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Normal Distribution Porbability

  1. May 18, 2014 #1
    I have ##\bar{X}## ~ ##N(\mu , 9/25)##

    I have ##E[X] = \mu##
    ##Var[X] = 9/25##
    ##SD[X] = 3/5 = 0.6##

    An interval for ##\bar{X}## has been recorded: ##\bar{X} \pm 1.05##.

    I asked to find ##P(\bar{X} > \mu + 1.05)##

    I can "normalize" the distribution through:

    ##Z = \frac{\bar{X} - \mu}{0.6}## ~ ##N(0,1)##

    I'm confused by this next step:

    ##P(\bar{X} > \mu + 1.05) = P(Z > \frac{1.05}{0.6} = 1.75)##

    I'm not sure how you go from the first probability to the other. Could any help please?
     
  2. jcsd
  3. May 18, 2014 #2

    Ray Vickson

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    The events ##\{\bar{X} > \mu + 1.05\}## and ##\{Z > 1.05/0.6\}## are the same:
    [tex] \{ \bar{X} > \mu + 1.05\} = \{\bar{X} - \mu > 1.05 \}
    = \{ (\bar{X} - \mu)/0.6 > 1.05/0.6 \}[/tex]
     
    Last edited: May 18, 2014
  4. May 18, 2014 #3
    Oh I see, I was being a bit silly there, thanks!
     
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