Binomial:
P[x={10,17,25,40}|550, .04] = {0.0017997, 0.051547, 0.066349, 0.00012953}
Poisson:
P[x={10,17,25,40}|550*0.04] = {{0.0020417, 0.051956, 0.06538, 0.00017013}
Consider a bank which has a set of 550 loans. The bank estimates, that 4 out of
100 loans will get problems each year. Calculate the probability that 10,17,25 and
40 loans will get problems next year (use the binomial distribution and the poisson
distribution and compare the results).
I tried to use the binomial formula:
where n=550
k=10
p=0.04
My result is 0.18% which in my opinion makes not much sense as for 550 Loans there should be 22 defaults...
for the Poisson distribution I got 2.77^(-21)
I put in the figures in the following formula:
where k=10
and lambda=0.04
Thx in advance
Binomial:
P[x={10,17,25,40}|550, .04] = {0.0017997, 0.051547, 0.066349, 0.00012953}
Poisson:
P[x={10,17,25,40}|550*0.04] = {{0.0020417, 0.051956, 0.06538, 0.00017013}
There are a few different order statistics formulas that include one for a maximum, one for a minimum and one for a distribution that is so many places from the minimum or the maximum.
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