{"id":222,"date":"2023-08-04T00:49:55","date_gmt":"2023-08-04T00:49:55","guid":{"rendered":"https:\/\/statorials.org\/tr\/bernoulli-dagilimi\/"},"modified":"2023-08-04T00:49:55","modified_gmt":"2023-08-04T00:49:55","slug":"bernoulli-dagilimi","status":"publish","type":"post","link":"https:\/\/statorials.org\/tr\/bernoulli-dagilimi\/","title":{"rendered":"Bernoulli da\u011f\u0131l\u0131m\u0131"},"content":{"rendered":"

Bu makalede Bernoulli da\u011f\u0131l\u0131m\u0131n\u0131n ne oldu\u011fu ve form\u00fcl\u00fcn\u00fcn ne oldu\u011fu a\u00e7\u0131klanmaktad\u0131r. Ek olarak Bernoulli da\u011f\u0131l\u0131m\u0131n\u0131n \u00f6zelliklerini ve anlam\u0131n\u0131 daha iyi anlaman\u0131z\u0131 sa\u011flayacak \u00e7\u00f6z\u00fclm\u00fc\u015f bir al\u0131\u015ft\u0131rmay\u0131 bulacaks\u0131n\u0131z. <\/p>\n

<\/span> Bernoulli da\u011f\u0131l\u0131m\u0131 nedir?<\/span><\/h2>\n

\u0130kili da\u011f\u0131l\u0131m<\/strong> olarak da bilinen Bernoulli da\u011f\u0131l\u0131m\u0131<\/strong> , yaln\u0131zca iki sonuca sahip olabilen ayr\u0131 bir de\u011fi\u015fkeni temsil eden bir olas\u0131l\u0131k da\u011f\u0131l\u0131m\u0131d\u0131r: “ba\u015far\u0131” veya “ba\u015far\u0131s\u0131zl\u0131k”.<\/p>\n

Bernoulli da\u011f\u0131l\u0131m\u0131nda “ba\u015far\u0131” bekledi\u011fimiz sonu\u00e7tur ve 1 de\u011ferine sahipken, “ba\u015far\u0131s\u0131zl\u0131k” sonucu beklenenin d\u0131\u015f\u0131nda bir sonu\u00e7tur ve 0 de\u011ferine sahiptir. ba\u015far\u0131\u201d p<\/em> , \u201cba\u015far\u0131s\u0131zl\u0131k\u201d sonucunun olas\u0131l\u0131\u011f\u0131 q=1-p’dir<\/em> .<\/p>\n<\/p>\n

\"\\begin{array}{c}X\\sim<\/p>\n<\/p>\n

Bernoulli da\u011f\u0131l\u0131m\u0131 ismini \u0130svi\u00e7reli istatistik\u00e7i Jacob Bernoulli’den alm\u0131\u015ft\u0131r.<\/p>\n

\u0130statistikte, Bernoulli da\u011f\u0131l\u0131m\u0131n\u0131n esas olarak tek bir uygulamas\u0131 vard\u0131r: yaln\u0131zca iki olas\u0131 sonucun (ba\u015far\u0131 ve ba\u015far\u0131s\u0131zl\u0131k) oldu\u011fu deneylerin olas\u0131l\u0131klar\u0131n\u0131 tan\u0131mlamak. Dolay\u0131s\u0131yla Bernoulli da\u011f\u0131l\u0131m\u0131n\u0131 kullanan bir deneye Bernoulli testi veya Bernoulli deneyi denir. <\/p>\n

<\/span> Bernoulli da\u011f\u0131l\u0131m form\u00fcl\u00fc<\/span><\/h2>\n

E\u011fer p<\/em> , “ba\u015far\u0131” sonucunun ortaya \u00e7\u0131kma olas\u0131l\u0131\u011f\u0131 ise, Bernoulli da\u011f\u0131l\u0131m\u0131n\u0131n olas\u0131l\u0131\u011f\u0131 p’nin<\/em> x’e<\/em> y\u00fckseltilmesiyle 1-p’nin<\/em> 1-x’e<\/em> y\u00fckseltilmesine e\u015fittir. B\u00f6ylece Bernoulli da\u011f\u0131l\u0131m\u0131n\u0131n olas\u0131l\u0131klar\u0131 a\u015fa\u011f\u0131daki form\u00fcl kullan\u0131larak hesaplanabilir<\/strong> : <\/p>\n

