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Sweeting example 14.8

Discussion in 'SP9' started by yellowvonpen, Jul 24, 2016.

  1. yellowvonpen

    yellowvonpen Member

    Can anyone show me how to calculate the numbers in the table of example 14.8 on page 367? I'm getting numbers that are somewhat close, but not rounding close, so I'm not sure if what I'm doing is correct.
    For example:
    f(d80)=exp(-lambda)*(lambda^20)/20! where lambda=q*250 where q=exp(-9.05 + 80*.08). I get 0.076, which isn't the 0.0888 listed.

    Thanks!
     
    yellowvonpen, Jul 24, 2016
    #1
  2. Simon James

    Simon James ActEd Tutor Staff Member

    Simon James, Jul 25, 2016
    #2
  3. yellowvonpen

    yellowvonpen Member

    I did check the errata but didn't find anything there for this exercise.
     
    yellowvonpen, Jul 25, 2016
    #3
  4. Simon James

    Simon James ActEd Tutor Staff Member

    OK, I will look at both of these for you on Monday
     
    Simon James, Jul 29, 2016
    #4
  5. Simon James

    Simon James ActEd Tutor Staff Member

    HI.

    As you spotted, it is not obvious from the formulae given, but Sweeting states he is using a log-linear model.
    So, \( ln\left ( q \right )=a+bx \). Note also he is using b=0.0815 (not 0.08 as given earlier!)
    I get q = exp(-9.05 + 80*.0815) = 0.79659, so lambda = q * 250 = 19.91476. Then I agree Sweeting's numbers.
     
    Simon James, Aug 8, 2016
    #5

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