{"id":836553,"date":"2021-08-15T14:20:43","date_gmt":"2021-08-15T17:20:43","guid":{"rendered":"https:\/\/www.estrategiaconcursos.com.br\/blog\/?p=836553"},"modified":"2021-08-17T10:51:37","modified_gmt":"2021-08-17T13:51:37","slug":"gabarito-sefaz-ce-estatistica","status":"publish","type":"post","link":"https:\/\/www.estrategiaconcursos.com.br\/blog\/gabarito-sefaz-ce-estatistica\/","title":{"rendered":"Gabarito Extraoficial SEFAZ-CE &#8211; Estat\u00edstica"},"content":{"rendered":"\n<p>Oie!!<\/p>\n\n\n\n<p>Como foi a prova da SEFAZ-CE? Gostou?<\/p>\n\n\n\n<p>A seguir, apresento os coment\u00e1rios para as quest\u00f5es 141 a 146. Atente-se aos enunciados, pois a ordem das quest\u00f5es em cada bloco pode ser diferente na sua prova.<\/p>\n\n\n\n<p><strong>Bloco de quest\u00f5es 141 \u2013 143<\/strong><\/p>\n\n\n\n<p><strong>Suponha que o n\u00famero di\u00e1rio (X) de transa\u00e7\u00f5es banc\u00e1rias registradas em determinada conta banc\u00e1ria se distribua conforme uma distribui\u00e7\u00e3o de Poisson. Com respeito ao total semanal de transa\u00e7\u00f5es banc\u00e1rias registradas nessa conta banc\u00e1ria, denotada como Y = X<sub>1<\/sub>&nbsp;+ X<sub>2<\/sub>&nbsp;+ X<sub>3&nbsp;<\/sub>+ X<sub>4<\/sub>&nbsp;+ X<sub>5<\/sub>, em que {X<sub>1<\/sub>,&#8230;, X<sub>5<\/sub>} representa uma amostra aleat\u00f3ria simples retirada de uma distribui\u00e7\u00e3o de Poisson com m\u00e9dia igual a 5 transa\u00e7\u00f5es por dia, julgue os seguintes itens.<\/strong><\/p>\n\n\n\n<p><strong>Quest\u00e3o 141<\/strong>. O desvio padr\u00e3o da vari\u00e1vel aleat\u00f3ria Y \u00e9 igual a 5 transa\u00e7\u00f5es por semana.<\/p>\n\n\n\n<p><strong>Coment\u00e1rios:<\/strong><\/p>\n\n\n\n<p>O enunciado informa que X segue distribui\u00e7\u00e3o de Poisson com m\u00e9dia de 5 transa\u00e7\u00f5es por dia e que Y \u00e9 a soma de 5 amostras de X. Ou seja, a vari\u00e1vel Y representa o n\u00famero de transa\u00e7\u00f5es por semana (considerando-se os 5 dias \u00fateis).&nbsp;<\/p>\n\n\n\n<p>A soma de 5 vari\u00e1veis com distribui\u00e7\u00e3o de Poisson e par\u00e2metro 5 segue distribui\u00e7\u00e3o de Poisson com par\u00e2metro:<\/p>\n\n\n\n<figure class=\"wp-block-image size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/08\/15140635\/image-268.png\" alt=\"\" class=\"wp-image-836555\" width=\"197\" height=\"25\" \/><\/figure>\n\n\n\n<p>Em uma distribui\u00e7\u00e3o de Poisson, a vari\u00e2ncia \u00e9 igual ao par\u00e2metro:<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"127\" height=\"25\" src=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/08\/15140700\/image-269.png\" alt=\"\" class=\"wp-image-836556\" \/><\/figure>\n\n\n\n<p>Sabendo que o desvio padr\u00e3o \u00e9 a raiz da vari\u00e2ncia, ent\u00e3o o desvio padr\u00e3o de Y \u00e9:<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"173\" height=\"33\" src=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/08\/17103715\/image-329.png\" alt=\"\" class=\"wp-image-838257\" \/><\/figure>\n\n\n\n<p>Portanto, o desvio padr\u00e3o da vari\u00e1vel Y \u00e9, de fato, igual a 5 transa\u00e7\u00f5es por semana.<\/p>\n\n\n\n<p><strong>Gabarito preliminar: Certo<\/strong><\/p>\n\n\n\n<p><strong>Quest\u00e3o 142<\/strong>.&nbsp;<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"598\" height=\"25\" src=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/08\/15140919\/image-271.png\" alt=\"\" class=\"wp-image-836559\" srcset=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/08\/15140919\/image-271.png 598w, https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/08\/15140919\/image-271-300x13.png 300w\" sizes=\"auto, (max-width: 598px) 100vw, 598px\" \/><\/figure>\n\n\n\n<p><strong>Coment\u00e1rios:<\/strong><\/p>\n\n\n\n<p>O item pede a probabilidade P(Y = 0) que \u00e9 a probabilidade de n\u00e3o haver transa\u00e7\u00f5es em determinada semana. Para isso, \u00e9 necess\u00e1rio que n\u00e3o haja transa\u00e7\u00f5es em todos os 5 dias, ou seja, n\u00e3o ter transa\u00e7\u00e3o no primeiro E n\u00e3o ter transa\u00e7\u00e3o no segundo e assim at\u00e9 o quinto do dia. Portanto, temos a&nbsp;<strong>interse\u00e7\u00e3o<\/strong>&nbsp;desses eventos.