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Thread: Normal Distribution

  1. #1
    Super Member
    Sep 2007

    Normal Distribution

    The wording confuses me. How can this question be done?


    Q: The time taken by a bus to travel between Town A and Town B can be modeled by a normal distribution with Mean 180 minutes and Standard Deviation 12 minutes. A passenger on the bus needs to be in Town B t minutes after the bus leaves Town A. Find the value of t such that there is only a 1 \% chance the passenger does not arrive in time.


    I understand standardizing should be done but not aware how it should be done? The answer is 207.9.
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  2. #2
    Super Member
    Oct 2007
    London / Cambridge
    Hello air

    I assume your taking the S1 exam this month.

    let the variable of time taken be T

    you want P(T < t ) = 0.99

    transform the variable to Z

    giving P(Z < \frac{t - 180}{12} ) = 0.99

    now you need to look in your normal distribution table the value of z which gives 0.99. the table I have only has  \Phi (2.32)  = 0.9898 and \Phi (2.34) = 0.9904 . If your are taking S1 your not expected to interpolate so I'll just take 2.32

    giving \frac{t - 180}{12}  = 2.32

    giving t = 207.84
    Last edited by bobak; Jan 1st 2008 at 11:45 AM. Reason: Horrific errors thanks jhevon
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