{"id":1265,"date":"2017-11-07T09:00:32","date_gmt":"2017-11-07T14:00:32","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/matg1013\/?p=1265"},"modified":"2026-04-05T19:53:03","modified_gmt":"2026-04-06T00:53:03","slug":"s1eva2007tii_t1-distribucion-binomial-acumulada","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s1eva10\/s1eva2007tii_t1-distribucion-binomial-acumulada\/","title":{"rendered":"s1Eva2007TII_T1 Distribuci\u00f3n binomial acumulada"},"content":{"rendered":"\n

Ejercicio<\/strong>: 1Eva2007TII_T1 Distribuci\u00f3n binomial acumulada<\/a><\/p>\n\n\n\n

Dado F<\/strong>=0.4, dado que n<\/strong>=5 y k<\/strong>=1<\/p>\n\n\n F = \\sum_{t=0}^{k} \\binom{n}{t} p^t (1-p)^{n-t}<\/span>\n\n\n\n

La f\u00f3rmula para el ejercicio se convierte en:<\/p>\n\n\n F = \\Bigg( \\begin{array}{c} 5 \\\\ 0 \\end{array} \\Bigg) p ^0 (1-p)^{(5-0)} + \\Bigg( \\begin{array}{c} 5 \\\\ 1 \\end{array} \\Bigg) p ^1 (1-p)^{(5-1)} = 0.4<\/span>\n\n\n\n

