Hoy en día, cualquier dispositivo con Android (teléfono móvil o tablet) tiene distintos sensores, así como GPS. Estas cualidades le hacen ser versátiles en cuanto a la toma de datos (ruido, vibración, luminiscencia...) y su conexión con la posición GPS.
El problema reside en la inexactitud del GPS, pudiendo tener errores de varios metros. De la necesidad de reducir estos errores y de la posibilidad de reducirlos es el argumento de éste y los siguientes artículos.
Today, any Android device (mobile phone or tablet) has different sensors and GPS. These qualities make it versatile in terms of data collection (noise, vibration, luminescence ...) and its connection with the GPS position.
The problem is the inaccuracy of GPS, can have errors of several meters. The need to reduce these errors and the possibility of reducing them is the argument of this and the following articles.
Today, any Android device (mobile phone or tablet) has different sensors and GPS. These qualities make it versatile in terms of data collection (noise, vibration, luminescence ...) and its connection with the GPS position.
The problem is the inaccuracy of GPS, can have errors of several meters. The need to reduce these errors and the possibility of reducing them is the argument of this and the following articles.
Primera consideración. / First consideration.
No queriendo entrar de lleno en formulaciones matemáticas complejas he preferido ir paso a paso. Ya llegaremos al resultado final. Hoy veremos el problema del arquero y de reiteración con ejemplos de grupos de bolas.
I do not want to start with complex mathematical formulations. I prefer to go step by step. We will reach the final result later. Today we will see the problem of the goalkeeper and reiteration with examples of groups of balls.
El problema del arquero / The problem of the goalkeeper
El enunciado es el siguiente: Supongamos dos arqueros. El primero, campeón olímpico, El segundo, un principiante.
Si nos formulan la siguiente pregunta: Con un intento el primero y 10 intentos el segundo ¿Por cual deberíamos apostar? Seguramente nos decantaríamos por el primero.
Si la pregunta fuera: Con un intento el primero y 10.000 intentos el segundo ¿Por cual deberíamos apostar? Seguramente cambiaríamos la respuesta por el segundo.
Pero... la pregunta difícil sería ¿A partir de cuantos intentos del segundo deberíamos decantarnos por éste?
Y otra pregunta difícil sería: si el error medio del primer arquero al centro de la diana es de 5 cm, dado un error medio X del segundo arquero ¿Cuantos intentos debería hacer el segundo para que el error medio fuera de 5cm?
Son preguntas más difíciles de responder de lo que pudiéramos pensar en un primer momento. Por ese motivo vamos antes a resolver un problema más sencillo con una estadística más discreta.
The statement is as follows: We have two archers. The first, Olympic champion, The second, a beginner.
If you ask the following question: With one try the first and 10 tries the second Why should we bet? Surely we would opt for the first.
If the question were: With one try the first and 10,000 tries the second Why should we bet? Surely we would change the answer for the second.
But ... the difficult question would be: From how many attempts of the second should we opt for this one?
And another difficult question would be: if the average error of the first goalkeeper at the center of the target is 5 cm, given an average error X of the second goalkeeper How many attempts should the second make for the average error to be 5cm?
These are more difficult questions to answer than we might think at first. For this reason we are going to solve a simpler problem with a more discrete statistic.
The statement is as follows: We have two archers. The first, Olympic champion, The second, a beginner.
If you ask the following question: With one try the first and 10 tries the second Why should we bet? Surely we would opt for the first.
If the question were: With one try the first and 10,000 tries the second Why should we bet? Surely we would change the answer for the second.
But ... the difficult question would be: From how many attempts of the second should we opt for this one?
And another difficult question would be: if the average error of the first goalkeeper at the center of the target is 5 cm, given an average error X of the second goalkeeper How many attempts should the second make for the average error to be 5cm?
These are more difficult questions to answer than we might think at first. For this reason we are going to solve a simpler problem with a more discrete statistic.
El problema con grupos de bolas / The problem with groups of balls
Sea un número de bolas pequeño. por ejemplo 3. para identificarlas las numeraremos: {1,2,3}
La primera pregunta será: ¿Cual será la probabilidad de sacar, al menos, un "1"?
Es demasiado sencillo: P = 0.3333333 (1/3) (Una posibilidad entre tres posibles).
With a small number of balls. for example 3. we can identify them with numbers: {1,2,3}
The first question will be: What will be the probability of getting at least a "1"?
It's too simple: P = 0.3333333 (1/3) (A possibility among three possibilities)
With a small number of balls. for example 3. we can identify them with numbers: {1,2,3}
The first question will be: What will be the probability of getting at least a "1"?
It's too simple: P = 0.3333333 (1/3) (A possibility among three possibilities)
Compliquemos el problema, Si realizamos el experimento dos veces ¿cual será la probabilidad de obtener, al menos un "1"? (Poco a poco el problema se irá pareciendo al del arquero con el numero de intentos)
Será más fácil ver la respuesta si vemos todo el conjunto de posibilidades:
Let's complicate the problem. If we perform the experiment twice, what will be the probability of obtaining, at least a "1"? (The problem will be similar to that of the archer with the number of attempts)
It will be easier to see the answer if we see the whole set of possibilities.
