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Thread: probability involving z-score

  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?

    $\displaystyle z=\frac{299.5-300}{\frac{3}{\sqrt{n}}}$ 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 $\displaystyle P(\bar{x}<299.5)$ or $\displaystyle P(x<299.5)$?
    Last edited by Jskid; Jul 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?

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

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

    Can you see the difference?
    Last edited by pickslides; Jul 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
    $\displaystyle P(\bar{X}<299.5) \implies P\left(Z<\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}}\right)$

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

    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 $\displaystyle \frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}}=\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{1}}}=\frac{X-\mu}{\frac{\sigma}{1}}$
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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 $\displaystyle \frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}}=\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{1}}}=\frac{X-\mu}{\frac{\sigma}{1}}$
    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 $\displaystyle n>1$ for $\displaystyle X$ and the standard tranformation would apply given it was normal.
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