{"id":49120,"date":"2016-09-12T20:20:53","date_gmt":"2016-09-12T23:20:53","guid":{"rendered":"https:\/\/www.estrategiaconcursos.com.br\/blog\/?p=49120"},"modified":"2018-09-21T17:54:02","modified_gmt":"2018-09-21T20:54:02","slug":"gabarito-extraoficial-tcepr-estatistica","status":"publish","type":"post","link":"https:\/\/www.estrategiaconcursos.com.br\/blog\/gabarito-extraoficial-tcepr-estatistica\/","title":{"rendered":"Gabarito extraoficial TCE\/PR &#8211; Estat\u00edstica"},"content":{"rendered":"<p>Caros alunos,<\/p>\n<p style=\"text-align: justify;\">Atendendo aos pedidos de voc\u00eas, vejam abaixo a resolu\u00e7\u00e3o das quest\u00f5es de Estat\u00edstica\u00a0da prova de Analista do TCE\/PR 2016, da banca CESPE, com o meu gabarito extra oficial. Utilizei o caderno de provas da \u00c1rea Administra\u00e7\u00e3o, ok?<\/p>\n<p style=\"text-align: justify;\"><em><strong>***Atualiza\u00e7\u00e3o: hoje, 13\/09, foi publicado o gabarito preliminar do Cespe. Entendo que n\u00e3o h\u00e1 recurso, pois todos os gabaritos bateram com a minha resolu\u00e7\u00e3o extra oficial***<\/strong><\/em><\/p>\n<p style=\"text-align: justify;\"><strong>CESPE &#8211; TCE\/PR &#8211; 2016)\u00a0Em um levantamento feito&#8230;<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p style=\"text-align: justify;\">Suponha que n empresas seguem o padr\u00e3o, de modo que 17 &#8211; n empresas n\u00e3o o seguem. A m\u00e9dia \u00e9 calculada assim:<\/p>\n<p style=\"text-align: justify;\">M\u00e9dia = (n.1 + (17-n).0) \/ 17<\/p>\n<p style=\"text-align: justify;\">0,8 = n \/ 17<\/p>\n<p style=\"text-align: justify;\">n = 0,8 x 17<\/p>\n<p style=\"text-align: justify;\">n = 13,6<\/p>\n<p style=\"text-align: justify;\">Portanto, em m\u00e9dia 13,6 das 17 empresas seguem o padr\u00e3o e 17 &#8211; 13,6 = 3,4 n\u00e3o seguem. Podemos calcular a vari\u00e2ncia amostral assim:<\/p>\n<p style=\"text-align: justify;\">Soma dos valores = 13,6&#215;1 + 3,4&#215;0 = 13,6<\/p>\n<p style=\"text-align: justify;\">Soma dos quadrados dos valores = 13,6&#215;1^2 + 3,4&#215;0^2 = 13,6<\/p>\n<p style=\"text-align: justify;\">Logo,<\/p>\n<p>Var(X) = (soma dos quadrados &#8211; (1\/n).(soma dos valores)^2) \/ (n &#8211; 1)<\/p>\n<p>Var(X) = (13,6 &#8211; (1\/17).(13,6)^2) \/ (17 &#8211; 1)<\/p>\n<p>Var(X) = (13,6 &#8211; 10,88) \/ 16 = 0,17<\/p>\n<p style=\"text-align: justify;\"><strong>Resposta: B (vari\u00e2ncia entre 0,15 e 0,18)<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>CESPE &#8211; TCE\/PR &#8211; 2016) Com base na figura antecedente&#8230;<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p style=\"text-align: justify;\">Veja que temos um diagrama Box-Plot. O primeiro quartil (parte inferior do ret\u00e2ngulo) \u00e9 0,6, o terceiro quartil (parte superior do ret\u00e2ngulo) \u00e9 0,8, a mediana (tra\u00e7o horizontal no interior do ret\u00e2ngulo) est\u00e1 mais pr\u00f3ximo de 0,8 do que de 0,7. At\u00e9 aqui podemos eliminar as alternativas A (n\u00e3o \u00e9 curva de frequ\u00eancia), B (Q1 \u00e9 0,6) e C (mediana acima de 0,7).<\/p>\n<p style=\"text-align: justify;\">A alternativa D est\u00e1 correta. Temos um ponto isolado na altura de 0,2, que sugere a exist\u00eancia de uma observa\u00e7\u00e3o que destoa das demais.<\/p>\n<p style=\"text-align: justify;\">A alternativa E est\u00e1 errada pois, \u00a0no Box-Plot, os tra\u00e7os que vemos em 0,3 e 0,9 n\u00e3o necessariamente representam o m\u00ednimo e o m\u00e1ximo da distribui\u00e7\u00e3o. Eles s\u00e3o, na verdade, o limite inferior e limite superior do gr\u00e1fico, cujo c\u00e1lculo \u00e9 feito assim:<\/p>\n<p>Limite inferior: \u00e9 o maior valor entre a menor observa\u00e7\u00e3o realizada e o resultado da express\u00e3o Q1 &#8211; 1,5(Q3 &#8211; Q1).<\/p>\n<p>Limite superior: \u00a0\u00e9 o menor valor entre a maior observa\u00e7\u00e3o realizada e o resultado da express\u00e3o Q3 + 1,5 (Q3 &#8211; Q1).<\/p>\n<p>Repare que o valor 0,2 \u00e9 menor que o limite inferior do gr\u00e1fico.<\/p>\n<p><strong>Resposta: D (a figura sugere a exist\u00eancia de uma observa\u00e7\u00e3o destoante)<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>CESPE &#8211; TCE\/PR &#8211; 2016)\u00a0Se satisfa\u00e7\u00e3o no trabalho e sa\u00fade&#8230;<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p style=\"text-align: justify;\">Se as vari\u00e2ncias da satisfa\u00e7\u00e3o e da sa\u00fade s\u00e3o 8 e 2, respectivamente, ent\u00e3o seus desvios padr\u00f5es s\u00e3o raiz(8) e raiz(2).