(* Content-type: application/vnd.wolfram.cdf.text *) (*** Wolfram CDF File ***) (* http://www.wolfram.com/cdf *) (* CreatedBy='Mathematica 9.0' *) (*************************************************************************) (* *) (* The Mathematica License under which this file was created prohibits *) (* restricting third parties in receipt of this file from republishing *) (* or redistributing it by any means, including but not limited to *) (* rights management or terms of use, without the express consent of *) (* Wolfram Research, Inc. For additional information concerning CDF *) (* licensing and redistribution see: *) (* *) (* www.wolfram.com/cdf/adopting-cdf/licensing-options.html *) (* *) (*************************************************************************) (*CacheID: 234*) (* Internal cache information: NotebookFileLineBreakTest NotebookFileLineBreakTest NotebookDataPosition[ 1063, 20] NotebookDataLength[ 259828, 7427] NotebookOptionsPosition[ 240661, 6810] NotebookOutlinePosition[ 243744, 6907] CellTagsIndexPosition[ 243442, 6897] WindowFrame->Normal*) (* Beginning of Notebook Content *) Notebook[{ Cell[CellGroupData[{ Cell["\<\ Distribuciones continuas de probabilidad. Aplicaci\[OAcute]n al calculo de \ intervalos de confianza en poblaciones normales\ \>", "Title", Evaluatable->False, FontFamily->"Arial"], Cell[TextData[{ "por ", StyleBox["Guillermo S\[AAcute]nchez", FontSlant->"Italic"], " " }], "Subtitle"], Cell[TextData[ButtonBox["http://diarium.usal.es/guillermo", BaseStyle->"Hyperlink", ButtonData->{ URL["http://diarium.usal.es/guillermo"], None}, ButtonNote->"http://diarium.usal.es/guillermo"]], "Subsubtitle"], Cell[TextData[{ "Universidad de Salamanca\n", ButtonBox["http://www.usal.es", BaseStyle->"Hyperlink", ButtonData:>{"www.usal.es", None}] }], "Subsubtitle"], Cell[TextData[{ "Actualizado : Corresponde a un tutorial desarrollado por el autor en el a\ \[NTilde]o 2000. Se ha vuelto a ejecutar con la versi\[OAcute]n 8 y 9 de ", StyleBox["Mathematica", FontSlant->"Italic"], " sin practicamente modificar el contenido." }], "Text", FontSize->12], Cell[CellGroupData[{ Cell["Introducci\[OAcute]n", "Section", FontFamily->"Arial"], Cell["\<\ En esta pr\[AAcute]ctica se describen y dan ejemplos de las distribuciones \ continuas de mayor empleo en Control Estadistico de Calidad (CEC). Asimismo \ se muestra el uso de estas para la determinaci\[OAcute]n del intervalo de \ confianza. \ \>", "Text", FontFamily->"Arial"], Cell[TextData[{ "Recordemos que una", StyleBox[" distribuci\[OAcute]n de probabilidad", FontWeight->"Bold"], " es un modelo matem\[AAcute]tico que relaciona el valor de la variable con \ la probabilidad de ocurrencia de dicho valor en la poblaci\[OAcute]n \ considerada P{x =", Cell[BoxData[ RowBox[{" ", "x"}]]], "} = p(", Cell[BoxData["x"]], "). Cuando el parametro toma forma de funci\[OAcute]n continua en un \ determinado intervalo estamos ante una ", StyleBox["Distribuci\[OAcute]n Continua.", FontSlant->"Italic"] }], "Text", Evaluatable->False, FontFamily->"Arial"], Cell[CellGroupData[{ Cell[TextData[{ "Ejemplo 1.- Sea una variable aleatoria X el contenido real de una garrafa \ de aceite, en Litros. La distribuci\[OAcute]n de probabilidad se asume que es \ de la forma: ", StyleBox["f", FontSlant->"Italic"], "(x)=1/1.5 en el intervalo 15.5\[LessEqual]x\[LessEqual]17.0 (es decir es \ continua en dicho intervalo). Esta distribuci\[OAcute]n es llamada ", StyleBox["distribuci\[OAcute]n uniforme. ", FontSlant->"Italic"], "La probabilidad de que el contenido de una deterenada garrafa sea menor o \ igual a 16.0 L se calcula sencillamente como:" }], "Subsubsection", Evaluatable->False, FontFamily->"Arial", FontSize->10], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"\n", RowBox[{ SubsuperscriptBox["\[Integral]", "15.5", "16.0"], RowBox[{ FractionBox["1", "1.5"], RowBox[{"\[DifferentialD]", "x"}]}]}]}]], "Input"], Cell[BoxData["0.3333333333333333`"], "Output"] }, Open ]], Cell[TextData[{ "Representa graficamente la distribuci\[OAcute]n anterior, a calcular su \ media y varianza [ Recuerda que la media porresponde a la esperanza \ matematica E[X] = ", Cell[BoxData[ RowBox[{ SubsuperscriptBox["\[Integral]", "a", "b"], RowBox[{"x", " ", "p", RowBox[{"(", "x", ")"}], RowBox[{"\[DifferentialD]", "x"}]}]}]]], " y que ", Cell[BoxData[ RowBox[{ RowBox[{"V", "[", "X", "]"}], " ", "=", " ", RowBox[{ RowBox[{"E", "[", SuperscriptBox["X", "2"], "]"}], " ", "-", " ", SuperscriptBox[ RowBox[{"(", RowBox[{"E", "[", "X", "]"}], ")"}], "2"]}]}]]], "]." }], "Text"], Cell["\<\ Las distribuci\[OAcute]n continuas mas utilizadas en CEC son la Normal, \ Exponencial, T-Student, Gamma y Weibull.