Let's complicate the problem. If we perform the experiment twice, what will be the probability of obtaining, at least a "1"? (The problem will be similar to that of the archer with the number of attempts)
It will be easier to see the answer if we see the whole set of possibilities.
5 posiblidades entre 9 = 2/9= 0.55555555
5 possibilities and 9 options = 2/9 = 0.55555555
5 possibilities and 9 options = 2/9 = 0.55555555
Si realizamos el experimento tres veces ¿cual será la probabilidad de obtener, al menos un "1"?
Veamos de nuevo el conjunto de posibilidades:
If we perform the experiment three times, what will be the probability of obtaining, at least a "1"?
Let's look again at the set of possibilities:
Let's look again at the set of possibilities:
19 posiblidades entre 27 = 19/27= 0,7037037
19 possibilities and 27 options = 19/27 = 0.7037037
19 possibilities and 27 options = 19/27 = 0.7037037
Ahora podemos encontrar la clave. Para el numerador:
Now we can find the key. For the numerator:
Now we can find the key. For the numerator:
Y, para el denominador, sencillamente, potencias de 3
And, for the denominator, simply, powers of 3
And, for the denominator, simply, powers of 3
Así, de P1 a P8: So, from P1 to P8:
P1 = 0,333333333
P2 = 0,555555556
P3 = 0,703703704
P4 = 0,802469136
P5 = 0,868312757
P6 = 0,912208505
P7 = 0,941472337
P8 = 0,960981558
En este punto podemos hacer un poco de magia, aplicando teoría de conjuntos La fórmula de unión de dos sucesos sería:
At this point we can do some magic, applying set theory. The joining formula of two events would be:
At this point we can do some magic, applying set theory. The joining formula of two events would be:
P(A∪B) = P(A) + P(B) - P(A∩B)
P2 = P1+P1 - P1·P1 = 2P1 - P12 = 2·0,33333-0,333332 = 0,555555
P4 = 2P2 - P22 = 2·0,55555-0,555552 = 0,802469136
P8 = 2P4 - P42 = 2·0,812469-0,8124692 = 0,960981558
Pm = 2P√m - P√m2
Esto nos permite encontrar la relación entre el comjunto de "m" intentos uno anterior "√m"
This allows us to find the relation between the set of "m" attempts one previous "√m"
This allows us to find the relation between the set of "m" attempts one previous "√m"
Más difícil será encontrar la función directa:
More difficult will be to find the direct function:
More difficult will be to find the direct function:
Pn =1-(1-1/3)n
ó
Pn =1-(1-P1)n
ó la que a mí, estéticamente, más me gusta
or the one that, aesthetically, I like the most
or the one that, aesthetically, I like the most
1- Pn = (1-P1)n
Ésta última ecuación resulta bastante lógica si en nuestro problema en lugar de enunciarlo como ¿Cual es la probabilidad del suceso tal en un intento, o en dos,... o en "n"? lo enunciamos como ¿Cual es la probabilidad del no-suceso en un intento y en dos y, ... en "n"?
Esta forma de pensar no por el suceso Pn , si no por su opuesto On (= 1-Pn) la repetiremos en el futuro. (con el suceso opuesto la fórmula quedaría: On = O1n).
This last equation is quite logical if our problem instead of stating it: What is the probability of the event such in an attempt, or in two, ... or in "n"? we enunciate it as What is the probability of non-happening in one attempt and in two and, ... in "n"?
This way of thinking not by the event Pn, but by its opposite On (= 1-Pn) we will repeat it in the future. (with the opposite event the formula would be: On = O1n).
Volviendo al arquero / Going back to the goalkeeper
Supongamos que el arquero olímpico tiene una probabilidad del 95% de hacer diana y el segundo arquero tiene una probabilidad del 5%
Suppose that the Olympic goalkeeper has a 95% chance of making a target and the second goalkeeper has a 5% chance
Suppose that the Olympic goalkeeper has a 95% chance of making a target and the second goalkeeper has a 5% chance
Podremos apostar que a 59 intentos o más empieza a tener más posibilidades de diana el segundo arquero.
We can bet that in 59 attempts or more starts to be more likely to target the second goalkeeper.
También podríamos resolverlo analíticamente:
We could also solve it analytically:
We could also solve it analytically:
0.95 = 1-(1-0.05)n
1-0.95=(1-0.05)n
log(0.05)=n log(0.95)
n=log(0.05)/log(0.95)=58.4039
Primera conclusión: / First conclusion:
La primera conclusión que podemos tener es que por mucho error que tengamos en una medición. Si ésta se toma muchas veces, este error podría minimizarse hasta donde queramos.
En próximas entregas veremos como encontrarle utilidad a ésto. Sobre todo, con el GPS. y también veremos que otros condicionantes tendrán que cumplirse.
The first conclusion we can have is that no matter how much error we have in a measurement. If this is taken many times, this error could be minimized as far as we want.
In future installments we will see how to find utility to this. Above all, with GPS. and we will also see that other conditions will have to be fulfilled.
Continúa en: / Continue on:
https://carreteras-laser-escaner.blogspot.com/2017/09/estadistica-cartografica-el-centrado.html
Véase también: / See also:
https://carreteras-laser-escaner.blogspot.com/2017/11/toma-de-datos-utilizando-la-estadistica.html
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