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Sweeting example 14.8

Y

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!
 
I did check the errata but didn't find anything there for this exercise.
 
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.
 
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