\ \>", "Text"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[" Distribuciones continuas", "Section", Evaluatable->False, FontFamily->"Arial", FontSize->12, FontWeight->"Bold"], Cell["\<\ El Mathematica dispone probablemente mas funciones de distribuci\[OAcute]n \ incorporada que ningun otro programa:\ \>", "Text", Evaluatable->False, FontFamily->"Arial", FontSize->10, FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Outline"->False, "Shadow"->False, "Underline"->False}], Cell[TextData[{ ButtonBox["Continuous Distributions", BaseStyle->"Link", ButtonData->"paclet:tutorial/ContinuousDistributions"], ": tutorial/ContinuousDistributions" }], "Text", CellID->537095617], Cell[CellGroupData[{ Cell["Distribucion Normal", "Subsection"], Cell[TextData[{ "Es probablemente la distribuci\[OAcute]n mas importante. Entre otras muchas \ propiedades, en muestreo estadistico se utiliza la siguiente: \nTomados n \ valores ", Cell[BoxData[ SubscriptBox["x", "i"]]], " de una poblacion no necesariamente normal la media muestral ", Cell[BoxData[ OverscriptBox["x", "_"]]], " sigue aproximadamente una distribuci\[OAcute]n de media \[Mu] y varianza ", Cell[BoxData[ FractionBox[ SuperscriptBox["\[Sigma]", "2"], "n"]]], ", es decir ", Cell[BoxData[ OverscriptBox["x", "_"]]], " \[Tilde] N(\[Mu], ", Cell[BoxData[ FractionBox[ SuperscriptBox["\[Sigma]", "2"], "n"]]], ")" }], "Text"], Cell["\<\ Para obtener obtener informaci\[OAcute]n de ella podemos utilizar la ayuda en \ la forma habitual . Por ejemplo\ \>", "SmallText"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"?", "NormalDistribution*"}]], "Input"], Cell[BoxData[ RowBox[{ StyleBox["\<\"\!\(\*RowBox[{\\\"NormalDistribution\\\", \\\"[\\\", \ RowBox[{StyleBox[\\\"\[Mu]\\\", \\\"TR\\\"], \\\",\\\", StyleBox[\\\"\[Sigma]\ \\\", \\\"TR\\\"]}], \\\"]\\\"}]\) represents a normal (Gaussian) \ distribution with mean \!\(\*StyleBox[\\\"\[Mu]\\\", \\\"TR\\\"]\) and \ standard deviation \!\(\*StyleBox[\\\"\[Sigma]\\\", \ \\\"TR\\\"]\).\\n\!\(\*RowBox[{\\\"NormalDistribution\\\", \\\"[\\\", \\\"]\\\ \"}]\) represents a normal distribution with zero mean and unit standard \ deviation.\"\>", "MSG"], "\[NonBreakingSpace]", ButtonBox[ StyleBox["\[RightSkeleton]", "SR"], Active->True, BaseStyle->"Link", ButtonData->"paclet:ref/NormalDistribution"]}]], "Print", "PrintUsage", CellTags->"Info3589381481-7242388"] }, Open ]], Cell["\<\ Ojo: Observese que aplica el convenio N(media, desviacion estandar) en vez \ del mas usual de N(Media, varianza), por eso cuando se emplee algun programa \ de calculo por primera vez es conveniente consultar el criterio que aplica en \ algunos terminos\ \>", "SmallText"], Cell[CellGroupData[{ Cell[BoxData[{ RowBox[{"PDF", "[", RowBox[{ RowBox[{"NormalDistribution", "[", RowBox[{"m", ",", "\[Sigma]"}], "]"}], ",", "x"}], "]"}], "\n", RowBox[{"CDF", "[", RowBox[{ RowBox[{"NormalDistribution", "[", RowBox[{"m", ",", "\[Sigma]"}], "]"}], ",", "x"}], "]"}], "\n", RowBox[{"Mean", "[", RowBox[{"NormalDistribution", "[", RowBox[{"m", ",", "\[Sigma]"}], "]"}], "]"}], "\n", RowBox[{"Variance", "[", RowBox[{"NormalDistribution", "[", RowBox[{"m", ",", "\[Sigma]"}], "]"}], "]"}]}], "Input"], Cell[BoxData[ FractionBox[ SuperscriptBox["\[ExponentialE]", RowBox[{"-", FractionBox[ SuperscriptBox[ RowBox[{"(", RowBox[{ RowBox[{"-", "m"}], "+", "x"}], ")"}], "2"], RowBox[{"2", " ", SuperscriptBox["\[Sigma]", "2"]}]]}]], RowBox[{ SqrtBox[ RowBox[{"2", " ", "\[Pi]"}]], " ", "\[Sigma]"}]]], "Output"], Cell[BoxData[ RowBox[{ FractionBox["1", "2"], " ", RowBox[{"Erfc", "[", FractionBox[ RowBox[{"m", "-", "x"}], RowBox[{ SqrtBox["2"], " ", "\[Sigma]"}]], "]"}]}]], "Output"], Cell[BoxData["m"], "Output"], Cell[BoxData[ SuperscriptBox["\[Sigma]", "2"]], "Output"] }, Open ]], Cell[CellGroupData[{ Cell["\<\ Ejemplo 2.-Calcular para una distribucion Normal -de media 7 y desviaci\ \[OAcute]n standard 2 -, a) El valor de la funci\[OAcute]n densidad y de la \ funci\[OAcute]n de distribuci\[OAcute]n para x=5, b) asimetria y c) \ curtosis. La forma de utilizar estas funciones es similar a la de las \ distribuciones discretas, se proceder\[AAcute] como sigue:\ \>", "Subsubsection", Evaluatable->False, FontFamily->"Arial", FontSize->10], Cell["\<\ Nota: Para calcular la densidad de probabilidad y la Distribucion (acumulada) \ de probabilidad se utiliza, respectivamente PDF y CDF, cuyo significado es \ facil obtener directamente de la ayuda\ \>", "SmallText", Evaluatable->False], Cell[CellGroupData[{ Cell[BoxData[{ RowBox[{ RowBox[{"PDF", "[", RowBox[{ RowBox[{"NormalDistribution", "[", RowBox[{"3", ",", "2"}], "]"}], ",", "5"}], "]"}], "//", "N"}], "\n", RowBox[{ RowBox[{"CDF", "[", RowBox[{ RowBox[{"NormalDistribution", "[", RowBox[{"3", ",", "2"}], "]"}], ",", "5"}], "]"}], "//", "N"}], "\n", RowBox[{"Skewness", "[", RowBox[{"NormalDistribution", "[", RowBox[{"3", ",", "2"}], "]"}], "]"}], "\n", RowBox[{"Kurtosis", "[", RowBox[{"NormalDistribution", "[", RowBox[{"3", ",", "2"}], "]"}], "]"}]}], "Input", 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3.