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Thread: Distribution and probabilty - parent and sample

  1. #1
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    Exclamation Distribution and probabilty - parent and sample

    Ok I am completely lost, can anyone help me out? I have done the first question, and the second one is based on it.

    The first question said:
    The parent distribution of x is normal with mean 80 and standard deviation 9.
    Based on samples of size n=25, find the mean and standard error of the sampling distribution of $\displaystyle \bar{x}$.

    I said: $\displaystyle \mu_{\bar{x}}$ = 80
    $\displaystyle \sigma_{\bar{x}}$ = 9/5

    The second question is where I am lost.

    It says find:
    A. P(62 < x < 80)
    B. P(71 < $\displaystyle \bar{x}$ < 77)
    C. x' such that P($\displaystyle \bar{x}$ > x') = .05
    Last edited by rba_mandy; Mar 18th 2009 at 09:14 PM. Reason: Wrong symbol >
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  2. #2
    MHF Contributor matheagle's Avatar
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    Is n=25 or 225? Because $\displaystyle \sigma_{\bar{x}}={\sigma \over\sqrt{n}}$.

    Or, did you mean $\displaystyle \sigma_{\bar{x}}={9\over 5}$
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  3. #3
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    n=25

    it is the number in the sample
    $\displaystyle \sigma\bar{x}$ = 9/5 (thats what I got) sorry
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  4. #4
    MHF Contributor matheagle's Avatar
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    $\displaystyle P(62<X<80)=P\biggl({62-80\over 9}<Z<{80-80\over 9}\biggr)=P(-2<Z<0)$.

    $\displaystyle P(71<\bar X<77)=P\biggl({71-80\over 9/5}<Z<{77-80\over 9/5}\biggr)$.

    Can you finish these? I'll check on you later.

    $\displaystyle .05=P(\bar X<a)=P\biggl(Z<{a-80\over 9/5}\biggr)$ so $\displaystyle {a-80\over 9/5}=-1.645$.
    Last edited by matheagle; Mar 18th 2009 at 09:03 PM.
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  5. #5
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    P[(-5) < z < (-5/3)]

    It was supposed to be P($\displaystyle \bar{x}$ > x') SORRY

    P($\displaystyle \bar{x}$ > $\displaystyle x'-80\over 9/5$) ?

    Can you tell me what formula you used to find that last one?

    Did you use these for the first two?
    Z= $\displaystyle X-\mu\over \sigma$ and Z=$\displaystyle \bar{x}-\mu\over \sigma / \sqrt n$
    Last edited by rba_mandy; Mar 18th 2009 at 09:25 PM.
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  6. #6
    MHF Contributor matheagle's Avatar
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    Let X be a random variable with mean $\displaystyle \mu$ and standard deviation $\displaystyle \sigma$.
    Then $\displaystyle Z={X-\mu\over \sigma}$ has mean zero and standard deviation 1.
    And if X was a normal random variable, then Z is a standard normal random variable and you can use your Z table.
    We did that for both X and $\displaystyle \bar X$.
    Last edited by matheagle; Mar 19th 2009 at 03:59 PM.
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  7. #7
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    Smile

    Thank you SO much for your help. It helped me a lot on my test. Thanks a hundred times. You are a really nice person
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