<\/p>\n<p style=\"text-align: justify;\">A correla\u00e7\u00e3o \u00e9 calculada assim:<\/p>\n<p style=\"text-align: justify;\">correla\u00e7\u00e3o(X,Y) = covari\u00e2ncia(X,Y) \/ [desvio padr\u00e3o X . desvio padr\u00e3o Y]<\/p>\n<p style=\"text-align: justify;\">Aplicando essa f\u00f3rmula ao nosso caso:<\/p>\n<p style=\"text-align: justify;\">correla\u00e7\u00e3o (satisfa\u00e7\u00e3o, sa\u00fade) = 3 \/ [raiz(8) . raiz(2)] = 3 \/ raiz(16) = 3 \/ 4 = 0,75<\/p>\n<p style=\"text-align: justify;\"><strong>Resposta: C (0,75)<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>CESPE &#8211; TCE\/PR &#8211; 2016) Se X for uma vari\u00e1vel aleat\u00f3ria normal&#8230;<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p style=\"text-align: justify;\">Vamos avaliar cada alternativa de resposta. A letra A est\u00e1 errada pois a transforma\u00e7\u00e3o padr\u00e3o \u00e9 feita assim:<\/p>\n<p style=\"text-align: justify;\">Z = (X &#8211; m\u00e9dia) \/ desvio padr\u00e3o<\/p>\n<p style=\"text-align: justify;\">Na letra A foi utilizada a vari\u00e2ncia no lugar do desvio padr\u00e3o nesta f\u00f3rmula.<\/p>\n<p style=\"text-align: justify;\">O coeficiente de varia\u00e7\u00e3o \u00e9 dado por:<\/p>\n<p style=\"text-align: justify;\">CV = desvio padr\u00e3o \/ m\u00e9dia = raiz(0,4) \/ 0,8 = 0,63 \/ 0,8 = 0,79<\/p>\n<p style=\"text-align: justify;\">Vemos que a letra B tamb\u00e9m est\u00e1 errada.<\/p>\n<p style=\"text-align: justify;\">A moda da distribui\u00e7\u00e3o normal \u00e9 o mesmo valor da m\u00e9dia, ou seja, 0,8, o que torna a letra C errada.<\/p>\n<p style=\"text-align: justify;\">A alternativa D\u00a0\u00e9 o nosso gabarito. Como X \u00e9 uma distribui\u00e7\u00e3o cont\u00ednua, a probabilidade de qualquer valor isolado \u00e9 igual a ZERO. Ou seja, P(X=0,8) = P(X=0,1) = 0. Nesse tipo de distribui\u00e7\u00e3o n\u00f3s calculamos a probabilidade de um intervalo, e n\u00e3o de um valor espec\u00edfico. Tanto que utilizamos fun\u00e7\u00f5es de densidade de probabilidade, e n\u00e3o fun\u00e7\u00f5es de probabilidade (que usamos nas distribui\u00e7\u00f5es discretas).<\/p>\n<p style=\"text-align: justify;\">Na letra E, veja que os valores 0,7 e 0,9 s\u00e3o equidistantes em rela\u00e7\u00e3o \u00e0 m\u00e9dia 0,8, que \u00e9 o eixo de simetria da curva normal. Portanto, a probabilidade de termos valores abaixo de 0,7 \u00e9 igual \u00e0 probabilidade de termos valores acima de 0,9, o que torna esta alternativa errada.<\/p>\n<p style=\"text-align: justify;\"><strong>Resposta: D \u00a0(P(X=0,8) = P(X=0,1))<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Caros alunos, Atendendo aos pedidos de voc\u00eas, vejam abaixo a resolu\u00e7\u00e3o das quest\u00f5es de Estat\u00edstica\u00a0da prova de Analista do TCE\/PR 2016, da banca CESPE, com o meu gabarito extra oficial. Utilizei o caderno de provas da \u00c1rea Administra\u00e7\u00e3o, ok? ***Atualiza\u00e7\u00e3o: hoje, 13\/09, foi publicado o gabarito preliminar do Cespe. Entendo que n\u00e3o h\u00e1 recurso, pois [&hellip;]<\/p>\n","protected":false},"author":48,"featured_media":43262,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"post_tipo":"article","footnotes":""},"categories":[],"tags":[1847],"tax_estado":[],"class_list":["post-49120","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","tag-concurso-tce-pr"],"acf":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v27.7 (Yoast SEO v27.7) - https:\/\/yoast.com\/product\/yoast-seo-premium-wordpress\/ -->\n<title>Gabarito extraoficial TCE\/PR - Estat\u00edstica<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/blog-estrategia.mystagingwebsite.com\/gabarito-extraoficial-tcepr-estatistica\/\" \/>\n<meta property=\"og:locale\" content=\"pt_BR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Gabarito extraoficial TCE\/PR - Estat\u00edstica\" \/>\n<meta property=\"og:description\" content=\"Caros alunos, Atendendo aos pedidos de voc\u00eas, vejam abaixo a resolu\u00e7\u00e3o das quest\u00f5es de Estat\u00edstica\u00a0da prova de Analista do TCE\/PR 2016, da banca CESPE, com o meu gabarito extra oficial. 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Utilizei o caderno de provas da \u00c1rea Administra\u00e7\u00e3o, ok? ***Atualiza\u00e7\u00e3o: hoje, 13\/09, foi publicado o gabarito preliminar do Cespe. 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