- La tasa de fallo de una lampara electrica es de ", Cell[BoxData[ SuperscriptBox["10", RowBox[{"-", "4"}]]]], " ", Cell[BoxData[ SuperscriptBox["h", RowBox[{"-", "1"}]]]], " . \[DownQuestion]Cual es la duraci\[OAcute]n media estimada?." }], "Subsubsection", Evaluatable->False, FontFamily->"Arial", FontSize->10], Cell[TextData[{ "\nEn este caso \[Lambda]=", Cell[BoxData[ SuperscriptBox["10", RowBox[{"-", "4"}]]]], " y por tanto" }], "SmallText", Evaluatable->False], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{"Mean", "[", RowBox[{"ExponentialDistribution", "[", SuperscriptBox["10", RowBox[{"-", "4"}]], "]"}], "]"}], " ", RowBox[{"(*", RowBox[{"en", " ", "horas"}], "*)"}]}]], "Input"], Cell[BoxData["10000"], "Output"] }, Open ]], Cell["\<\ La probabilicad de fallo en func\[OAcute]n del tiempo , para un periodo \ [0,T], con T =100000, podemos representarla como sigue: \ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Plot", "[", RowBox[{ RowBox[{"CDF", "[", RowBox[{ 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La tasa de fallo de cada equipo esta dado por una \ distribuci\[OAcute]n exponencial con una tasa de fallo de ", Cell[BoxData[ SuperscriptBox["10", RowBox[{"-", "4"}]]]], " ", Cell[BoxData[ SuperscriptBox["h", RowBox[{"-", "1"}]]]], " . El conjunto ser\[AAcute] una distribuci\[OAcute]n gamma con \ \[Alpha]=2, y \[Beta]=1/\[Lambda]=", Cell[BoxData[ SuperscriptBox["10", "4"]]], ". El tiempo medio de fallo esperado para el conjunto es:" }], "Subsubsection", Evaluatable->False, FontFamily->"Arial", FontSize->10], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Mean", "[", RowBox[{"GammaDistribution", "[", RowBox[{"2", ",", " ", SuperscriptBox["10", "4"]}], "]"}], "]"}]], "Input"], Cell[BoxData["20000"], "Output"] }, Open ]] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Distribucion Weibull", "Subsection"], Cell["\<\ Esta funci\[OAcute]n es muy versatil pues una apropiada elecci\[OAcute]n de \ los parametros alfa y beta permite ajustar datos experimentales a esta funci\ \[OAcute]n.\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Information", "[", RowBox[{"\"\\"", ",", RowBox[{"LongForm", "\[Rule]", "False"}]}], "]"}]], "Input", PageWidth->Infinity], Cell[BoxData[ RowBox[{ StyleBox["\<\"\!\(\*RowBox[{\\\"WeibullDistribution\\\", \\\"[\\\", \ RowBox[{StyleBox[\\\"\[Alpha]\\\", \\\"TR\\\"], \\\",\\\", StyleBox[\\\"\ \[Beta]\\\", \\\"TR\\\"]}], 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Ejemplo: el valor de z para una normal N(0,1) por debajo del cual se \ encuentra el 95% de la poblaci\[OAcute]n es z(0.95) = 1.64." }], "Text"], Cell[TextData[{ Cell[BoxData[ FormBox[ RowBox[{"t", "(", RowBox[{"m", " ", ";", " ", "\[Alpha]"}], ")"}], TraditionalForm]]], " = El percentil \[Alpha] de la ", Cell[BoxData[ FormBox["t", TraditionalForm]]], " Student con ", Cell[BoxData[ FormBox["m", TraditionalForm]]], " grados de libertad. Ejemplo: el valor de ", Cell[BoxData[ FormBox["t", TraditionalForm]]], " para \[Alpha] = 95% y ", Cell[BoxData[ FormBox["m", TraditionalForm]]], " = 5 es ", Cell[BoxData[ FormBox[ RowBox[{"t", "(", RowBox[{"5", " ", ";", " ", "0.95"}], ")"}], TraditionalForm]]], " = 2.01." }], "Text"], Cell[TextData[{ Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"t", "'"}], RowBox[{"(", RowBox[{"m", " ", ",", " ", RowBox[{"\[Delta]", ";", " ", "\[Alpha]"}]}], ")"}]}], TraditionalForm]]], " = El percentil \[Alpha] de la ", Cell[BoxData[ FormBox["t", TraditionalForm]]], " Student no central con ", Cell[BoxData[ FormBox["m", TraditionalForm]]], " grados de libertad y un parametro de descentralizacion \[Delta] . Su \ interpretaci\[OAcute]n es mas compleja que en los casos anteriores. " }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["T-Student No central", "Subsection"], Cell["\<\ Vamos a necesitar la T-Student No central. La informaci\[OAcute]n disponible \ en al ayuda es la siguiente:\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"?", StyleBox["NoncentralStudentTDistribution", "MR"]}]], "Input"], Cell[BoxData[ RowBox[{ StyleBox["\<\"\!\(\*RowBox[{\\\"NoncentralStudentTDistribution\\\", \\\"[\\\ \", RowBox[{StyleBox[\\\"\[Nu]\\\", \\\"TR\\\"], \\\",\\\", StyleBox[\\\"\ \[Delta]\\\", \\\"TR\\\"]}], \\\"]\\\"}]\) represents a noncentral Student \!