{"id":909071,"date":"2021-11-23T08:30:00","date_gmt":"2021-11-23T11:30:00","guid":{"rendered":"https:\/\/www.estrategiaconcursos.com.br\/blog\/?p=909071"},"modified":"2021-11-22T14:09:03","modified_gmt":"2021-11-22T17:09:03","slug":"distribuicoes-discretas-para-iss-bh-estatistica","status":"publish","type":"post","link":"https:\/\/www.estrategiaconcursos.com.br\/blog\/distribuicoes-discretas-para-iss-bh-estatistica\/","title":{"rendered":"Resumo das Distribui\u00e7\u00f5es Discretas para ISS-BH \u2013 Estat\u00edstica"},"content":{"rendered":"\n<p>Ol\u00e1, pessoal. Tudo certo? No artigo de hoje veremos o Resumo das Distribui\u00e7\u00f5es Discretas para ISS-BH.<\/p>\n\n\n\n<p>Veremos as seguintes distribui\u00e7\u00f5es:<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>Uniforme<\/li><li>Bernoulli<\/li><li>Binomial<\/li><li>Geom\u00e9trica<\/li><li>Hipergeom\u00e9trica<\/li><li>Poisson<\/li><\/ul>\n\n\n\n<p>Preparado (a)? Vamos l\u00e1<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Distribui\u00e7\u00e3o Uniforme<\/h2>\n\n\n\n<p>Iniciemos o Resumo das Distribui\u00e7\u00f5es Discretas para ISS-BH pela Distribui\u00e7\u00e3o Uniforme.<\/p>\n\n\n\n<p>Na <strong><span class=\"has-inline-color has-vivid-cyan-blue-color\">Distribui\u00e7\u00e3o Uniforme<\/span><\/strong> todos os valores s\u00e3o equiprov\u00e1veis, ou seja, todos os elementos t\u00eam a mesma chance de serem sorteados.<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"243\" height=\"98\" src=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/11\/15105555\/image-312.png\" alt=\"Resumo das Distribui\u00e7\u00f5es Discretas para ISS-BH \u2013 Estat\u00edstica\" class=\"wp-image-909072\"\/><figcaption>Distribui\u00e7\u00e3o Uniforme<\/figcaption><\/figure><\/div>\n\n\n\n<ul class=\"wp-block-list\"><li><strong><span class=\"has-inline-color has-vivid-green-cyan-color\">Esperan\u00e7a matem\u00e1tica<\/span><\/strong><\/li><\/ul>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"196\" height=\"72\" src=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/11\/15105603\/image-313.png\" alt=\"\" class=\"wp-image-909073\"\/><figcaption>Esperan\u00e7a matem\u00e1tica &#8211; Distribui\u00e7\u00e3o Uniforme<\/figcaption><\/figure><\/div>\n\n\n\n<p>Como todos os valores t\u00eam a mesma probabilidade, a esperan\u00e7a matem\u00e1tica \u00e9 calculada da mesma forma de a m\u00e9dia \u201cconvencional\u201d.<strong><\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\"><li><strong><span class=\"has-inline-color has-vivid-green-cyan-color\">Vari\u00e2ncia<\/span><\/strong><\/li><\/ul>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"359\" height=\"85\" src=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/11\/15105617\/image-314.png\" alt=\"\" class=\"wp-image-909074\" srcset=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/11\/15105617\/image-314.png 359w, https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/11\/15105617\/image-314-300x71.png 300w\" sizes=\"auto, (max-width: 359px) 100vw, 359px\" \/><figcaption>Vari\u00e2ncia &#8211; Distribui\u00e7\u00e3o Uniforme<\/figcaption><\/figure><\/div>\n\n\n\n<p>Tamb\u00e9m \u00e9 uma f\u00f3rmula j\u00e1 conhecida, a m\u00e9dia dos quadrados menos o quadrado da m\u00e9dia.