Thanks professor, you have the best explanation for the median. Intuitively, I would do ratios since median value doesn’t shift, but hats off for using LOTP.
@analystprep Жыл бұрын
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@aquiljones99574 ай бұрын
for the alternative solution starting at 4:25 when we calculate the probability that the policy experiences exactly 3 losses we take the probability of 3 losses from both distributions and multiply by their respective weights. Aren't there other outcomes such as 2 losses from sub policy 1 and 1 from sub policy 2, or 1 loss from sub policy 1(binomial) and 2 losses from sub policy 2(poisson)?
@rudyerickson38304 ай бұрын
the models are exclusive events
@rgg_48 Жыл бұрын
Can we get an alternative way to find the value associated with a z-score of .375? I ask this because the SOA table provided for exam p doesn't list the z-scores below .50.
@yaweli2968 Жыл бұрын
You mean the Z score associated with the probability of 0.375. Take the phi inverse but don’t forget to negate it. Like -phi_inverse(1-0.375). Check for the Z score associated with 0.625 and negate it. This is because the norm. Distribution is symmetric and so are the Z scores with their associated probability bands.
@Alexanderp488 Жыл бұрын
Can anyone explain on the very last question how standardizing the .375 leads to a -.32?
@AaronRiffie Жыл бұрын
If you look up a Z score of -.32 on the distribution table it is .37448. Maybe you were looking up .375 as the z score? I did that at first by mistake.