\ \(\*StyleBox[\\\"t\\\", \\\"TI\\\"]\) distribution with \!\(\*StyleBox[\\\"\ \[Nu]\\\", \\\"TR\\\"]\) degrees of freedom and noncentrality parameter \ \!\(\*StyleBox[\\\"\[Delta]\\\", \\\"TR\\\"]\).\"\>", "MSG"], "\[NonBreakingSpace]", ButtonBox[ StyleBox["\[RightSkeleton]", "SR"], Active->True, BaseStyle->"Link", ButtonData->"paclet:ref/NoncentralStudentTDistribution"]}]], "Print", \ "PrintUsage", CellTags->"Info3589381486-7242388"] }, Open ]], Cell["\<\ La funci\[OAcute]n de densidad de esta distribuci\[OAcute]n es\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"PDF", "[", RowBox[{ RowBox[{ StyleBox["NoncentralStudentTDistribution", "MR"], StyleBox["[", "MR"], RowBox[{ StyleBox["n", "MR"], StyleBox[",", "MR"], "\[Delta]"}], "]"}], ",", "x"}], "]"}]], "Input"], Cell[BoxData[ FractionBox[ RowBox[{ SuperscriptBox["2", "n"], " ", SuperscriptBox["\[ExponentialE]", RowBox[{"-", FractionBox[ SuperscriptBox["\[Delta]", "2"], "2"]}]], " ", SuperscriptBox["n", RowBox[{"1", "+", FractionBox["n", "2"]}]], " ", SuperscriptBox[ RowBox[{"(", RowBox[{"n", "+", SuperscriptBox["x", "2"]}], ")"}], RowBox[{ FractionBox["1", "2"], " ", RowBox[{"(", RowBox[{ RowBox[{"-", "1"}], "-", "n"}], ")"}]}]], " ", RowBox[{"Gamma", "[", FractionBox[ RowBox[{"1", "+", "n"}], "2"], "]"}], " ", RowBox[{"HermiteH", "[", RowBox[{ RowBox[{ RowBox[{"-", "1"}], "-", "n"}], ",", RowBox[{"-", FractionBox[ RowBox[{"x", " ", "\[Delta]"}], RowBox[{ SqrtBox["2"], " ", SqrtBox[ RowBox[{"n", "+", SuperscriptBox["x", "2"]}]]}]]}]}], "]"}]}], "\[Pi]"]], "Output"] }, Open ]], Cell[TextData[{ "Su representaci\[OAcute]n gr\[AAcute]fica es para ", Cell[BoxData[ FormBox["n", TraditionalForm]]], " = 10, y \[Delta] = 2" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{"Plot", "[", RowBox[{ RowBox[{"PDF", "[", RowBox[{ RowBox[{ StyleBox["NoncentralStudentTDistribution", "MR"], StyleBox["[", "MR"], StyleBox[ RowBox[{"10", ",", "2"}], "MR"], "]"}], ",", "x"}], "]"}], ",", " ", RowBox[{"{", RowBox[{"x", ",", "0", ",", "10"}], "}"}]}], "]"}], " "}]], "Input"], Cell[BoxData[ GraphicsBox[{{}, {}, {Hue[0.67, 0.6, 0.6], LineBox[CompressedData[" 1:eJwVl3VcVE8XxglBpaQ7dlkkJERQ0B/oGRRBBUFaKaUEJCREUqVBQikBKaUU JCWUbkQ694LYpIQgpfS++/51P9/PzJ0z9zzPOXcGb3lX14aKgoKinZKC4v/P 9IeWtCesI88pFw2cV/37BtqGcZpGOHvQ2VCW+4gvANKHhiAczhsqHn3hDbpa AOlN0UI0uMdwXEPIMCO3ABzPejj9E3oOrx4tjTldL4ThHbuEP0L5IMElKx35 sQgOfp4Q+y1UDdotSWca7UthrKneY/FXNcROqxkyBJXCm1fJrQtva8A9fTHp flopaLlp3Zy/UAfSLsllWf2lkHi4JnHWthESIg4YGCu+Bda0j1Qroy3wnGWP 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llamaremos intervalo de confianza para ", Cell[BoxData[ FormBox[ OverscriptBox["y", RowBox[{"_", " "}]], TraditionalForm]]], " con nivel de confianza (1 - \[Alpha]) aquel que tiene una probabilidad \ de (1 - \[Alpha])100% de contener la verdadera media \[Mu]. Los \ l\[IAcute]mites generalmente suelen ser:\na) Bilateral centrado ", Cell[BoxData[ FormBox[ OverscriptBox["y", RowBox[{"_", " "}]], TraditionalForm]]], " que consisten en calcular ", Cell[BoxData[ FormBox[ OverscriptBox["y", RowBox[{"_", " "}]], TraditionalForm]]], "\[PlusMinus] k \[Sigma] tal que la verdadera media est\[AAcute] \ comprendida en el intervalo con una probabilidad 100", Cell[BoxData[ FormBox["p", TraditionalForm]]], "%.\nb) Unilateral superior que consisten en calcular ", Cell[BoxData[ FormBox[ OverscriptBox["y", RowBox[{"_", " "}]], TraditionalForm]]], "+ k \[Sigma] tal que la verdadera media \[Mu] est\[AAcute] por debajo ", Cell[BoxData[ FormBox[ OverscriptBox["y", RowBox[{"_", " "}]], TraditionalForm]]], "+ k \[Sigma] con una probabilidad 100", Cell[BoxData[ FormBox["p", TraditionalForm]]], "%. \nc) Unilateral inferior que consisten en calcular ", Cell[BoxData[ FormBox[ OverscriptBox["y", RowBox[{"_", " "}]], TraditionalForm]]], "- k \[Sigma] tal que la verdadera media \[Mu] est\[AAcute] por encima de ", Cell[BoxData[ FormBox[ OverscriptBox["y", RowBox[{"_", " "}]], TraditionalForm]]], "- k \[Sigma] con una probabilidad 100", Cell[BoxData[ FormBox["p", TraditionalForm]]], "% . " }], "Text"], Cell[TextData[{ "Si tomamos ", Cell[BoxData[ FormBox["n", TraditionalForm]]], " muestras de una poblaci\[OAcute]n, y de cada muestra calculamos la media ", Cell[BoxData[ FormBox[ SubscriptBox["y", "i"], TraditionalForm]]], ", las expresiones espec\[IAcute]ficas para los l\[IAcute]mites de los \ distintos intervalos de ", Cell[BoxData[ FormBox[ OverscriptBox["y", "_"], TraditionalForm]]], " , que denotamos por ", Cell[BoxData[ FormBox[ SubscriptBox["Y", "p"], TraditionalForm]]], " para un nivel de confianza (1 - \[Alpha])100% son las