Los valores de las combinatorias se calculan como:<\/p>\n\n\n\n

>>> import scipy.special as sts\n>>> sts.comb(5,0,repetition=False)\n1.0\n>>> sts.comb(5,1,repetition=False)\n5.0\n>>> <\/code><\/pre>\n\n\n (1-p)^{5} + 5p (1-p)^{4} = 0.4<\/span>\n\n\n\n
La expresi\u00f3n para el ejercicio se convierte en:<\/p>\n\n\n f(p) = (1-p)^{5} + 5p (1-p)^{4} - 0.4<\/span>\n\n\n\n
como referencia se revisa la gr\u00e1fica para f(p)<\/p>\n\n\n\n
\"s1eiit2007t1<\/figure>\n\n\n f(p) = (1-p)^5 + 5p(1-p)^4 - 0.4 <\/span>\n\n\n = (1-p)^4 (1 - p + 5p) - 0.4 <\/span>\n\n\n = (1-p)^4 (1 + 4p) - 0.4 <\/span>\n\n\n = (1-p)^2 (1-p)^2 (1 + 4p) - 0.4 <\/span>\n\n\n = (1-2p+p^2) (1-2p+p^2) (1 + 4p) - 0.4 <\/span>\n\n\n = (1 - 4p + 6p^2 - 4p^3 +p^4 ) (1 + 4p) - 0.4<\/span>\n\n\n = 1 - 10p^2 + 20p^3 + 15p^4 + 4p^5 - 0.4<\/span>\n\n\n f(p) = 0.6 - 10p^2 + 20p^3 + 15p^4 + 4p^5<\/span>\n\n\n\n
y su derivada:<\/p>\n\n\n f'(p) = - 20p + 60p^2 + 60p^3 +20p^4 <\/span>\n\n\n\n
con lo que se puede desarrollar el m\u00e9todo Newton-Raphson.<\/p>\n\n\n\n
Verificando el polinomio  obtenido a partir de la expresi\u00f3n inicial usando Sympy:<\/p>\n\n\n\n
>>> import sympy as sp\n>>> p = sp.Symbol('p')\n>>> poli = (1-p)**5 + 5*p*((1-p)**4) - 0.4\n>>> pol = poli.expand()\n>>> pol\n4*p**5 - 15*p**4 + 20*p**3 - 10*p**2 + 0.6\n>>> pol.diff(p,1)\n20*p**4 - 60*p**3 + 60*p**2 - 20*p<\/code><\/pre>\n\n\n\n
A partir de la gr\u00e1fica, un punto inicial cercano a la ra\u00edz es X0 = 0.2<\/p>\n\n\n\n
itera = 0<\/p>\n\n\n f(0.2) = 0.6 - 10(0.2)^2 + 20(0.2)^3 + 15(0.2)^4 + 4(0.2)^5 = 0.3373 <\/span>\n\n\n f'(0.2) = - 20(0.2) + 60(0.2)^2 + 60(0.2)^3 +20(0.2)^4 = -2.048 <\/span>\n\n\n x_1 = 0.2 -\\frac{0.3373}{-2.048} = 0.3647<\/span>\n\n\n errado = |0.2 - 0.36469| = 0.1647<\/span>\n\n\n\n
itera = 1<\/p>\n\n\n f(0.36469) = 0.6 - 10(0.36469)^2 + 20(0.36469)^3 + 15(0.36469)^4 + 4(0.36469)^5 <\/span>\n\n\n = 0.0005566 <\/span>\n\n\n f'(0.36469) = - 20(0.36469) + 60(0.36469)^2 + 60(0.36469)^3 +20(0.36469)^4 <\/span>\n\n\n = -1.8703 <\/span>\n\n\n x_1 = 0.36469 -\\frac{0.0005566}{-1.8703} <\/span>\n\n\n = 0.36499<\/span>\n\n\n errado = |0.36469 - 0.36499| = 0.000297<\/span>\n\n\n\n
itera = 3<\/p>\n\n\n f(0.36499) = 0.6 - 10(0.36499)^2 + 20(0.36499)^3 + 15(0.36499)^4 + 4(0.36499)^5 <\/span>\n\n\n = 1.6412291237166698e-07 <\/span>\n\n\n f'(0.36499) = - 20(0.36499) + 60(0.36499)^2 + 60(0.36499)^3 +20(0.36499)^4 <\/span>\n\n\n = -1.869204616112814 <\/span>\n\n\n x_1 = 0.36469 -\\frac{1.6412291237166698e-07}{ -1.8692} <\/span>\n\n\n = 0.36499 <\/span>\n\n\n errado = |0.36499 - 0.36499| = 8.7804e-08<\/span>\n\n\n\n
verificar con ra\u00edz: 0.3649852264049102<\/em><\/strong><\/p>\n\n\n\n
Instrucciones para la gr\u00e1fica<\/p>\n\n\n
\n# 1Eva2007TII_T1 Distribuci\u00f3n binomial acumulada\nimport numpy as np\nimport matplotlib.pyplot as plt\nimport scipy.special as sts\n \nfp = lambda p: (1-p)**5 + 5*p*((1-p)**4) - 0.4\n \na = 0\nb = 1\nmuestras = 21\n \n# PROCEDIMIENTO\nxi = np.linspace(a,b,muestras)\np_i = fp(xi)\n \n# SALIDA\nplt.plot(xi,p_i)\nplt.axhline(0)\nplt.show()\n\n<\/pre><\/div>\n\n\n
\n\n\n\n
Algoritmo en Python<\/h2>\n\n\n\n
el resultado usando el algoritmo es:<\/p>\n\n\n\n
i ['xi', 'fi', 'dfi', 'xnuevo', 'tramo']\n0 [ 0.2     0.3373 -2.048   0.3647  0.1647]\n1 [ 3.6469e-01  5.5668e-04 -1.8703e+00  3.6499e-01  2.9764e-04]\n2 [ 3.6499e-01  1.6412e-07 -1.8692e+00  3.6499e-01  8.7804e-08<\/strong>]\nraiz en:  0.3649852264049102<\/strong><\/code><\/pre>\n\n\n\n
con error de: 8.780360960525257e-08<\/p>\n\n\n\n
Algoritmo en Python para Newton-Raphson<\/h2>\n\n\n
\n# 1Eva2007TII_T1 Distribuci\u00f3n binomial acumulada\nimport numpy as np\nimport matplotlib.pyplot as plt\n \ndef newton_raphson(fx,dfx,x0, tolera, iteramax=100,\n                   vertabla=False, precision=4):\n    '''fx y dfx en forma num\u00e9rica lambda\n    xi es el punto inicial de b\u00fasqueda\n    '''\n    itera=0\n    tramo = abs(2*tolera)\n    xi = x0 \n    if vertabla==True:\n        print('m\u00e9todo de Newton-Raphson')\n        print('i', ['xi','fi','dfi', 'xnuevo', 'tramo'])\n        np.set_printoptions(precision)\n    while (tramo>=tolera):\n        fi = fx(xi)\n        dfi = dfx(xi)\n        xnuevo = xi - fi\/dfi\n        tramo = abs(xnuevo-xi)\n        if vertabla==True:\n            print(itera,np.array([xi,fi,dfi,xnuevo,tramo]))\n        xi = xnuevo\n        itera = itera + 1\n \n    if itera>=iteramax:\n        xi = np.nan\n        print('itera: ',itera,\n              'No converge,se alcanz\u00f3 el m\u00e1ximo de iteraciones')\n \n    return(xi)\n \n# INGRESO\nfx  = lambda p: 4*p**5 - 15*p**4 + 20*p**3 - 10*p**2 + 0.6\ndfx = lambda p: 20*p**4 - 60*p**3 + 60*p**2 - 20*p\n\nx0 = 0.2\ntolera = 0.0000001\n \n# PROCEDIMIENTO\nxi = newton_raphson(fx,dfx,x0, tolera, vertabla=True)\n# SALIDA\nprint('ra\u00edz en: ', xi)\n\n# GRAFICA\nimport matplotlib.pyplot as plt\na = 0\nb = 1\nmuestras = 21\n \nxj = np.linspace(a,b,muestras)\nfj = fx(xj)\nplt.plot(xj,fj, label='f(x)')\nplt.plot(xi,0, 'o')\nplt.axhline(0)\nplt.xlabel('x')\nplt.ylabel('f(x)')\nplt.grid()\nplt.legend()\nplt.show()\n<\/pre><\/div>","protected":false},"excerpt":{"rendered":"
Ejercicio: 1Eva2007TII_T1 Distribuci\u00f3n binomial acumulada Dado F=0.4, dado que n=5 y k=1 La f\u00f3rmula para el ejercicio se convierte en: Los valores de las combinatorias se calculan como: La expresi\u00f3n para el ejercicio se convierte en: como referencia se revisa la gr\u00e1fica para f(p) y su derivada: con lo que se puede desarrollar el m\u00e9todo […]<\/p>\n","protected":false},"author":8043,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"wp-custom-template-entrada-mn-ejemplo","format":"standard","meta":{"footnotes":""},"categories":[44],"tags":[58,54],"class_list":["post-1265","post","type-post","status-publish","format-standard","hentry","category-mn-s1eva10","tag-ejemplos-python","tag-mnumericos"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/1265","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/users\/8043"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/comments?post=1265"}],"version-history":[{"count":6,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/1265\/revisions"}],"predecessor-version":[{"id":23816,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/1265\/revisions\/23816"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=1265"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=1265"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=1265"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}