<\/p>\n\n\n\n<p><strong><span class=\"has-inline-color has-vivid-purple-color\">Exemplo<\/span>:<\/strong> FGV\/2015 &#8211; TJ\/RO Suponha que o n\u00famero de advogados atendidos por um diretor de vara, por dia, \u00e9 uma vari\u00e1vel aleat\u00f3ria distribu\u00edda uniformemente entre 11 e 25, inclusive. Ent\u00e3o, se em um dia qualquer, at\u00e9 certo hor\u00e1rio, 18 advogados foram atendidos, a probabilidade de que mais de 23 sejam atendidos naquele dia \u00e9:<\/p>\n\n\n\n<p>Casos totais: 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25<\/p>\n\n\n\n<p>Os casos que queremos dividido pelos casos poss\u00edveis<\/p>\n\n\n\n<p>24 ou 25 (mais de 23) \/ 18, 19, 20, 21, 22, 23, 24, 25 (18 j\u00e1 foram atendidos)<\/p>\n\n\n\n<p>2 \/ 8 = 0,25<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Distribui\u00e7\u00e3o de Bernoulli<\/h2>\n\n\n\n<p>A <span class=\"has-inline-color has-vivid-cyan-blue-color\"><strong>Distribui\u00e7\u00e3o de Bernoulli<\/strong> <\/span>trata-se de um experimento com 2 poss\u00edveis, realizados uma \u00fanica vez.<\/p>\n\n\n\n<p><strong><span class=\"has-inline-color has-vivid-green-cyan-color\">Condi\u00e7\u00f5es<\/span>:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>2 resultados poss\u00edveis: <strong>sucesso<\/strong> (X = 1) ou <strong>fracasso<\/strong> (X = 0)<\/li><li>Experimento realizado uma <strong>\u00fanica vez<\/strong><\/li><\/ul>\n\n\n\n<p>Assim a probabilidade de sucesso P[X = 1]&nbsp; = p e a probabilidade do fracasso ser\u00e1 seu complemento P[X = 0] = q = 1 \u2013 p<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li><strong><span class=\"has-inline-color has-vivid-green-cyan-color\">Esperan\u00e7a matem\u00e1tica<\/span><\/strong><\/li><\/ul>\n\n\n\n<p class=\"has-text-align-center\">E (X) = P<\/p>\n\n\n\n<p><strong><span class=\"has-inline-color has-vivid-green-cyan-color\">Vari\u00e2ncia<\/span><\/strong><\/p>\n\n\n\n<p class=\"has-text-align-center\">v (X) = p x q <\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Distribui\u00e7\u00e3o Binomial<\/h2>\n\n\n\n<p>A <span class=\"has-inline-color has-vivid-cyan-blue-color\"><strong>Distribui\u00e7\u00e3o Binomial<\/strong> <\/span>\u00e9 basicamente a Distribui\u00e7\u00e3o de Bernoulli repetida v\u00e1rias vezes, ou seja, teremos dois resultados poss\u00edveis, por\u00e9m v\u00e1rios <strong>ensaios independentes<\/strong>.<\/p>\n\n\n\n<p>Para utilizarmos a Distribui\u00e7\u00e3o Binomial, temos que garantir que os ensaios sejam independentes, assim observe.<\/p>\n\n\n\n<p><strong><span class=\"has-inline-color has-vivid-green-cyan-color\">Condi\u00e7\u00f5es para ensaio independente:<\/span><\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>Sele\u00e7\u00e3o com reposi\u00e7\u00e3o; ou<\/li><li>Popula\u00e7\u00e3o infinita<\/li><\/ul>\n\n\n\n<p>Dissemos que a distribui\u00e7\u00e3o binominal \u00e9 a repeti\u00e7\u00e3o do ensaio de Bernoulli, assim veja que a esperan\u00e7a matem\u00e1tica e a vari\u00e2ncia s\u00e3o a mesma da Distribui\u00e7\u00e3o de Bernoulli vezes o n\u00famero de ensaios (n).<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li><strong><span class=\"has-inline-color has-vivid-green-cyan-color\">Esperan\u00e7a matem\u00e1tica<\/span><\/strong><\/li><\/ul>\n\n\n\n<p class=\"has-text-align-center\">E (X) = n x p<\/p>\n\n\n\n<p><strong><span class=\"has-inline-color has-vivid-green-cyan-color\">Vari\u00e2ncia<\/span><\/strong><\/p>\n\n\n\n<p class=\"has-text-align-center\">V (X) = n x p x q<\/p>\n\n\n\n<p><strong><span class=\"has-inline-color has-vivid-purple-color\">Exemplo<\/span>:<\/strong> (2018 \u2013 C\u00e2mara de Goi\u00e2nia) Considere uma vari\u00e1vel aleat\u00f3ria X com distribui\u00e7\u00e3o binomial e par\u00e2metros p = 1\/3 e n = 4. Qual \u00e9 a probabilidade de X = 2?