New answers tagged probability-or-statistics
0
If a problem can easily be solved analytically, you should not use the PC. You learn more from the analytical calculation.
Given height: h and the acceleration: g, the distance after time t is:
s[t_]= 1/2 g t^2
That means the maximal falling time is:
tmax= Sqrt[2h/g]
If we now calculate the mean of s[t] over t:
Mean s = 1/tmax Integrate[1/2 g t^2,{t,0,tmax}...
4
These are all quite fancy solutions so far. For beginners' sake, I'd like to add a somewhat more elementary one that uses somewhat more universal tools:
ages = {14, 15, 16, 22, 24, 25};
counts = {1, 1, 3, 2, 2, 5};
ages.counts/Total[counts]
21
The . (a.k.a. Dot) computes this sum
Sum[ages[[i]] counts[[i]], {i, 1, Length[ages]}]
and Total computes this ...
10
values = {14, 15, 16, 22, 24, 25};
weights = {1, 1, 3, 2, 2, 5};
You can also use WeightedData:
Mean @ WeightedData[values, weights]
21
or use Mean with EmpiricalDistribution:
Mean @ EmpiricalDistribution[weights -> values]
21
Both methods also work with symbolic input:
values = Array[Subscript[x, #] &, 5];
weights = Array[Subscript[w, #] &, ...
3
Using your formula
Clear["Global`*"]
ages = {14, 15, 16, 22, 24, 25};
counts = {1, 1, 3, 2, 2, 5};
Evaluate[n /@ ages] = counts;
?n
mean = Sum[j*n[j], {j, ages}]/Total[counts]
(* 21 *)
12
You need EmpiricalDistribution:
data = {14, 15, 16, 22, 24, 25};
weights = {1, 1, 3, 2, 2, 5};
\[ScriptCapitalD] = EmpiricalDistribution[weights -> data];
Expectation[x, x \[Distributed] \[ScriptCapitalD]]
(* 21 *)
3
You mainly need TransformedDistribution:
xrule = DSolve[{x'[t] == g t, x[0] == 0}, x, t][[1]]
(* {x -> Function[{t}, (g t^2)/2]} *)
grule = Solve[x[T] == h /. xrule, g][[1]]
(* {g -> (2 h)/T^2} *)
dist = TransformedDistribution[x[t] /. xrule,
t \[Distributed] UniformDistribution[{0, T}]];
Assuming[{T > 0, h > 0},...
3
If you want to see "how normal" your data appears to be, you first need to decide on a metric (or metrics) that characterize the departures from normality that are important to you. You also really need to know what values of those metrics imply "non-normality".
I'd argue that many folks don't know they need to consider both the kind of ...
1
I never played roulette, so check my assumptions: There are 37 wholes, 18 red 18 black, 1 green. If I set on red and the ball stop on red, I get twice my money back, otherwise I lose all. With this assumptions:
We mark "win" by 1 and loose by "0". Then playing n times the numbers of "wins" is:
Count[RandomChoice[{18, 19} -> {...
3
The answer turns out to be quite simple at the end:
(5 (956877309536 + 243 Sqrt[2] (-3217542976 + 1024176615 ArcCos[5983/6561])))/2654208
0
TransformedDistribution[Max[E^(x-1)-κ,0],x\[Distributed] NormalDistribution[μ,σ]]//PDF[#,y]&
1/2 DiracDelta[y+κ] (1+Erf[(1-μ+Log[y+κ])/(Sqrt[2] σ)]) (-1+UnitStep[-y]) (UnitStep[-y] (-1+UnitStep[-κ])-UnitStep[-κ])-DiracDelta[y] (1+Erf[(1-μ+Log[κ])/(Sqrt[2] σ)]) (-1+UnitStep[-κ])-(E^(-((1-μ+Log[y+κ])^2/(2 σ^2))) (-1+UnitStep[-y]) (1+UnitStep[-y] (-1+...