\"Bernoulli<\/figure>\n

Bernoulli da\u011f\u0131l\u0131m\u0131nda x’in<\/em> de\u011ferinin yaln\u0131zca 0 (ba\u015far\u0131s\u0131zl\u0131k) veya 1 (ba\u015far\u0131) olabilece\u011fini unutmay\u0131n.<\/p>\n

\u00d6te yandan \u00f6nceki form\u00fcl a\u015fa\u011f\u0131daki e\u015fde\u011fer ifade kullan\u0131larak da yaz\u0131labilir: <\/p>\n<\/p>\n

\"<\/p>\n<\/p>\n

<\/span> Bernoulli da\u011f\u0131l\u0131m\u0131 \u00f6rne\u011fi<\/span><\/h2>\n

Art\u0131k Bernoulli da\u011f\u0131l\u0131m\u0131n\u0131n tan\u0131m\u0131n\u0131 ve form\u00fcl\u00fcn\u00fcn ne oldu\u011funu bildi\u011fimize g\u00f6re, Bernoulli da\u011f\u0131l\u0131m\u0131n\u0131n somut bir \u00f6rne\u011fini g\u00f6relim.<\/p>\n

    \n
  • Bir oyunu kazanmak i\u00e7in bir oyuncunun zar atmas\u0131 ve 2 almas\u0131 gerekir, aksi takdirde ba\u015fka bir oyuncu oyunu kazan\u0131r ve dolay\u0131s\u0131yla oyun kaybedilir. Ba\u015far\u0131 ve ba\u015far\u0131s\u0131zl\u0131k olas\u0131l\u0131\u011f\u0131n\u0131 hesaplay\u0131n.<\/li>\n<\/ul>\n

    Bir zar\u0131n alt\u0131 olas\u0131 sonucu vard\u0131r (1, 2, 3, 4, 5, 6), dolay\u0131s\u0131yla bu durumda deneyin \u00f6rnek uzay\u0131 \u015f\u00f6yledir:<\/p>\n<\/p>\n

    \"\\Omega=\\{1,2,3,4,5,6\\}\"<\/p>\n<\/p>\n

    Bizim durumumuzda, ba\u015far\u0131n\u0131n tek durumu iki numaray\u0131 elde etmektir, dolay\u0131s\u0131yla Laplace kural\u0131n\u0131 uygularken ba\u015far\u0131 olas\u0131l\u0131\u011f\u0131 birin olas\u0131 sonu\u00e7lar\u0131n toplam say\u0131s\u0131na b\u00f6l\u00fcnmesine e\u015fittir (6):<\/p>\n<\/p>\n

    \"p=\\cfrac{1}{6}=0,1667\"<\/p>\n<\/p>\n

    \u00d6te yandan, zar at\u0131ld\u0131\u011f\u0131nda ba\u015fka bir say\u0131 ortaya \u00e7\u0131karsa, oyuncu oyunu kaybedece\u011fi i\u00e7in deneyin sonucu ba\u015far\u0131s\u0131zl\u0131k olarak kabul edilecektir. Dolay\u0131s\u0131yla bu olas\u0131l\u0131k, daha \u00f6nce hesaplanan olas\u0131l\u0131\u011f\u0131n bir eksi\u011fine e\u015fittir:<\/p>\n<\/p>\n

    \"q=1-p=1-\\cfrac{1}{6}=\\cfrac{5}{6}=0,8333\"<\/p>\n<\/p>\n

    K\u0131saca bu deneyin Bernoulli da\u011f\u0131l\u0131m\u0131 a\u015fa\u011f\u0131daki ifadeyle tan\u0131mlan\u0131r:<\/p>\n<\/p>\n

    \"<\/p>\n<\/p>\n

    A\u015fa\u011f\u0131da g\u00f6rebilece\u011finiz gibi Bernoulli da\u011f\u0131l\u0131m\u0131n\u0131n olas\u0131l\u0131klar\u0131 yukar\u0131da g\u00f6r\u00fclen form\u00fcl\u00fcn uygulanmas\u0131yla da bulunabilir: <\/p>\n<\/p>\n

    \"P[X=x]=p^x\\cdot<\/p>\n<\/p>\n

    \"\\displaystyle<\/p>\n<\/p>\n

    \"\\displaystyle<\/p>\n<\/p>\n

    <\/span> Bernoulli da\u011f\u0131l\u0131m\u0131n\u0131n \u00f6zellikleri<\/span><\/h2>\n

    A\u015fa\u011f\u0131da Bernoulli da\u011f\u0131l\u0131m\u0131n\u0131n en \u00f6nemli \u00f6zellikleri verilmi\u015ftir.<\/p>\n

