## Inverse Poisson Distribution...

I'm wondering if there is an available command that can evaluate the number of terms required to produce a desired outcome.

Specifically, I am interested in determining the probability of a Poisson distribution, given the parameter (mean) value and the probability outcome. I can obtain the desired result using trial and error / brute force, but I am curious to know if there is a more efficient way.

Suppose that, lambda = 2.6 and the cumulative sum of the probabilities is 95%. I know that I must add the first 6 terms for P(x) in the series (x=0,1, ..,5) to sum to 0.95. Each term ...  P(x=0)= 0.07, P(x=1)=0.19, and so on.

However, how can we know that desired 95% outcome can be determined from the first 5 terms without trial & error?

## Bug in Statistics:-CDF

Maple 2016

Let us consider

```with(Statistics):
X1 := RandomVariable(Normal(0, 1)):
X2 := RandomVariable(Normal(0, 1)):
X3 := RandomVariable(Uniform(0, 1)):
X4 := RandomVariable(Uniform(0, 1)):
Z := max(X1, X2, X3, X4); CDF(Z, t);

int((1/2)*(_t0*Heaviside(_t0-1)-_t0*Heaviside(_t0)-Heaviside(1-_t0)*Heaviside(-_t0)+Heaviside(-_t0)+Heaviside(1-_t0)-1)*(1+erf((1/2)*_t0*2^(1/2)))*(2^(1/2)*Heaviside(_t0-1)*exp(-(1/2)*_t0^2)*_t0-2^(1/2)*Heaviside(_t0)*exp(-(1/2)*_t0^2)*_t0-2^(1/2)*Heaviside(-_t0)*Heaviside(1-_t0)*exp(-(1/2)*_t0^2)-Pi^(1/2)*undefined*erf((1/2)*_t0*2^(1/2))*Dirac(_t0)-Pi^(1/2)*undefined*erf((1/2)*_t0*2^(1/2))*Dirac(_t0-1)+2^(1/2)*Heaviside(-_t0)*exp(-(1/2)*_t0^2)+2^(1/2)*Heaviside(1-_t0)*exp(-(1/2)*_t0^2)-Pi^(1/2)*undefined*Dirac(_t0)-Pi^(1/2)*undefined*Dirac(_t0-1)+Pi^(1/2)*Heaviside(_t0-1)*erf((1/2)*_t0*2^(1/2))-Pi^(1/2)*Heaviside(_t0)*erf((1/2)*_t0*2^(1/2))-exp(-(1/2)*_t0^2)*2^(1/2)+Pi^(1/2)*Heaviside(_t0-1)-Pi^(1/2)*Heaviside(_t0))/Pi^(1/2), _t0 = -infinity .. t)```

whereas Mma 11 produces the correct piecewise expression (see that here screen15.11.16.docx).

`Edit. Mma output.`

## Determine the distribution function on resulting o...

Hello! Prompt please as in Maple can determine the distribution function of the resulting histogram distribution? I know about cdf function, but how to act in relation to the histogram do not know.

Histogram:

```restart;
with(stats);
with(stats[statplots]);
data2 := [30, 30.5, 31, 31.5, 32, 32.5, 32.6, 33, 33.1, 33.3, 33.6, 34, 35, 36];
histogram(data2, area = count);```

In other words, I need smoothing the histogram, get graph and get on it to obtain an analytical expression.

## What is the expected behaviour of cos(RandomVariab...

Hi all,

The following command:

limit(CDF(cos(RandomVariable(Normal(0,1))),x),x=infinity);

returns erf(Pi/sqrt(2)) instead of 1, as expected for a CDF.

Does someone has an explanation for this strange behaviour?

Best regards,

Régis

## Cumulative Distribution Error help...

I'm trying to write the cumulative distribution function into maple but I seem to get an error. The code is:

>with(Statistics):

>N:=RandomVariable(Normal(0,1)):

>CDF(N,t,inert=true);

I get an error message "Error, (in Statistics:-CDF) unexpected parameters: inert = true".

It used to work, maybe because I was using a newer version of Maple.

## Computable Document Format - wasn't Maple first?

If I'm not mistaken Maple already had technical interactive documents.  And Mathematica introduced cdf only in mid 2011, however why is there so much hype about cdf and media coverage about it replacing pdf's.  Of course it differs by one letter and probably done so by design.  And if pdf is so popular, then so might cdf be in the future.  Pretty slick and trick marketing by Wolfram if you ask me.

Also Maple released the maple Player in...

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