeEm)
    look at the normal tables backwards

    you are inverting probability of 0.95

    (the minus is because the "cut" is on the left of the mean)
    (Original post by mk_98)
    You do 1-0.95 which is 0.05. The z value for 0.05 is 1.6449. Because what you're looking for is on the left hand side, the value is negative.
    Guys can you check this :/
    Thanks for your help!
    Attached Images
     
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    (Original post by Aty100)
    Guys can you check this :/
    Thanks for your help!
    the is no INVERSE PHI of 0.2 in standard tables

    you only do 1 minus if you are inverting probabilities under 0.5

    however this self corrects since INVERSE PHI of 0.8 is indeed 0.8416

    the issue is that this is minus if you draw a picture since the "cut" is on the left.
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    (Original post by TeeEm)
    the is no INVERSE PHI of 0.2 in standard tables

    you only do 1 minus if you are inverting probabilities under 0.5

    however this self corrects since INVERSE PHI of 0.8 is indeed 0.8416

    the issue is that this is minus if you draw a picture since the "cut" is on the left.
    Sorry to hijack but TeeEm could you please have a look at this question?
    http://www.thestudentroom.co.uk/show....php?t=3708387

    Again sorry OP for jacking this thread.
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    (Original post by TeeEm)
    the is no INVERSE PHI of 0.2 in standard tables

    you only do 1 minus if you are inverting probabilities under 0.5

    however this self corrects since INVERSE PHI of 0.8 is indeed 0.8416

    the issue is that this is minus if you draw a picture since the "cut" is on the left.
    Oh sorry.
    Thanks so much
    So the final answer is correct then ?
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    (Original post by Aty100)
    Oh sorry.
    Thanks so much
    So the final answer is correct then ?
    No because you need -0.8416

    the "cut" is on the left

    you do not do NORMAL without a diagram....
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    the mean is 2500 hours, the standard deviation is 30 hours. you need to look up the z value for 0.8... this will be about 0.84

    0.84 = (2500 - a)/30

    rearrange to find a.

    we put 2500 - a instead of a - 2500 because the value of a must lie to the left of the mean.
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    (Original post by TeeEm)
    No because you need -0.8416

    the "cut" is on the left

    you do not do NORMAL without a diagram....
    But that gives the same answer :/
    I accidentally forgot to put it in the attachment when I wrote it
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    (Original post by the bear)
    the mean is 2500 hours, the standard deviation is 30 hours. you need to look up the z value for 0.8... this will be about 0.84

    0.84 = (2500 - a)/30

    rearrange to find a.

    we put 2500 - a instead of a - 2500 because the value of a must lie to the left of the mean.
    So the answer is -2525 ?
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    (Original post by Aty100)
    But that gives the same answer :/
    I accidentally forgot to put it in the attachment when I wrote it
    the answer you will get from these workings is incorrect as it will produce a value over 2500 which cannot be right...

    you need -0.8416
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    (Original post by Aty100)
    So the answer is -2525 ?
    0.84*30 = 2500 - a

    so a should be a positive number ?
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    (Original post by the bear)
    0.84*30 = 2500 - a

    so a should be a positive number ?
    So a = 2475? I used -0.8416
 
 
 
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