siguientes:" }], "Text"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Varianza conocida", "Subsection"], Cell[TextData[{ "Si conocemos la varianza de la poblaci\[OAcute]n, \[Sigma], por el teorema \ central del l\[IAcute]mite, sabemos que el error relativo en la determinaci\ \[OAcute]n de \[Mu] mediante la media muestral ", Cell[BoxData[ FormBox[ OverscriptBox["y", RowBox[{"_", " "}]], TraditionalForm]]], " es: ", Cell[BoxData[ FormBox["z", TraditionalForm]]], " = (", Cell[BoxData[ FormBox[ OverscriptBox["y", RowBox[{"_", " "}]], TraditionalForm]]], "- \[Mu])/(\[Sigma]/", Cell[BoxData[ FormBox[ SqrtBox["n"], TraditionalForm]]], ") es una variable normal est\[AAcute]ndar de lo que se deduce los \ siguientes l\[IAcute]mites para los intervalos" }], "Text"], Cell[CellGroupData[{ Cell["L\[IAcute]mite unilateral inferior", "Subsubsection"], Cell[BoxData[ RowBox[{ SubscriptBox["Y", "p"], " ", "=", " ", RowBox[{ OverscriptBox["y", "_"], " ", "-", " ", RowBox[{ FractionBox[ FormBox[ RowBox[{" ", RowBox[{"z", "(", RowBox[{"1", "-", "\[Alpha]"}], ")"}]}], TraditionalForm], SqrtBox["n"]], "\[Sigma]"}]}]}]], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["L\[IAcute]mite unilateral superior", "Subsubsection"], Cell[BoxData[ RowBox[{ SubscriptBox["Y", "p"], " ", "=", " ", RowBox[{ OverscriptBox["y", "_"], " ", "+", " ", RowBox[{ FractionBox[ FormBox[ RowBox[{" ", RowBox[{"z", "(", RowBox[{"1", "-", "\[Alpha]"}], ")"}]}], 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utilizarmos la \ distribuci\[OAcute]n ", Cell[BoxData[ FormBox["t", TraditionalForm]]], " que la que se verifica ", Cell[BoxData[ FormBox["t", TraditionalForm]]], " = (", Cell[BoxData[ FormBox[ OverscriptBox["y", RowBox[{"_", " "}]], TraditionalForm]]], "- \[Mu])/(", Cell[BoxData[ FormBox["s", TraditionalForm]]], "/", Cell[BoxData[ FormBox[ SqrtBox["n"], TraditionalForm]]], "), donde ", Cell[BoxData[ FormBox["s", TraditionalForm]]], " es la desviaci\[OAcute]n est\[AAcute]ndar muestral, con la que se obtienne \ los siguientes l\[IAcute]mites para los intervalos" }], "Text"], Cell[CellGroupData[{ Cell["L\[IAcute]mite unilateral inferior", "Subsubsection"], Cell[BoxData[ RowBox[{ SubscriptBox["Y", "p"], " ", "=", " ", RowBox[{ OverscriptBox["y", "_"], " ", "-", " ", RowBox[{ FractionBox[ FormBox[ RowBox[{" ", RowBox[{"t", "(", RowBox[{ RowBox[{"n", "-", "1"}], ",", " ", RowBox[{"1", "-", "\[Alpha]"}]}], ")"}]}], TraditionalForm], SqrtBox["n"]], "s"}]}]}]], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["L\[IAcute]mite unilateral superior", "Subsubsection"], Cell[BoxData[ RowBox[{ SubscriptBox["Y", "p"], " ", "=", " ", RowBox[{ OverscriptBox["y", "_"], " ", "+", " ", RowBox[{ FractionBox[ FormBox[ RowBox[{" ", RowBox[{"t", "(", RowBox[{ RowBox[{"n", "-", "1"}], ",", " ", RowBox[{"1", "-", "\[Alpha]"}]}], ")"}]}], TraditionalForm], SqrtBox["n"]], "s"}]}]}]], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["L\[IAcute]mite bilateral inferior", "Subsubsection"], Cell[BoxData[ RowBox[{ SubscriptBox["Y", "p"], " ", "=", " ", RowBox[{ OverscriptBox["y", "_"], " ", "-", " ", RowBox[{ FractionBox[ FormBox[ RowBox[{" ", RowBox[{"t", "(", RowBox[{ RowBox[{"n", "-", "1"}], ",", " ", RowBox[{"1", "-", RowBox[{"\[Alpha]", "/", "2"}]}]}], ")"}]}], TraditionalForm], SqrtBox["n"]], "s"}]}]}]], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["L\[IAcute]mite bilateral superior", "Subsubsection"], Cell[BoxData[ RowBox[{ SubscriptBox["Y", "p"], " ", "=", " ", RowBox[{ OverscriptBox["y", "_"], " ", "+", " ", RowBox[{ FractionBox[ FormBox[ RowBox[{" ", RowBox[{"t", "(", RowBox[{ RowBox[{"n", "-", "1"}], ",", " ", RowBox[{"1", "-", RowBox[{"\[Alpha]", "/", "2"}]}]}], ")"}]}], TraditionalForm], SqrtBox["n"]], "s"}]}]}]], "Text"] }, Open ]] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Intervalos de tolerancia ", "Section"], Cell["\<\ Lo haremos en una sesi\[OAcute]n nueva por ello salimos de la actual\ \>", "Text"], Cell[BoxData[ RowBox[{"Quit", "[", "]"}]], "Input"], Cell[CellGroupData[{ Cell["Conceptos", "Subsection"], Cell[TextData[{ "Definimos el 100", Cell[BoxData[ FormBox["p", TraditionalForm]]], "-\[EAcute]simo percentil al la valor ", Cell[BoxData[ FormBox[ SubscriptBox["Y", "p"], TraditionalForm]]], " por debajo del cual est\[AAcute] el 100", Cell[BoxData[ FormBox["p", TraditionalForm]]], " % de la poblaci\[OAcute]n, o, lo que es equivalente, aquel valor para el \ que las observaciones de muestra aleatoria de la poblaci\[OAcute]n cae con \ una probabilidad ", Cell[BoxData[ FormBox["p", TraditionalForm]]], ". En ocasiones nos interesa estimar el valor de ", Cell[BoxData[ FormBox[ SubscriptBox["Y", "p"], TraditionalForm]]], " asociandole