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Math Help - need help on probability

  1. #1
    littlemissy
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    need help on probability

    My Solution::
    b i) no answer
    b ii) no answer
    b iii) no answer
    Last edited by littlemissy; July 12th 2007 at 08:39 AM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by littlemissy View Post
    From the long experience, a staff in casino found that the duration a visitor would spend in the casino per visit is normally distributed with the mean 90 minutes and standard deviation 20mins

    1)If one visitor was selected randomly, find the probability that he would spend
    a) more than 100minutes
    b) between 80 minutes and 130 minutes
    ...
    You should have a table of the standard normal distribution.

    Let z=(t-90)/20, where t is the time a punter spends in the casino in minutes, then z has a standard normal distribution.

    a) That t>100 corresponds to z>(100-90)/20=0.5

    If you have a table of the cumulative standard normal you can look z=0.5 up on that table to tell you p(z<0.5)=0.6915 but p(z>0.5)=1-p(z<0.5)=0.3085.

    b) That 80<t<130 corresponds to (80-90)/20<z<(130-90)/20 or -0.5<z<2

    If you have a table of the cumulative standard normal you can look z=-0.5 up on that table to tell you p(z<-0.5)=0.3085 and z=2 to tell you p(z<2)=0.9772, then p(-0.5<z<2)=p(z<2)-p(z<-0.5)=0.9772-0.3085=0.6687

    RonL
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  3. #3
    Newbie
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    Jul 2007
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    Quention 2 is hard for me....sorry cant help
    Last edited by badboychow; July 11th 2007 at 03:56 AM.
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