       Re: Re: binomial distribution

• To: mathgroup at smc.vnet.net
• Subject: [mg16967] Re: [mg16915] Re: [mg16895] binomial distribution
• From: BobHanlon at aol.com
• Date: Thu, 8 Apr 1999 02:32:55 -0400
• Sender: owner-wri-mathgroup at wolfram.com

```Arnold,

If you need precision, the evaluated form of the function can also provide
it. Just use an exact (rational) number for p (i.e., 1/5 instead of 0.2 in
this case).

pts = PrBinomial[#,84,1/5]& /@ Range;
ListPlot[pts,PlotJoined->True,PlotRange->All];

This will slow it down.

I suspect that the preferred method would depend on the specific application.

For the problem given, a bigger mistake that I made was that I should have
incremented the values of k by two rather than one, i.e., I should have used
{k, 0, 83, 2}.  This would have still hit all of the permissable integers
except 84 (i.e., c = n, for which case, PrBinomial[n, n, p] = 1) as one of
the starting values for the FindRoot.

I could have also combined the Round step with the FindRoot step by using
Cases rather than Select:

soln = Cases[c /.
Table[FindRoot[eqn, {c, k, k + 1}], {k, 0, 83, 2}],
x_?IntegerQQ -> Round[x] ]

Bob Hanlon

In a message dated 4/7/99 12:29:17 PM, arnoldk at cam.wits.ac.za writes:

>While using Evaluate may be fast I should point out that its highly
inaccurate.
>For example observe the zigzag effect from your function of
>
>lst= PrBinomial[#,84,0.2]& /@ Range
>ListPlot[%,PlotJoined->True,PlotRange->All]
>
>Using greater accuracy one gets the correct monotonicity of the graph:
>
>PrBinomial2[c_, n_, p_] :=
>  N[Sum[Binomial[n, k]*(1 - p)^k*p^(n - k),
>    {k, 0, c}],25]
>
>lst= PrBinomial2[#,84,0.2]& /@ Range
>
>ListPlot[%,PlotJoined->True,PlotRange->All]
>

```

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