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Thread: Normal distribution with multiple values, probability

  1. #1
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    Normal distribution with multiple values, probability

    I'm not sure how to go about b,c,d here :

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    answer = 0.1587

    I get (ish) that for part a I can standardise the difference between the desired value and the given mean with

    z = [x - \mu] / (\sigma)

    (not sure how to render latex...)

    And then I can look that up in the Z table, but I'm not sure what to do if i want more than one?

    It seems that I have something such as

    P(a > 51) :and: P(b > 51) :and: P(c > 51) :and: P(d > 51)

    where a,b,c,d are four bags of nuts, but this doesn't seem like a very sensible approach and doesn't give the answer (which is 0.1587)

    Cheers
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  2. #2
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    Re: Normal distribution with multiple values, probability

    OK - I think what I need to do is use sigma / sqrt(n)

    in the denominator for

    (x - mu) / [ sigma / sqrt(n) ]

    I can't explain the reasoning too well, but this gives the answer.

    So if anyone has any insights that would be appreciated, thanks
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  3. #3
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    Re: Normal distribution with multiple values, probability

    Hey silverpen.

    That is correct and the reason has to do with standardizing the normal distribution.

    Do you know what subtracting the mean and dividing by the standard deviation does to the distribution if it's normal?
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  4. #4
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    Re: Normal distribution with multiple values, probability

    cheers Chiro - I'm not sure other than saying that it standardises it?

    I mean, If i have a normal dist. with \mu = 6 , \sigma = 9 and I standardise it then i would expect for the distribution to be \mu = 0 and \sigma = 1 (after standardising), ?

    That's all i can reallly think of though
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  5. #5
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    Re: Normal distribution with multiple values, probability

    That's correct and that is why you do this and then look at a Normal table because it has a normal probability of N(0,1) not N(mu.sigma^2)
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