un intervalo de confianza, (1 - \[Alpha]) 100%. Ejemplo: \ Deseamos calcular el valor por debajo del cual est\[AAcute] el 95% de la \ poblaci\[OAcute]n con un nivel de confianza del 95%. Problemas similares \ consisten en calcular l\[IAcute]mites centrales dentro de los cuales est\ \[AAcute] el 100", Cell[BoxData[ FormBox["p", TraditionalForm]]], "% de la poblaci\[OAcute]n, y l\[IAcute]mites inferiores por encima del cual \ est\[AAcute] 100", Cell[BoxData[ FormBox["p", TraditionalForm]]], "% de la poblaci\[OAcute]n. Asimismo un l\[IAcute]mite superior de confianza \ (1 - \[Alpha]) 100% para el percentil 100", Cell[BoxData[ FormBox["p", TraditionalForm]]], "-\[EAcute]simo es equivalente al l\[IAcute]mite superior de tolerancia por \ debajo del cual est\[AAcute] al menos una proporci\[OAcute]n ", Cell[BoxData[ FormBox["p", TraditionalForm]]], " de la poblaci\[OAcute]n con una probabilidad 1 - \[Alpha]. Una \ interpretaci\[OAcute]n similar cabe del l\[IAcute]mite inferior de tolerancia." }], "Text"], Cell[TextData[{ "Las expresiones espec\[IAcute]ficas para obtener los l\[IAcute]mites de \ tolerancia ", Cell[BoxData[ FormBox[ SubscriptBox["Y", "p"], TraditionalForm]]], " son las siguientes" }], "Text"], Cell[CellGroupData[{ Cell["L\[IAcute]mite unilateral inferior", "Subsubsection"], Cell[BoxData[ RowBox[{ SubscriptBox["Y", "p"], " ", "=", " ", RowBox[{ OverscriptBox["y", "_"], " ", "-", " ", RowBox[{ FractionBox[ FormBox[ RowBox[{ RowBox[{"t", "'"}], RowBox[{"(", RowBox[{ RowBox[{"n", "-", "1"}], ",", RowBox[{ RowBox[{ RowBox[{"-", " ", RowBox[{"z", "(", "p", ")"}]}], SqrtBox[ RowBox[{" ", "n"}]]}], ";", " ", RowBox[{"1", "-", "\[Alpha]"}]}]}], ")"}]}], TraditionalForm], SqrtBox["n"]], "s"}]}]}]], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["L\[IAcute]mite unilateral superior", "Subsubsection"], Cell[BoxData[ RowBox[{ SubscriptBox["Y", "p"], " ", "=", " ", RowBox[{ 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Una aproximaci\[OAcute]n excelente a (13) puede obtenerse para el \ caso bilateral, que es el mas complicado aplicando la siguiente aproximaci\ \[OAcute]n" }], "Text"], Cell[BoxData[ RowBox[{ SubscriptBox["k", RowBox[{"n", ",", "\[Alpha]", ",", "p"}]], "\[TildeTilde]", RowBox[{ RowBox[{"z", "[", RowBox[{ RowBox[{"(", RowBox[{"1", "+", "p"}], ")"}], "/", "2"}], "]"}], " ", SuperscriptBox[ RowBox[{"(", FractionBox[ RowBox[{"(", RowBox[{"n", "-", "1"}], ")"}], RowBox[{ SuperscriptBox["\[Chi]", "2"], RowBox[{"(", RowBox[{ RowBox[{"n", "-", "1"}], ";", "\[Alpha]"}], ")"}]}]], ")"}], RowBox[{"1", "/", "2"}]], RowBox[{"(", RowBox[{"1", "+", FractionBox["1", RowBox[{"2", "n"}]]}], ")"}]}]}]], "Text"], Cell[TextData[{ "Expresado en el lenguaje del ", StyleBox["Mathematica", FontSlant->"Italic"], " es" }], "Text"], Cell[BoxData[ RowBox[{ RowBox[{"g2", "[", RowBox[{"n_", ",", "\[Alpha]_", ",", "p_"}], "]"}], " ", ":=", RowBox[{"Module", "[", RowBox[{ RowBox[{"{", RowBox[{"z", ",", "ji2"}], "}"}], 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" }], "Text"], Cell[TextData[{ "b) La poblaci\[OAcute]n ser\[AAcute] aceptada si se verifica que ", Cell[BoxData[ FormBox[ SubscriptBox["p", "LS"], TraditionalForm]]], "-", Cell[BoxData[ FormBox[ SubscriptBox["p", "LI"], TraditionalForm]]], " =", Cell[BoxData[ FormBox[ RowBox[{" ", RowBox[{ RowBox[{"p", " ", "\[GreaterEqual]", " ", "P"}], ","}]}], TraditionalForm]]], " siendo ", Cell[BoxData[ FormBox["P", TraditionalForm]]], " la proporci\[OAcute]n de la poblaci\[OAcute]n que se considera aceptable. \ " }], "Text"], Cell[TextData[{ "Creamos una sentencia que nos automatice el anterior proceso (para P = 0.95 \ y alfa = 0.05), donde mean es la media de la muestra ", Cell[BoxData[ OverscriptBox["x", "_"]]], ", size su tama\[NTilde]o ", Cell[BoxData[ FormBox["n", TraditionalForm]]], " , uL le LS, uL el LI y sigma la ", Cell[BoxData[ FormBox["s", TraditionalForm]]], " de la meustra" }], "Text"], Cell["\<\ p[mean_, size_, uL_, lL_, sigma_] := Module[{pLS, pLI}, pLS = FindRoot[g1[size, 0.05, pS] == (uL - mean)/sigma, {pS, 0.5, \ 0.99}][[1,2]]; pLI = FindRoot[g1[size, 0.05, 1 - pI] == (mean - lL)/sigma, {pI, 0.01, 0.2}][[1,2]];{pLS, pLI, pLS - pLI >= 0.95}]\ \>", "Input"] }, Open ]], Cell[CellGroupData[{ Cell["Ejemplo", "Subsubsection"], Cell[TextData[{ " Sea ", Cell[BoxData[ RowBox[{ OverscriptBox["x", "_"], "="}]]], "10.1, s = 0.3, LS = 10.8 , LI = 9.3, n = 50, \[DownQuestion]Es rechazable \ la muestra para un alfa = 0.05 y un P = 0.95?.