      \n
    • Bernoulli da\u011f\u0131l\u0131m\u0131 yaln\u0131zca 1 (ba\u015far\u0131l\u0131) veya 0 (ba\u015far\u0131s\u0131z) de\u011ferini alabilir.<\/li>\n<\/ul>\n

      \"x=\\{0\\<\/p>\n<\/p>\n

        \n
      • Bernoulli da\u011f\u0131l\u0131m\u0131n\u0131n ortalamas\u0131 \u201cba\u015far\u0131\u201d sonucunun ortaya \u00e7\u0131kma olas\u0131l\u0131\u011f\u0131na e\u015fde\u011ferdir.<\/li>\n<\/ul>\n

        \"E[X]=p\"<\/p>\n<\/p>\n

          \n
        • Bernoulli da\u011f\u0131l\u0131m\u0131n\u0131n varyans\u0131, \u201cba\u015far\u0131\u201d ve \u201cba\u015far\u0131s\u0131zl\u0131k\u201d sonucunun ortaya \u00e7\u0131kma olas\u0131l\u0131klar\u0131n\u0131n \u00e7arp\u0131lmas\u0131yla hesaplanabilir. Veya e\u015fde\u011fer olarak varyans p<\/em> \u00e7arp\u0131 1-p’dir<\/em> .<\/li>\n<\/ul>\n

