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Math Help - normal distribution

  1. #1
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    normal distribution

    hi, in my lecture notes, it says:

    given, standard deviation of 0.1
     P(Z< \frac{xbar -0.3}{0.1/\sqrt{n}}) = 0.95
     \frac{xbar -0.3}{0.1/\sqrt{n}} = 1.645

    this is a dumb question but how did they get to the second step?
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  2. #2
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    Quote Originally Posted by Dgphru View Post
    hi, in my lecture notes, it says:

    given, standard deviation of 0.1
     P(Z< \frac{xbar -0.3}{0.1/\sqrt{n}}) = 0.95
     \frac{xbar -0.3}{0.1/\sqrt{n}} = 1.645

    this is a dumb question but how did they get to the second step?
    By using tables or technology you should be able to establish that for a standard normal variate, Pr(Z < 1.645) = 0.95 (correct to 4 decimal places). I hope you see the connection.
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