\nLa soluci\[OAcute]n aplicando \ la sentencia anterior es que con dicha muestra podriamos aceptar la poblaci\ \[OAcute]n" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"p", "[", RowBox[{"10.1", ",", " ", "50", ",", "10.8", ",", " ", "9.3", ",", "0.3"}], "]"}]], "Input"], Cell[BoxData[ RowBox[{"{", RowBox[{"0.970028070629612`", ",", "0.015146602794094677`", ",", "True"}], "}"}]], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Aclaraci\[OAcute]n.", "Subsubsection"], Cell["\<\ Este criterio da valores muy pr\[OAcute]ximos al caso bilateral, aunque algo \ mas conservadores como era de esperar. 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Si a cada valor ", Cell[BoxData[ FormBox[ SubscriptBox["X", "i"], TraditionalForm]]], " le restamos la media \[Mu] y lo dividimos entre la varianza \[Sigma] y lo \ elevamos al cuadrado obtendremos (2) que es una distribuci\[OAcute]n de las \ mismas caracteristicas que (1), es decir tendremos una distribuci\[OAcute]n ", Cell[BoxData[ FormBox[ SuperscriptBox[ SubscriptBox["\[Chi]", "n"], "2"], TraditionalForm]]] }], "Text"], Cell[BoxData[ RowBox[{"Y", " ", "=", RowBox[{ RowBox[{ RowBox[{ RowBox[{ SuperscriptBox[ RowBox[{"(", FractionBox[ RowBox[{ SubscriptBox["X", "1"], "-", "\[Mu]"}], "\[Sigma]"], ")"}], "2"], "+", SuperscriptBox[ RowBox[{"(", FractionBox[ RowBox[{ SubscriptBox["X", "2"], "-", "\[Mu]"}], "\[Sigma]"], ")"}], "2"], "+"}], " ", "..."}], " ", "+", SuperscriptBox[ RowBox[{"(", FractionBox[ RowBox[{ SubscriptBox["X", "n"], "-", "\[Mu]"}], "\[Sigma]"], ")"}], "2"]}], "=", RowBox[{ RowBox[{ UnderoverscriptBox["\[Sum]", RowBox[{"i", "=", "1"}], "n"], SuperscriptBox[ RowBox[{"(", 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RowBox[{"n", ",", " ", RowBox[{ RowBox[{"(", RowBox[{"1", "+", "\[Gamma]"}], ")"}], "/", "2"}]}]], TraditionalForm]]], " para lo que proceemos como sigue:" }], "Text"], Cell[BoxData[ RowBox[{"\[Gamma]", " ", "=", RowBox[{ RowBox[{"P", RowBox[{"(", RowBox[{"a", "\[LessEqual]", RowBox[{ FractionBox["1", SuperscriptBox["\[Sigma]", "2"]], RowBox[{ UnderoverscriptBox["\[Sum]", RowBox[{"i", "=", "1"}], "n"], SuperscriptBox[ RowBox[{"(", RowBox[{ SubscriptBox["X", "i"], "-", "\[Mu]"}], ")"}], "2"]}]}], "\[LessEqual]", "b"}], ")"}]}], "=", "\[IndentingNewLine]", RowBox[{"P", RowBox[{"(", RowBox[{ RowBox[{ FractionBox["1", "b"], RowBox[{ UnderoverscriptBox["\[Sum]", RowBox[{"i", "=", "1"}], "n"], SuperscriptBox[ RowBox[{"(", RowBox[{ SubscriptBox["X", "i"], "-", "\[Mu]"}], ")"}], "2"]}]}], "\[LessEqual]", SuperscriptBox["\[Sigma]", "2"], "\[LessEqual]", RowBox[{ FractionBox["1", "a"], RowBox[{ UnderoverscriptBox["\[Sum]", RowBox[{"i", "=", "1"}], "n"], SuperscriptBox[ RowBox[{"(", RowBox[{ SubscriptBox["X", "i"], "-", "\[Mu]"}], ")"}], "2"]}]}]}], ")"}]}]}]}]], "Text"], Cell[TextData[{ "Sustituimos ", Cell[BoxData[ FormBox["a", TraditionalForm]]], " y ", Cell[BoxData[ FormBox["b", TraditionalForm]]], " por sus valores y reagruparmos los terminos para obtener el intervalo de \ confianza para ", Cell[BoxData[ SuperscriptBox["\[Sigma]", "2"]]] }], "Text"], Cell[BoxData[ RowBox[{"\[IndentingNewLine]", RowBox[{ RowBox[{ RowBox[{ FractionBox["1", SubscriptBox[ SuperscriptBox["\[Chi]", "2"], RowBox[{"n", ",", " ", RowBox[{ RowBox[{"(", RowBox[{"1", "+", "\[Gamma]"}], ")"}], "/", "2"}]}]]], RowBox[{ UnderoverscriptBox["\[Sum]", RowBox[{"i", "=", "1"}], "n"], SuperscriptBox[ RowBox[{"(", RowBox[{ SubscriptBox["X", "i"], "-", "\[Mu]"}], ")"}], "2"]}]}], "\[LessEqual]", SuperscriptBox["\[Sigma]", "2"], "\[LessEqual]", RowBox[{ FractionBox["1", SubscriptBox[ SuperscriptBox["\[Chi]", "2"], RowBox[{"n", ",", " ", RowBox[{ RowBox[{"(", RowBox[{"1", "-", "\[Gamma]"}], ")"}], "/", 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TraditionalForm]]], "/", Cell[BoxData[ SuperscriptBox["\[Sigma]", "2"]]], " es una distribuci\[OAcute]n ", Cell[BoxData[ FormBox[ SuperscriptBox[ SubscriptBox["\[Chi]", RowBox[{"n", "-", "1"}]], "2"], TraditionalForm]]], " y se procedede forma similar al caso anterior obteniendose " }], "Text"], Cell[BoxData[ RowBox[{"\[IndentingNewLine]", RowBox[{ FractionBox[ RowBox[{ RowBox[{"(", RowBox[{"n", "-", "1"}], ")"}], FormBox[ SuperscriptBox[ SubscriptBox["s", "c"], "2"], TraditionalForm]}], SubscriptBox[ SuperscriptBox["\[Chi]", "2"], RowBox[{"n", ",", " ", RowBox[{ RowBox[{"(", RowBox[{"1", "+", "\[Gamma]"}], ")"}], "/", "2"}]}]]], "\[LessEqual]", SuperscriptBox["\[Sigma]", "2"], "\[LessEqual]", FractionBox[ RowBox[{ RowBox[{"(", RowBox[{"n", "-", "1"}], ")"}], FormBox[ SuperscriptBox[ SubscriptBox["s", "c"], "2"], TraditionalForm]}], SubscriptBox[ SuperscriptBox["\[Chi]", "2"], RowBox[{"n", ",", " ", RowBox[{ RowBox[{"(", RowBox[{"1", "-", "\[Gamma]"}], ")"}], "/", "2"}]}]]]}]}]], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["\<\ Determinaci\[OAcute]n del intervalo