<\/p>\n\n\n\n<p>P (X = 2) = C4,2 x (1\/3)\u00b2 x (2\/3)\u00b2<\/p>\n\n\n\n<p>P (X = 2) = C4,2 x (1\/3)\u00b2 x (2\/3)\u00b2<\/p>\n\n\n\n<p>P (X = 2) = 6 x 1\/9 x 4\/9<\/p>\n\n\n\n<p>P (X = 2) = 8\/27<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Distribui\u00e7\u00e3o Geom\u00e9trica<\/h2>\n\n\n\n<p>A <strong><span class=\"has-inline-color has-vivid-cyan-blue-color\">distribui\u00e7\u00e3o geom\u00e9trica<\/span><\/strong> tamb\u00e9m se baseia nos ensaios de Bernoulli, entretanto a distribui\u00e7\u00e3o corresponde ao <strong>n\u00famero de ensaios at\u00e9 o primeiro sucesso<\/strong>. Ou seja, v\u00e1rios fracassos at\u00e9 o primeiro sucesso &#8211; F + F + F &#8230;. P<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-full is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/11\/15105721\/image-319.png\" alt=\"Resumo das Distribui\u00e7\u00f5es Discretas para ISS-BH \u2013 Estat\u00edstica\" class=\"wp-image-909079\" width=\"295\" height=\"46\"\/><figcaption>Distribui\u00e7\u00e3o Geom\u00e9trica<\/figcaption><\/figure><\/div>\n\n\n\n<ul class=\"wp-block-list\"><li><strong><span class=\"has-inline-color has-vivid-green-cyan-color\">Esperan\u00e7a matem\u00e1tica<\/span><\/strong><\/li><\/ul>\n\n\n\n<p class=\"has-text-align-center\">E (X) = 1\/<strong><span class=\"has-inline-color has-vivid-red-color\">p<\/span><\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\"><li><strong><span class=\"has-inline-color has-vivid-green-cyan-color\">Vari\u00e2ncia<\/span><\/strong><\/li><\/ul>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"186\" height=\"69\" src=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/11\/15105746\/image-321.png\" alt=\"\" class=\"wp-image-909081\"\/><figcaption>Vari\u00e2ncia &#8211; Distribui\u00e7\u00e3o Geom\u00e9trica<\/figcaption><\/figure><\/div>\n\n\n\n<p><strong><span class=\"has-inline-color has-vivid-purple-color\">Exemplo<\/span>:<\/strong> (2018 \u2013 IPM &#8211; Adaptada) Sabe-se que a distribui\u00e7\u00e3o geom\u00e9trica pode ser interpretada como uma sequ\u00eancia de ensaios de Bernoulli, independentes, at\u00e9 a ocorr\u00eancia do primeiro sucesso. Qual o valor da m\u00e9dia e da vari\u00e2ncia de uma distribui\u00e7\u00e3o geom\u00e9trica cujo par\u00e2metro \u00e9 p = 0,64 e tendo como parametriza\u00e7\u00e3o o n\u00famero de ensaios de Bernoulli at\u00e9 se obter um sucesso.<\/p>\n\n\n\n<p>A esperan\u00e7a matem\u00e1tica (m\u00e9dia) ser\u00e1:<\/p>\n\n\n\n<p>M\u00e9dia = 1 \/ 0,64<\/p>\n\n\n\n<p>M\u00e9dia = 1,56<\/p>\n\n\n\n<p>Enquanto a vari\u00e2ncia:<\/p>\n\n\n\n<p>VAR = (1 \u2013 0,64) \/ 0,64\u00b2<\/p>\n\n\n\n<p>VAR = 0,36 \/ 0,4096<\/p>\n\n\n\n<p>VAR = 0,88<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Distribui\u00e7\u00e3o Hipergeom\u00e9trica<\/h2>\n\n\n\n<p>Na <strong><span class=\"has-inline-color has-vivid-cyan-blue-color\">distribui\u00e7\u00e3o hipergeom\u00e9trica<\/span><\/strong> tamb\u00e9m temos dois resultados poss\u00edveis (sucesso ou fracasso), por\u00e9m a <strong>sele\u00e7\u00e3o ocorre sem reposi\u00e7\u00e3o<\/strong>, ou seja, os eventos n\u00e3o s\u00e3o independentes.<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"350\" height=\"112\" src=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/11\/15105839\/image-322.png\" alt=\"Resumo das Distribui\u00e7\u00f5es Discretas para ISS-BH \u2013 Estat\u00edstica\" class=\"wp-image-909086\" srcset=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/11\/15105839\/image-322.png 350w, https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/11\/15105839\/image-322-300x96.png 300w\" sizes=\"auto, (max-width: 350px) 100vw, 350px\" \/><figcaption>Distribui\u00e7\u00e3o Hipergeom\u00e9trica<\/figcaption><\/figure><\/div>\n\n\n\n<p>Onde,<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>N = total<\/li><li>S = total de &#8220;sucessos&#8221;<\/li><li>n = amostra<\/li><li>k = sucesso<\/li><\/ul>\n\n\n\n<p><strong>Exemplo:<\/strong> Suponha que haja N= 10 pe\u00e7as, no total, das quais S = 4 pe\u00e7as sejam defeituosas. Se retirarmos n = 3 pe\u00e7as, a probabilidade de encontrar k = 2 defeituosas \u00e9:<\/p>\n\n\n\n<p>P (X = 2) = C4,2 x C6,1 \/ C10,3<\/p>\n\n\n\n<p>P (X = 2) = 6 x 6 \/ 120<\/p>\n\n\n\n<p>P (X = 2) = 0,3 = 30%<\/p>\n\n\n\n<p>Tamb\u00e9m \u00e9 poss\u00edvel calcular sem utilizar f\u00f3rmulas<\/p>\n\n\n\n<p>Temos que selecionar duas pe\u00e7as defeituosas e outra boa, n\u00e3o \u00e9? Seria D D B<\/p>\n\n\n\n<p>P = (4 \/ 10) x (3 \/ 9) x (6 x 8)<\/p>\n\n\n\n<p>Entretanto pode ser em qualquer ordem e temos uma repeti\u00e7\u00e3o, assim devemos utilizar a permuta com repeti\u00e7\u00e3o<\/p>\n\n\n\n<p>P = (4 \/ 10) x (3 \/ 9) x (6 x 8) x 3 x 2! \/!<\/p>\n\n\n\n<p>P = 30%<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li><strong><span class=\"has-inline-color has-vivid-green-cyan-color\">Esperan\u00e7a matem\u00e1tica<\/span><\/strong><\/li><\/ul>\n\n\n\n<p class=\"has-text-align-center\">E (X) = n.p<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li><strong><span class=\"has-inline-color has-vivid-green-cyan-color\">Vari\u00e2ncia<\/span><\/strong><\/li><\/ul>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"284\" height=\"63\" src=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/11\/15105917\/image-323.png\" alt=\"\" class=\"wp-image-909088\"\/><figcaption>Vari\u00e2ncia &#8211; Distribui\u00e7\u00e3o Hipergeom\u00e9trica<\/figcaption><\/figure><\/div>\n\n\n\n<h2 class=\"wp-block-heading\">Distribui\u00e7\u00e3o de Poisson<\/h2>\n\n\n\n<p>Para finalizar o  Resumo das Distribui\u00e7\u00f5es Discretas para ISS-BH, vejamos sobre a Distribui\u00e7\u00e3o de Poisson.<\/p>\n\n\n\n<p>A Distribui\u00e7\u00e3o de Poisson \u00e9 utilizada como uma alternativa para quando &#8220;n&#8221; \u00e9 muito grande e &#8220;p&#8221; \u00e9 muito pequeno.<\/p>\n\n\n\n<p>Podemos dizer que se trata de uma aproxima\u00e7\u00e3o da Binomial para n -&gt; \u221e e p -&gt; 0<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"347\" height=\"96\" src=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/11\/15105929\/image-324.png\" alt=\"Resumo das Distribui\u00e7\u00f5es Discretas para ISS-BH \u2013 Estat\u00edstica\" class=\"wp-image-909089\" srcset=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/11\/15105929\/image-324.png 347w, https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/11\/15105929\/image-324-300x83.png 300w\" sizes=\"auto, (max-width: 347px) 100vw, 347px\" \/><figcaption>Distribui\u00e7\u00e3o de Poisson<\/figcaption><\/figure><\/div>\n\n\n\n<p><span class=\"has-inline-color has-luminous-vivid-amber-color\"><strong>Obs.:<\/strong> <\/span>O n\u00famero neperiano &#8220;e&#8221; \u00e9 aproximadamente 2,718<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li><strong><span class=\"has-inline-color has-vivid-green-cyan-color\">Esperan\u00e7a matem\u00e1tica<\/span><\/strong><\/li><\/ul>\n\n\n\n<p>A esperan\u00e7a matem\u00e1tica (m\u00e9dia) \u00e9 determinada pela demonstrada pela lambda.<\/p>\n\n\n\n<p class=\"has-text-align-center\">\u03bb = n x p<\/p>\n\n\n\n<p class=\"has-text-align-center\">E (X) = \u03bb<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li><strong><span class=\"has-inline-color has-vivid-green-cyan-color\">Vari\u00e2ncia<\/span><\/strong><\/li><\/ul>\n\n\n\n<p>Na Distribui\u00e7\u00e3o de Poisson o valor da lambda tamb\u00e9m demonstra o valor da vari\u00e2ncia.