<\/p>\n\n\n\n<p>A probabilidade da interse\u00e7\u00e3o corresponde ao <strong>produto<\/strong> das probabilidades, logo:<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"530\" height=\"25\" src=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/08\/15140937\/image-272.png\" alt=\"\" class=\"wp-image-836560\" srcset=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/08\/15140937\/image-272.png 530w, https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/08\/15140937\/image-272-300x14.png 300w\" sizes=\"auto, (max-width: 530px) 100vw, 530px\" \/><\/figure>\n\n\n\n<p>Assim, conclu\u00edmos que o item est\u00e1 <strong>errado<\/strong>.&nbsp;<\/p>\n\n\n\n<p>A t\u00edtulo de complementa\u00e7\u00e3o, a probabilidade de n\u00e3o haver transa\u00e7\u00f5es em determinado dia, sabendo que X segue distribui\u00e7\u00e3o de Poisson com par\u00e2metro igual a 5, \u00e9 dada por:<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"187\" height=\"108\" src=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/08\/15141011\/image-273.png\" alt=\"\" class=\"wp-image-836561\" \/><\/figure>\n\n\n\n<p>Logo, a probabilidade P(Y = 0) \u00e9:<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"412\" height=\"31\" src=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/08\/15141031\/image-274.png\" alt=\"\" class=\"wp-image-836562\" srcset=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/08\/15141031\/image-274.png 412w, https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/08\/15141031\/image-274-300x23.png 300w\" sizes=\"auto, (max-width: 412px) 100vw, 412px\" \/><\/figure>\n\n\n\n<p>Que \u00e9 o mesmo resultado que obter\u00edamos ao considerarmos que Y segue distribui\u00e7\u00e3o de Poisson com par\u00e2metro igual a 25.\u00a0<\/p>\n\n\n\n<p><strong>Gabarito preliminar: Errad<\/strong>o<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"633\" height=\"152\" src=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/08\/15143114\/image-286.png\" alt=\"\" class=\"wp-image-836587\" srcset=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/08\/15143114\/image-286.png 633w, https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/08\/15143114\/image-286-300x72.png 300w, https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/08\/15143114\/image-286-610x146.png 610w\" sizes=\"auto, (max-width: 633px) 100vw, 633px\" \/><\/figure>\n\n\n\n<p><strong>Coment\u00e1rios:<\/strong><\/p>\n\n\n\n<p>Pelo Teorema Central do Limite, a soma de vari\u00e1veis aleat\u00f3rias tende a uma distribui\u00e7\u00e3o normal. Logo, Y segue aproximadamente distribui\u00e7\u00e3o normal, com m\u00e9dia&nbsp;e desvio padr\u00e3o&nbsp;informados no item.<\/p>\n\n\n\n<p>Para obtermos a distribui\u00e7\u00e3o normal padr\u00e3o Z, utilizamos a seguinte transforma\u00e7\u00e3o:<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"94\" height=\"62\" src=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/08\/15141238\/image-278.png\" alt=\"\" class=\"wp-image-836566\" \/><\/figure>\n\n\n\n<p>Ou seja, n\u00e3o devemos dividir o desvio padr\u00e3o de Y pela raiz de 5 e por isso o item est\u00e1 errado.<\/p>\n\n\n\n<p><strong>Gabarito preliminar: Errado<\/strong><\/p>\n\n\n\n<p><strong>Bloco de quest\u00f5es 144 \u2013 146<\/strong><\/p>\n\n\n\n<p><strong>Uma amostra aleat\u00f3ria simples de tamanho igual a 4 foi retirada de uma popula\u00e7\u00e3o exponencial cuja fun\u00e7\u00e3o densidade de probabilidade \u00e9 dada por<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"227\" height=\"37\" src=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/08\/15141425\/image-280.png\" alt=\"\" class=\"wp-image-836568\" \/><\/figure>\n\n\n\n<p><strong>Em que A &gt; 0 \u00e9 o par\u00e2metro desconhecido.<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table><tbody><tr><td class=\"has-text-align-center\" data-align=\"center\"><strong>0,5<\/strong><\/td><td class=\"has-text-align-center\" data-align=\"center\"><strong>1,0<\/strong><\/td><td class=\"has-text-align-center\" data-align=\"center\"><strong>0,8<\/strong><\/td><td class=\"has-text-align-center\" data-align=\"center\"><strong>9,7<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p><strong>Com base nos valores mostrados no quadro anterior, que constituem uma realiza\u00e7\u00e3o dessa amostra aleat\u00f3ria simples, julgue os itens a seguir.