          \"Var(X)=p\\cdot<\/p>\n<\/p>\n

            \n
          • Bernoulli da\u011f\u0131l\u0131m\u0131n\u0131n modunun de\u011feri \u201cba\u015far\u0131\u201d ve \u201cba\u015far\u0131s\u0131zl\u0131k\u201d olas\u0131l\u0131klar\u0131na ba\u011fl\u0131d\u0131r. Dolay\u0131s\u0131yla bu t\u00fcr da\u011f\u0131t\u0131m\u0131n modu a\u015fa\u011f\u0131daki ifadeyle tan\u0131mlan\u0131r:<\/li>\n<\/ul>\n
             *** QuickLaTeX cannot compile formula:\n\\displaystyle Mo=\\left\\{\\begin{array}{ll}0 & \\text{si } q>p\\\\[2ex]0 \\ ;1 & \\text{si } q=p\\\\[2ex] 1 & \\text{si } q<ul><li> The formula for the probability function of a Bernoulli distribution is as follows:<\/li><\/ul>[latex] \\displaystyle P[X=x]= \\left\\{\\begin{array}{ll}1-p & \\text{si } x=0\\\\[2ex]p& \\text{si } x=1\\end{array}\\right.\n\n*** Error message:\nMissing $ inserted.\nleading text: \\displaystyle\nPlease use \\mathaccent for accents in math mode.\nleading text: ...> The formula for the probability function\n\\begin{array} on input line 8 ended by \\end{document}.\nleading text: \\end{document}\nImproper \\prevdepth.\nleading text: \\end{document}\nMissing $ inserted.\nleading text: \\end{document}\nMissing } inserted.\nleading text: \\end{document}\nMissing \\cr inserted.\nleading text: \\end{document}\nMissing $ inserted.\nleading text: \\end{document}\nYou can't use `\\end' in internal vertical mode.\nleading text: \\end{document}\n\\begin{array} on input line 8 ended by \\end{document}.\nleading text: \\end{document}\nMissing } inserted.\nleading text: \\end{document}\nEmergency stop.\n\n<\/pre>\n
              \n
            •  \u00d6te yandan Bernoulli da\u011f\u0131l\u0131m\u0131n\u0131n k\u00fcm\u00fclatif olas\u0131l\u0131k fonksiyonu a\u015fa\u011f\u0131daki ifadeyle tan\u0131mlan\u0131r:<\/li>\n<\/ul>\n
              \"<\/p>\n<\/p>\n
                \n
              •  Bir Bernoulli da\u011f\u0131l\u0131m\u0131n\u0131n asimetri katsay\u0131s\u0131 a\u015fa\u011f\u0131daki ifadeyle hesaplan\u0131r:<\/li>\n<\/ul>\n
                \"A=\\cfrac{q-p}{\\sqrt{p\\cdot<\/p>\n<\/p>\n
                  \n
                •  Benzer \u015fekilde Bernoulli da\u011f\u0131l\u0131m\u0131n\u0131n bas\u0131kl\u0131\u011f\u0131 p<\/em> parametresinin de\u011ferine ba\u011fl\u0131d\u0131r ve a\u015fa\u011f\u0131daki form\u00fcl\u00fcn uygulanmas\u0131yla bulunabilir: <\/li>\n<\/ul>\n
                  \"C=\\cfrac{3p^2-3p+1}{p(1-p)}\"<\/p>\n<\/p>\n
                  <\/span> Bernoulli da\u011f\u0131l\u0131m\u0131 ve binom da\u011f\u0131l\u0131m\u0131<\/span><\/h2>\n
                   Bu b\u00f6l\u00fcmde Bernoulli da\u011f\u0131l\u0131m\u0131 ile binom da\u011f\u0131l\u0131m\u0131 aras\u0131ndaki fark\u0131 g\u00f6rece\u011fiz \u00e7\u00fcnk\u00fc bunlar iki t\u00fcr ilgili olas\u0131l\u0131k da\u011f\u0131l\u0131m\u0131d\u0131r.<\/p>\n
                   Binom da\u011f\u0131l\u0131m\u0131,<\/strong> bir dizi Bernoulli denemesinden elde edilen “ba\u015far\u0131l\u0131” sonu\u00e7lar\u0131n say\u0131s\u0131n\u0131 sayar. Bu Bernoulli deneyleri ba\u011f\u0131ms\u0131z olmal\u0131 ancak ayn\u0131 ba\u015far\u0131 olas\u0131l\u0131\u011f\u0131na sahip olmal\u0131d\u0131r.<\/p>\n
                   Bu nedenle, binom da\u011f\u0131l\u0131m\u0131 Bernoulli da\u011f\u0131l\u0131m\u0131n\u0131 takip eden ve t\u00fcm\u00fc ayn\u0131 p<\/em> parametresi ile tan\u0131mlanan bir dizi de\u011fi\u015fkenin toplam\u0131d\u0131r<\/strong> .<\/p>\n<\/p>\n
                  \"\\begin{array}{c}X_i\\sim<\/p>\n<\/p>\n
                   Yani Bernoulli da\u011f\u0131l\u0131m\u0131nda yaln\u0131zca bir Bernoulli deneyi varken, binom da\u011f\u0131l\u0131m\u0131nda bir dizi Bernoulli deneyi var.<\/p>\n","protected":false},"excerpt":{"rendered":"
                  Bu makalede Bernoulli da\u011f\u0131l\u0131m\u0131n\u0131n ne oldu\u011fu ve form\u00fcl\u00fcn\u00fcn ne oldu\u011fu a\u00e7\u0131klanmaktad\u0131r. Ek olarak Bernoulli da\u011f\u0131l\u0131m\u0131n\u0131n \u00f6zelliklerini ve anlam\u0131n\u0131 daha iyi anlaman\u0131z\u0131 sa\u011flayacak \u00e7\u00f6z\u00fclm\u00fc\u015f bir al\u0131\u015ft\u0131rmay\u0131 bulacaks\u0131n\u0131z. Bernoulli da\u011f\u0131l\u0131m\u0131 nedir? \u0130kili da\u011f\u0131l\u0131m olarak da bilinen Bernoulli da\u011f\u0131l\u0131m\u0131 , yaln\u0131zca iki sonuca sahip olabilen ayr\u0131 bir de\u011fi\u015fkeni temsil eden bir olas\u0131l\u0131k da\u011f\u0131l\u0131m\u0131d\u0131r: “ba\u015far\u0131” veya “ba\u015far\u0131s\u0131zl\u0131k”. Bernoulli da\u011f\u0131l\u0131m\u0131nda […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[12],"tags":[],"class_list":["post-222","post","type-post","status-publish","format-standard","hentry","category-olasilik"],"yoast_head":"\n\u25b7 Bernoulli Da\u011f\u0131l\u0131m\u0131<\/title>\n<meta name=\"description\" content=\"Burada Bernoulli da\u011f\u0131l\u0131m\u0131n\u0131n ne oldu\u011funu, form\u00fcl\u00fcn\u00fc, Bernoulli da\u011f\u0131l\u0131m\u0131n\u0131n \u00f6zelliklerini ve somut bir \u00f6rne\u011fini bulacaks\u0131n\u0131z.