<\/p>\n\n\n\n<p class=\"has-text-align-center\">v (X) =  \u03bb <\/p>\n\n\n\n<p>Vale ressaltar os <strong><span class=\"has-inline-color has-vivid-green-cyan-color\">pressupostos da distribui\u00e7\u00e3o de Poisson.<\/span><\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>Homogeneidade (lambda) deve ser constante<\/li><li>Independ\u00eancia das ocorr\u00eancias em um intervalo, ou seja, independ\u00eancia dos eventos.<\/li><\/ul>\n\n\n\n<p><strong><span class=\"has-inline-color has-vivid-purple-color\">Exemplo<\/span><\/strong>: (FGV\/2014 \u2013 TJ\/BA) Suponha que a quantidade total de erros cometidos pelo judici\u00e1rio segue o padr\u00e3o de um Processo de Poisson com par\u00e2metro \u03bb = 4, relativo ao per\u00edodo de um ano. Ent\u00e3o a probabilidade de que sejam cometidos exatamente tr\u00eas erros, nos pr\u00f3ximos 18 meses, \u00e9 igual a:<\/p>\n\n\n\n<p>Se o lambda \u00e9 4 para um ano, logo para 18 meses (1 ano e meio)<\/p>\n\n\n\n<p>Lambda = 4 x 1,5 = 6<\/p>\n\n\n\n<p>Logo,<\/p>\n\n\n\n<p>P (X = 3) = (e^-4 x 4\u00b3) \/ 3<\/p>\n\n\n\n<p>Adotando 0,018 para e^-4<\/p>\n\n\n\n<p>P (X = 3) = (0,018 x 64) \/ 3<\/p>\n\n\n\n<p>P (X = 3) = 0,384<\/p>\n\n\n\n<p>P (X = 3) = 38,4<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Considera\u00e7\u00f5es Finais<\/h2>\n\n\n\n<p>Pessoal, chegamos ao final do  Resumo das Distribui\u00e7\u00f5es Discretas para ISS-BH. Espero que o artigo tenha sido \u00fatil para sua revis\u00e3o.<\/p>\n\n\n\n<p>Lembre-se que se trata apenas de um resumo, assim n\u00e3o deixe de conferir a aula na \u00edntegra para estudar o assunto de forma mais aprofundada (com exemplos, diversos exerc\u00edcios resolvidos e etc.), al\u00e9m de outros assuntos como por exemplo a <strong>Distribui\u00e7\u00e3o Binomial Negativa<\/strong>.<\/p>\n\n\n\n<p>Al\u00e9m disso, estat\u00edstica s\u00f3 se aprende de fato praticando, assim n\u00e3o deixe de realizar muitas quest\u00f5es em nosso Sistema de Quest\u00f5es.<\/p>\n\n\n\n<p class=\"has-text-align-center\"><a href=\"https:\/\/concursos.estrategiaeducacional.com.br\/\">Sistema de Quest\u00f5es (SQ) \u2013 Estrat\u00e9gia Concursos<\/a><\/p>\n\n\n\n<p>Gostou do artigo? 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Tudo certo? No artigo de hoje veremos o Resumo das Distribui\u00e7\u00f5es Discretas para ISS-BH. Veremos as seguintes distribui\u00e7\u00f5es: Uniforme Bernoulli Binomial Geom\u00e9trica Hipergeom\u00e9trica Poisson Preparado (a)? Vamos l\u00e1 Distribui\u00e7\u00e3o Uniforme Iniciemos o Resumo das Distribui\u00e7\u00f5es Discretas para ISS-BH pela Distribui\u00e7\u00e3o Uniforme. Na Distribui\u00e7\u00e3o Uniforme todos os valores s\u00e3o equiprov\u00e1veis, ou seja, todos os [&hellip;]<\/p>\n","protected":false},"author":1763,"featured_media":901957,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"post_tipo":"article","footnotes":""},"categories":[220183],"tags":[],"tax_estado":[219974],"class_list":["post-909071","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-fiscal-municipal","tax_estado-mg"],"acf":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v27.2 (Yoast SEO v27.2) - https:\/\/yoast.com\/product\/yoast-seo-premium-wordpress\/ -->\n<title>Resumo das Distribui\u00e7\u00f5es Discretas para ISS-BH \u2013 Estat\u00edstica<\/title>\n<meta name=\"description\" content=\"Veja neste artigo o Resumo das Distribui\u00e7\u00f5es Discretas para ISS-BH, tais como a Bernoulli, Poisson, Uniforme entre outras.\" \/>\n<meta name=\"robots\" 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