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

```

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