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Math Help - Help with finding probability of normal function...

  1. #1
    Senior Member Danneedshelp's Avatar
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    Help with finding probability of normal function...

    Let Y_{1},Y_{2},...,Y_{n} be independent, normal random variables, each with mean \mu and variance \sigma^{2}.

    a) Find the density function of U=Y^{bar}=\frac{1}{n}\sum_{i=1}^{n}Y_{i}.

    Solution to (a): So, U is normally distributed with mean \mu and variance \frac{\sigma^{2}}{n}.

    b) If \sigma^{2}=16 and n=25, what is the probability that the sample mean, Y^{bar}, takes on a value that is within one unit of the population mean, \mu? That is, find P(|Y^{bar}-\mu|\leq\\1).

    So, the way I have been trying to solve it is by finding the probability P(\mu-1\leq\\Y^{bar}\leq\mu+1), but I have had no luck. I get the feeling it is right in front of me. Should I use Tchebysheff's theorem?

    Thanks for the help

    oh, and Y^{bar} is supposed to represent y with a bar (_) over it...not sure how to do that.
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  2. #2
    MHF Contributor matheagle's Avatar
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    No, the point is that chebyshev's is a very weak inequality in this setting.

     P(|\bar Y-\mu|< 1) = P(-1<\bar Y-\mu< 1)

      = P(-1/(4/5)<{\bar Y-\mu\over \sigma/\sqrt{n}}< 1/(4/5))

      = P(-1.25<Z< 1.25)
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