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  1. #1
    Member Jskid's Avatar
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    [RESOLVED]probability involving z-score

    A bottling company uses a machine to fill plastic bottles with pop. Due to variations in the filling process, the pop amounts in all bottles are normally distributed with a mean of 300 ml and a standard deviation of 3 ml. One bottle is randomly selected, what is the probability that bottle contains less than 299.5 ml?

    z=\frac{299.5-300}{\frac{3}{\sqrt{n}}} I'm not sure what value n is, or if it's needed.

    I think I'm suppose to take n=1.
    For this question how I'm calculating the probability do I write P(\bar{x}<299.5) or P(x<299.5)?
    Last edited by Jskid; July 28th 2010 at 11:52 AM. Reason: resolved
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  2. #2
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    Quote Originally Posted by Jskid View Post
    A bottling company uses a machine to fill plastic bottles with pop. Due to variations in the filling process, the pop amounts in all bottles are normally distributed with a mean of 300 ml and a standard deviation of 3 ml. One bottle is randomly selected, what is the probability that bottle contains less than 299.5 ml?

    z=\frac{299.5-300}{\frac{3}{\sqrt{n}}} I'm not sure what value n is, or if it's needed.
    Where has n come from? \displaystyle z = \frac{x - \mu}{\sigma} and you have been told the values of mean, sd and x.
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  3. #3
    Member Jskid's Avatar
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    \sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}} where n is sample size.
    What I don't understand is a remeber doing similar questions but just used the S.D. \sigma
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    Quote Originally Posted by Jskid View Post
    \sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}} where n is sample size.
    What I don't understand is a remeber doing similar questions but just used the S.D. \sigma
    P(\bar{X}<299.5) \implies P\left(Z<\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}}\right)

    P(X<299.5) \implies P(Z<\frac{X-\mu}{\sigma})

    Can you see the difference?
    Last edited by pickslides; July 27th 2010 at 05:44 PM. Reason: Bad Latex
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  5. #5
    Member Jskid's Avatar
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    Quote Originally Posted by pickslides View Post
    P(\bar{X}<299.5) \implies P\left(Z<\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}}\right)

    P(X<299.5) \implies P(Z<\frac{X-\mu}{\sigma})

    Can you see the difference?
    I think I understand. The top equation is for the probability of a sample mean and the lower one is for the probability of getting a sample of a certain value? Does that mean the two formula are the same because \frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}}=\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{1}}}=\frac{X-\mu}{\frac{\sigma}{1}}
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    Quote Originally Posted by Jskid View Post
    I think I understand. The top equation is for the probability of a sample mean and the lower one is for the probability of getting a sample of a certain value? Does that mean the two formula are the same because \frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}}=\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{1}}}=\frac{X-\mu}{\frac{\sigma}{1}}
    I have never thought of it like that. But I am inclinded to say this is not true as you can be given a sample size n>1 for X and the standard tranformation would apply given it was normal.
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