David, Thanks for the problem. I really liked it. However poissinv([0.05 0.95],10) formula produces [5 15]. In the definition lowerlimit = 4 and upperlimit = 15. The poissinv function systematically produces one unit below the lower limits reported in the test suite.
It should be : The poissinv function systematically produces one unit above the lower limits reported in the test suite.
Hello, Mehmet OZC. Thanks for your kind words and interesting observation. . . . From the documentation: "X = poissinv(P,lambda) returns the smallest value X such that the Poisson cdf evaluated at X equals or exceeds P" — i.e. the associated probably would be ≥ P. . . . This won't generally satisfy the requirement in the present problem statement: "not be faced more than 5% of the time" — i.e. the associated probably must be ≤ 5%. That accounts for the difference that you noticed. . . .
Thanks for your kind response. If you don't mind, I want to ask another question. When mu=10 first six cumulative probs are : [0.0 0.0 0.0 0.0103 0.0293 0.0671]. so cumulative prob for 5 isn't below 0.05. If I was not interested in the problem, I would not ask questions. I had to ask this one.
No problem: keep asking and we will all learn from it! . . . In "EXAMPLE 1" and the corresponding test in the Test Suite: mu = 10; lowerLimit = 4. For this lower limit the corresponding probability Pr(N≤4) satisfies my requirement that Pr(N≤k) ≤ 5%. That's why lowerLimit = 4 is the correct solution here. Indeed Pr(N≤0) et cetera would also satisfy that part of the requirement, but implicitly [sorry if it wasn't clear to everyone] the _largest_ value of k that satisfies Pr(N≤k) ≤ 5% is of most relevance. . . . As you noticed, Pr(N≤5) does not satisfy my requirement that Pr(N≤k) ≤ 5%. The "poissinv" function reports a value of k=5 because it instead finds the _smallest_ value of k that satisfies Pr(N≤k) ≥ 5% [note flipped inequality]. That is not consistent with the condition specified in the problem statement. . . . On the other hand, for the upper limit both the Test Suite and the "poissinv" function find the smallest value of k that satisfies Pr(N≤k) ≥ 95%. That's why there was no difference in results for upperLimit. . . . Of course, there is freedom to define other requirements in a different Cody problem. The requirements in this problem have a nice symmetry and consistency in never allowing the probability in either of the 'tails' to exceed 5%. On the other hand, the sum of the two 'tails' will typically be considerably less than 10% when using the values of lowerLimit and upperLimit specified here (e.g. only 7.8% for mu=10). For some purposes that might be seen as 'too conservative': choosing instead lowerLimit=5 would increase the sum of the two 'tails' to 11.6% — although it doesn't satisfy the strict requirements of the present problem, notice that it is closer to 10% (in this instance). . . . I hope that some of the above has addressed your question.
Neat solution using some additional mathematical knowledge.
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