3
Directly translating the Wikipedia article linked in the comments:
poissonRandomNumber[λ_] :=
Module[{L = Exp[-λ], k = 0, p = 1, u},
While[p > L,
++k;
p = p*RandomReal[];
];
Return[k - 1]]
Histogram[Table[poissonRandomNumber[10], 20000]]
3
Just to summarize, @cvgmt has shown how to find all the possible outcomes using 3 dice with 6 sides on each die.
ndice = 3;
nsides=6;
outcomes = Tuples[Range[nsides], {ndice}];
@Anton has shown how to find those outcomes that have a given total.
We can use GatherBy to group all of the outcomes according to their totals. Then we count the number of outcomes ...
2
t = Input["Give Sum"];
Select[Tuples[Range[6], 3], Total[#] == t &]
3
nCows = 3
nCalves = RandomChoice[Range[6], nCows]
Counts[nCalves]
Edit:
This alternative answer is in response to a comment that reframed the question, so the two answers resond to different understandings of the question.
Catenate[Permutations /@ Select[IntegerPartitions[6], 3 == Length@# &]]
1
To reiterate (this was not explicit in the original of the question posed), the 2D constrained integration problem I am seeking to solve takes the form
Integrate[9081072000 (x - y)^2 (1 - 3 x + 3 y - 4 Sqrt[y - 2 x y])^2 (1 - 3 x +y - 2 Sqrt[y - 2 x y])^2 (2 y - 2 Sqrt[y - 2 x y])^2 (-1 + 2 x + 2 Sqrt[y - 2 x y])^2 (x - 3 y + 2 Sqrt[y - 2 x y])^2 Boole[x >...
1
You can use Manipulate, PopupMenu and InputField as follows:
functions = {Total, Min, Max, MinMax, Mean, Median, Commonest, Quartiles};
flabels = {"Suma Datos", "Dato Minimo", "Dato Maximo", "Rango", "Media Aritmetica",
"Mediana", "Moda", "Cuartiles"};
fl = Thread[functions -...
3
The probability distribution for count is
dist = ParameterMixtureDistribution[
BinomialDistribution[10, p],
p \[Distributed] BetaDistribution[1, 1]]
(* BetaBinomialDistribution[1, 1, 10] *)
The expected value of count is the Mean of dist
Mean[dist]
(* 5 *)
Probability[count > 1/2, count \[Distributed] dist]
(* 10/11 *)
5
Generating tuples and then machinating over them will get terribly inefficient very quickly (for example, try 10D20 with your code: you'll blow memory before getting anywhere).
Much easier, and efficient, to appeal directly to a probabilistic result.
Here's a function that given the number of dice and number of faces produces a grid with sums and various ...
2
The OP edit answer does what I need pretty snappily.
For reference, the function and its usage is
fn = With[{m = Sort@#, r = Range@Length@#},
Tr[m^Reverse@r/((Last@r - r + 1) (Last@r - r + 2) Append[
Reverse[FoldList[Times, Reverse[Rest@m]]], 1])]] &;
fn@maxlist
where $maxlist$ is a list of the endpoints (maximums) of your uniform ...
3
I'm not sure exactly what you're looking for. Anyway, here's an approximation to the mean of the order statistic that takes a second or two to compute on my laptop:
Sort[#][[n]] & /@ RandomVariate[dist, 10^6] // Mean
Of course you could take fewer or more draws depending on the accuracy you want.
3
Without filtering:
Total[Times @@@ Map[If[# == 6, 1/2, 1/10] &,
Flatten[Permutations /@
IntegerPartitions[18, {4}, Range[6]], 1], {2}]]
12/125
answered Sep 20 at 14:57
5
pd = EmpiricalDistribution[({1, 1, 1, 1, 1, 5}/10) -> Range[6]];
Probability[Total[Array[x, 4]] == 18, Thread[Array[x, 4] \[Distributed] pd]]
(* result 12/125 *)
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