de confianza de \[Sigma] de varias \ poblaciones normales que tienen en com\[UAcute]n la misma varianza \ \>", "Subsection"], Cell[TextData[{ "Sean varias poblaciones N(", Cell[BoxData[ FormBox[ SubscriptBox[ OverscriptBox["x", "_"], "1"], TraditionalForm]]], ", ", Cell[BoxData[ FormBox["\[Sigma]", TraditionalForm]]], "), N(", Cell[BoxData[ FormBox[ SubscriptBox[ OverscriptBox["x", "_"], "2"], TraditionalForm]]], ", ", Cell[BoxData[ FormBox["\[Sigma]", TraditionalForm]]], "), ..., N(", Cell[BoxData[ FormBox[ SubscriptBox[ OverscriptBox["x", "_"], "g"], TraditionalForm]]], ", ", Cell[BoxData[ FormBox["\[Sigma]", TraditionalForm]]], "), ..., N(", Cell[BoxData[ FormBox[ SubscriptBox[ OverscriptBox["x", "_"], "m"], TraditionalForm]]], ", ", Cell[BoxData[ FormBox["\[Sigma]", TraditionalForm]]], "), de cada una de las cuales se toma una muestra de tama\[NTilde]o ", Cell[BoxData[ FormBox["n", TraditionalForm]]], " con la que se determina la desviaci\[OAcute]n est\[AAcute]ndar de dicha \ muestra. Un ejemplo de este tipo es el caso de un proceso productivo del que \ vamos tomando periodicamente muestras aleatorias independientes del que \ medimos un parametro ", Cell[BoxData[ FormBox["p", TraditionalForm]]], ". Para dicho parametro de cada muestra tenemos su media y varianza ", Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox[ FormBox[ OverscriptBox["x", "_"], TraditionalForm], "g"], ",", " ", SuperscriptBox[ SubscriptBox["s", "g"], "2"]}], TraditionalForm]]], ". Pasado un tiempo tendremos las varianzas {", Cell[BoxData[ FormBox[ SuperscriptBox[ SubscriptBox["s", "1"], "2"], TraditionalForm]]], ",..., ", Cell[BoxData[ FormBox[ SuperscriptBox[ SubscriptBox["s", "m"], "2"], TraditionalForm]]], "}de ", Cell[BoxData[ FormBox["m", TraditionalForm]]], " muestras que asumimos tomadas con el proceso bajo control. Nuestro \ objetivo es calcular la dispersi\[OAcute]n natural ", Cell[BoxData[ FormBox["\[Sigma]", TraditionalForm]]], " del proceso, que suponemos desconocida, con un determinado intervalo de \ confianza. A los l\[IAcute]mites de dicho intervalos les denominamos L\ \[IAcute]mite Superior de Control (", Cell[BoxData[ FormBox[ SubscriptBox["LSC", "s"], TraditionalForm]]], ") y L\[IAcute]mite Inferior de Control (", Cell[BoxData[ FormBox[ SubscriptBox["LIC", "s"], TraditionalForm]]], "). Para cualquier muestra que tomemos con posterioridad mediremos su \ desviaci\[OAcute]n t\[IAcute]pica, ", Cell[BoxData[ FormBox[ SubscriptBox["s", "m"], TraditionalForm]]], ", si ", Cell[BoxData[ FormBox[ SubscriptBox["s", "m"], TraditionalForm]]], " est\[AAcute] comprendido en el intervalo (", Cell[BoxData[ FormBox[ SubscriptBox["LSC", "s"], TraditionalForm]]], ", ", Cell[BoxData[ FormBox[ SubscriptBox["LIC", "s"], TraditionalForm]]], ") consideramos que dicha muestra se ha tomado con el proceso est\[AAcute] \ bajo control, en caso contrario consideraremos que est\[AAcute] desajustado." }], "Text"], Cell[TextData[{ "Para determianar el intervalo de confianza para ", Cell[BoxData[ FormBox["s", TraditionalForm]]], " hemos de calcular la esperanza matematica E[s] y su dispersi\[OAcute]n \ D[s] " }], "Text"], Cell[CellGroupData[{ Cell["Esperanza matematica de s, E[s]", "Subsubsection"], Cell["\<\ Recordemos que la esperanza matematica para funciones continuas se define \ como \ \>", "Text"], Cell[BoxData[ RowBox[{ RowBox[{"E", "[", "z", "]"}], "=", RowBox[{ SubsuperscriptBox["\[Integral]", RowBox[{"-", "\[Infinity]"}], "\[Infinity]"], RowBox[{"z", " ", SubscriptBox["f", "z"], RowBox[{"(", "z", ")"}], RowBox[{"\[DifferentialD]", "z"}]}]}]}]], "Text"], Cell[TextData[{ "En nuestro caso ", Cell[BoxData[ FormBox["z", TraditionalForm]]], " es ", Cell[BoxData[ FormBox["s", TraditionalForm]]], ". Para determinar la fdd ", Cell[BoxData[ RowBox[{ SubscriptBox["f", "z"], RowBox[{"(", "z", ")"}]}]]], "tenemos en cuenta que ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[ RowBox[{"n", " "}], TraditionalForm], FormBox[ FractionBox[ SuperscriptBox["s", "2"], SuperscriptBox["\[Sigma]", "2"]], TraditionalForm], " "}], TraditionalForm]]], "sigue una distribuci\[OAcute]n ", Cell[BoxData[ FormBox[ SuperscriptBox[ SubscriptBox["\[Chi]", RowBox[{"n", "-", "1"}]], "2"], TraditionalForm]]], " y hacemos la transformaci\[OAcute]n ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ FormBox[ RowBox[{"n", " "}], TraditionalForm], FormBox[ FractionBox[ SuperscriptBox["s", "2"], SuperscriptBox["\[Sigma]", "2"]], TraditionalForm]}], " ", "=", " ", RowBox[{ RowBox[{"n", " ", FractionBox["y", SuperscriptBox["\[Sigma]", "2"]]}], "=", " ", "x"}]}], TraditionalForm]]], ". 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