Can some one help me evalute this integral or explain a bit
where Pa(P) is binomial distribution with a formulae
please help me evaluate this. the value for c = 2.
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ALi
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1What is the function $\varphi(p)$? – 5xum Feb 11 '14 at 08:29
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1And what is the value of $n$ in the definition of $P_a$? – 5xum Feb 11 '14 at 08:30
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Maybe $$P_a (p) = \sum _{k=0} ^n \binom{n}{k} p^k (1-p)^{n-k} ?$$ It is unclear what $\varphi (p)$ is though... – izœc Feb 11 '14 at 08:30
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n = 80 , c = 2, – ALi Feb 11 '14 at 10:42
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φ(p) is the distribution of P , whose values are given as 0.05-0.08,, i am also confused with this. i think its a defect rate in the lot , and author take it as binomial distribution. – ALi Feb 11 '14 at 10:47
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let me redefine th problem with actaul notations – ALi Feb 11 '14 at 10:47
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The integral you have can be simplified by simplifying the polynomial inside. It equals $$\int_{0.05}^{0.08}\left[(9.52 + 1698.8p)P_a(p) + (95.2 + 428p)(1-P_a(p))\right]\varphi(p) dp$$
If, in your definition of $P_a$, the values $c$ and $n$ are the same, then $P_a$ is equal to (see binomial expansion
$$P_a(p) = (p + (1-p))^n = 1^n = 1,$$
further simplifying your integral into $$\int_{0.05}^{0.08}\left[9.52 + 1698.8p\right]\varphi(p) dp,$$
which looks much less scary than what you originaly had.
5xum
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hey thanks a lot,, but n , c are not equal ,, n = 80 , c = 2 ,,without Pa(P) i can simplify to its current form as you did , But Pa(P) is not equal to one , Further Pa is function of P , do i need to integrate it. do you have any suggestion with this. thanks again – ALi Feb 11 '14 at 10:29
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