Re: binomial distribution
- To: mathgroup at smc.vnet.net
- Subject: [mg16915] Re: [mg16895] binomial distribution
- From: BobHanlon at aol.com
- Date: Tue, 6 Apr 1999 01:27:38 -0400
- Sender: owner-wri-mathgroup at wolfram.com
In a message dated 4/5/99 3:52:02 AM, robert.wright4 at virgin.net writes:
>Can someone explain how I can solve for 'c' or 'n' given the other variables
>in this equation: its the binomial form for calculation the operating
>characteristic in acceptance sampling. The problem is that 'c' and 'n'
>are
>discrete and therefore 'Findroot' or 'NSolve' do not work.
>
>The other problem is that it takes a long time to evaluate 'PrBinomial'
>for
>large 'c' and 'n'.... is there a better way of calculating?
>
>
>\!\(PrBinomial[c_, \ n_, p_] := \
> Sum[\ Binomial[n, k]\ \(\((1 - p)\)\^k\) p\^\(n - k\), {k, 0, c}] //
>N\)
>
Robert,
PrBinomial[c_, n_, p_] :=
Evaluate[Sum[Binomial[n, k]*(1 - p)^k*p^(n - k),
{k, 0, c}]]
Note the use of Evaluate.
x1 = PrBinomial[3,5,0.1];
FindRoot[PrBinomial[c,5,0.1]==x1, {c, 4, 5}]
{c\[Rule]3.}
FindRoot[PrBinomial[3,n,0.1]==x1, {n, 3, 4}]
{n\[Rule]5.}
x2 = PrBinomial[27,84,0.2];
If you know roughly what the solution should be
you can limit the search. If not, check all values:
IntegerQQ[x_] := Chop[x - Round[x]] == 0;
eqn = PrBinomial[c,84,0.2]==x2;
soln = Select[c /.Table[FindRoot[eqn, {c, k, k+1}], {k, 0, 83}],
IntegerQQ[#]&]
(* extraneous error messages removed *)
{6.,7.,10.,11.,14.,19.,17.,19.,19.,22.,25.,25.,27.,35.}
Rounding and eliminating duplicate solutions
soln = Union[Round[soln]]
{6,7,10,11,14,17,19,22,25,27,35}
Picking the best discrete solution
First[Sort[soln,
Abs[PrBinomial[#1,84,0.2]-x2] < Abs[PrBinomial[#2,84,0.2]-x2]&]]
27
Bob Hanlon