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/statorials.org\/tr\/bernoulli-dagilimi\/\" \/>\n<meta property=\"og:locale\" content=\"tr_TR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"\u25b7 Bernoulli Da\u011f\u0131l\u0131m\u0131\" \/>\n<meta property=\"og:description\" content=\"Burada Bernoulli da\u011f\u0131l\u0131m\u0131n\u0131n ne oldu\u011funu, form\u00fcl\u00fcn\u00fc, Bernoulli da\u011f\u0131l\u0131m\u0131n\u0131n \u00f6zelliklerini ve somut bir \u00f6rne\u011fini bulacaks\u0131n\u0131z.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/statorials.org\/tr\/bernoulli-dagilimi\/\" \/>\n<meta property=\"og:site_name\" content=\"Statorials\" \/>\n<meta property=\"article:published_time\" content=\"2023-08-04T00:49:55+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/statorials.org\/wp-content\/ql-cache\/quicklatex.com-384fd7d96d4d6584739b04a6e331b251_l3.png\" \/>\n<meta name=\"author\" content=\"Dr.benjamin anderson\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Yazan:\" \/>\n\t<meta name=\"twitter:data1\" content=\"Dr.benjamin anderson\" \/>\n\t<meta name=\"twitter:label2\" content=\"Tahmini okuma s\u00fcresi\" \/>\n\t<meta name=\"twitter:data2\" content=\"5 dakika\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"WebPage\",\"@id\":\"https:\/\/statorials.org\/tr\/bernoulli-dagilimi\/\",\"url\":\"https:\/\/statorials.org\/tr\/bernoulli-dagilimi\/\",\"name\":\"\u25b7 Bernoulli Da\u011f\u0131l\u0131m\u0131\",\"isPartOf\":{\"@id\":\"https:\/\/statorials.org\/tr\/#website\"},\"datePublished\":\"2023-08-04T00:49:55+00:00\",\"dateModified\":\"2023-08-04T00:49:55+00:00\",\"author\":{\"@id\":\"https:\/\/statorials.org\/tr\/#\/schema\/person\/365dc158a39a7c8ae256355451e3de48\"},\"description\":\"Burada Bernoulli da\u011f\u0131l\u0131m\u0131n\u0131n ne oldu\u011funu, form\u00fcl\u00fcn\u00fc, Bernoulli da\u011f\u0131l\u0131m\u0131n\u0131n \u00f6zelliklerini ve somut bir \u00f6rne\u011fini bulacaks\u0131n\u0131z.\",\"breadcrumb\":{\"@id\":\"https:\/\/statorials.org\/tr\/bernoulli-dagilimi\/#breadcrumb\"},\"inLanguage\":\"tr\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/statorials.org\/tr\/bernoulli-dagilimi\/\"]}]},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/statorials.org\/tr\/bernoulli-dagilimi\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Ev\",\"item\":\"https:\/\/statorials.org\/tr\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"Bernoulli da\u011f\u0131l\u0131m\u0131\"}]},{\"@type\":\"WebSite\",\"@id\":\"https:\/\/statorials.org\/tr\/#website\",\"url\":\"https:\/\/statorials.org\/tr\/\",\"name\":\"Statorials\",\"description\":\"\u0130statistik okuryazarl\u0131\u011f\u0131 rehberiniz!\",\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/statorials.org\/tr\/?s={search_term_string}\"},\"query-input\":\"required name=search_term_string\"}],\"inLanguage\":\"tr\"},{\"@type\":\"Person\",\"@id\":\"https:\/\/statorials.org\/tr\/#\/schema\/person\/365dc158a39a7c8ae256355451e3de48\",\"name\":\"Dr.benjamin anderson\",\"image\":{\"@type\":\"ImageObject\",\"inLanguage\":\"tr\",\"@id\":\"https:\/\/statorials.org\/tr\/#\/schema\/person\/image\/\",\"url\":\"https:\/\/statorials.org\/tr\/wp-content\/uploads\/2023\/10\/Dr.-Benjamin-Anderson-96x96.jpg\",\"contentUrl\":\"https:\/\/statorials.org\/tr\/wp-content\/uploads\/2023\/10\/Dr.-Benjamin-Anderson-96x96.jpg\",\"caption\":\"Dr.benjamin anderson\"},\"description\":\"Merhaba, ben Benjamin, emekli bir istatistik profes\u00f6r\u00fc ve Statorials \u00f6\u011fretmenine d\u00f6n\u00fc\u015ft\u00fcm. \u0130statistik alan\u0131ndaki kapsaml\u0131 deneyimim ve uzmanl\u0131\u011f\u0131mla, \u00f6\u011frencilerimi Statorials arac\u0131l\u0131\u011f\u0131yla g\u00fc\u00e7lendirmek i\u00e7in bilgilerimi payla\u015fmaya can at\u0131yorum. 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