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    well z = \frac{x-\mu}{\sqrt{\sigma^2}}, where \mu is the mean and \sigma^2 is the variance which you have already worked out. Rearranging this you get x = z\sqrt{\sigma^2} + \mu. Then just substitute in the values already calculated for the mean and variance and the critical value for z at 90% (which is 1.282) and you get x!
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    (Original post by sonofdot)
    well z = \frac{x-\mu}{\sqrt{\sigma^2}}, where \mu is the mean and \sigma^2 is the variance which you have already worked out. Rearranging this you get x = z\sqrt{\sigma^2} + \mu. Then just substitute in the values already calculated for the mean and variance and the critical value for z at 90% (which is 1.282) and you get x!
    i was told that z = x minus the mean over just the standard deviation, not the standard deviation squared?
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    (Original post by jermaindefoe)
    i was told that z = x minus the mean over just the standard deviation, not the standard deviation squared?
    It is (I put it was over the square root of the variance, which is the same as the standard deviation. Sorry if that confused you )
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    (Original post by sonofdot)
    It is (I put it was over the square root of the variance, which is the same as the standard deviation. Sorry if that confused you )
    how did you get 1.282 for the value of z :eek:

    sorry for being a pain in the arse but i want to learn this
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    (Original post by jermaindefoe)
    how did you get 1.282 for the value of z :eek:

    sorry for being a pain in the arse but i want to learn this
    From my normal distribution tables (I guess it depends on your exam board, but near the back of my OCR formula book there is a page headed "the normal distribution function" At the bottom, there are critical values for the normal distribution, which tells you the value of z for different "special" probabilities. At 0.90 (90%), the value for z is 1.282 )
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    hmmm i dont know if i have one of those i just had a table for the normal distribution that you look through for the values, it isn't for "special" probabilities though?
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    (Original post by jermaindefoe)
    hmmm i dont know if i have one of those i just had a table for the normal distribution that you look through for the values, it isn't for "special" probabilities though?
    well in that case, you can look through the values for the probability until you find the one closest to 0.9000, then read off the z-value for that probablility

    I've even drawn you a nice picture

    Name:  normal01.jpg
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    thanks very much for that

    so can i just clarify so i can write this up in my notes so i can revise for the exam with it

    when i need to find the area up to a certain point (x) i need to

    rearrange the forumla to make x the subject
    look through the table to find the closest possible to the boundary given (in this case 90%, 0.9)
    substitute this value with the others, such as the mean and the standard deviation, and work out from there?

    what would the rearranged formulae look like if it were to not use the variance, and just the SD instead? as there would be no square root would there, then i can remember that and the points above for the exam

    rep comming ur way
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    (Original post by jermaindefoe)
    thanks very much for that

    so can i just clarify so i can write this up in my notes so i can revise for the exam with it

    when i need to find the area up to a certain point (x) i need to

    rearrange the forumla to make x the subject
    look through the table to find the closest possible to the boundary given (in this case 90%, 0.9)
    substitute this value with the others, such as the mean and the standard deviation, and work out from there?

    what would the rearranged formulae look like if it were to not use the variance, and just the SD instead? as there would be no square root would there, then i can remember that and the points above for the exam

    rep comming ur way
    Yep, those notes all look fine, and thanks for the rep
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    what would the rearranged formulae look like if it were to not use the variance, and just the SD instead?
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    (Original post by jermaindefoe)
    what would the rearranged formulae look like if it were to not use the variance, and just the SD instead?
    x = z\sigma + \mu where \sigma is the standard deviation
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    cheers mate

    you dont no how much i appreciate that! i am struggling with the statistics a little in computer science as the lectures move so fast, having explanations like that make it so much easier to understand
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    well that makes the answer

    17035.2

    which is too much isnt it
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    (Original post by jermaindefoe)
    well that makes the answer

    17035.2

    which is too much isnt it
    Are you sure?

    x = 1.282\sigma + \mu



x = 1.282 \times \sqrt{7840} + 7000



x = 113.5 + 7000



x = 7113.5
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    just seen what i am doing wrong i need to square root the variance
 
 
 

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