<\/strong><\/p>\n\n\n\n<p><strong>Quest\u00e3o 144<\/strong>. A estimativa da m\u00e9dia populacional obtida pelo crit\u00e9rio de m\u00ednimos quadrados ordin\u00e1rios \u00e9 igual a 3.<\/p>\n\n\n\n<p><strong>Coment\u00e1rios:<\/strong><\/p>\n\n\n\n<p>O estimador de m\u00ednimos quadrados para a m\u00e9dia populacional \u00e9 a <strong>m\u00e9dia amostral<\/strong>:<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"314\" height=\"60\" src=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/08\/15141441\/image-281.png\" alt=\"\" class=\"wp-image-836569\" srcset=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/08\/15141441\/image-281.png 314w, https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/08\/15141441\/image-281-300x57.png 300w\" sizes=\"auto, (max-width: 314px) 100vw, 314px\" \/><\/figure>\n\n\n\n<p>Ou seja, de fato, \u00e9 igual a 3.<\/p>\n\n\n\n<p><strong>Gabarito preliminar: Certo.<\/strong><\/p>\n\n\n\n<p><strong>Quest\u00e3o 145<\/strong>. Pelo crit\u00e9rio de m\u00e1xima verossimilhan\u00e7a, a estimativa do par\u00e2metro A \u00e9 igual a 3.<\/p>\n\n\n\n<p><strong>Coment\u00e1rios:<\/strong><\/p>\n\n\n\n<p>O estimador de m\u00e1xima verossimilhan\u00e7a para o par\u00e2metro da distribui\u00e7\u00e3o exponencial \u00e9 o inverso da m\u00e9dia amostral (vale lembrar que a distribui\u00e7\u00e3o exponencial indica o tempo entre ocorr\u00eancias):<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"92\" height=\"53\" src=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/08\/15141530\/image-282.png\" alt=\"\" class=\"wp-image-836570\" \/><\/figure>\n\n\n\n<p>Ou seja, \u00e9 diferente de 3.<\/p>\n\n\n\n<p><strong>Gabarito preliminar: Errado.<\/strong><\/p>\n\n\n\n<p><strong>Quest\u00e3o 146<\/strong>. A estimativa de m\u00e1xima verossimilhan\u00e7a para o desvio padr\u00e3o populacional \u00e9 igual a 3.<\/p>\n\n\n\n<p><strong>Coment\u00e1rios:<\/strong><\/p>\n\n\n\n<p>Em uma distribui\u00e7\u00e3o exponencial, o desvio padr\u00e3o \u00e9 o inverso do par\u00e2metro (assim como a m\u00e9dia da distribui\u00e7\u00e3o):<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"64\" height=\"52\" src=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/08\/15141636\/image-283.png\" alt=\"\" class=\"wp-image-836571\" \/><\/figure>\n\n\n\n<p>No item anterior, vimos que pela estimativa da m\u00e1xima verossimilhan\u00e7a para A \u00e9 \u00c2 = 1\/3, logo:<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"115\" height=\"66\" src=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/08\/15141708\/image-284.png\" alt=\"\" class=\"wp-image-836572\" \/><\/figure>\n\n\n\n<p>Ou seja, a estimativa para o desvio padr\u00e3o \u00e9, de fato, igual a 3.<\/p>\n\n\n\n<p><strong>Gabarito preliminar: Certo.<\/strong><strong><\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Oie!! Como foi a prova da SEFAZ-CE? Gostou? A seguir, apresento os coment\u00e1rios para as quest\u00f5es 141 a 146. Atente-se aos enunciados, pois a ordem das quest\u00f5es em cada bloco pode ser diferente na sua prova. Bloco de quest\u00f5es 141 \u2013 143 Suponha que o n\u00famero di\u00e1rio (X) de transa\u00e7\u00f5es banc\u00e1rias registradas em determinada conta [&hellip;]<\/p>\n","protected":false},"author":2036,"featured_media":789170,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"post_tipo":"article","footnotes":""},"categories":[1983],"tags":[8,661],"tax_estado":[219969],"class_list":["post-836553","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-fiscal-estadual","tag-recurso","tag-recursos","tax_estado-ce"],"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>Gabarito Extraoficial SEFAZ-CE - 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:\/\/www.estrategiaconcursos.com.br\/blog\/gabarito-sefaz-ce-estatistica\/\" \/>\n<meta property=\"og:locale\" content=\"pt_BR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Gabarito Extraoficial SEFAZ-CE - Estat\u00edstica\" \/>\n<meta property=\"og:description\" content=\"Oie!! Como foi a prova da SEFAZ-CE? Gostou? A seguir, apresento os coment\u00e1rios para as quest\u00f5es 141 a 146. Atente-se aos enunciados, pois a ordem das quest\u00f5es em cada bloco pode ser diferente na sua prova. 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