{"id":42,"date":"2023-08-06T06:39:02","date_gmt":"2023-08-06T06:39:02","guid":{"rendered":"https:\/\/statorials.org\/tr\/hipergeometrik-dagilim-1\/"},"modified":"2023-08-06T06:39:02","modified_gmt":"2023-08-06T06:39:02","slug":"hipergeometrik-dagilim-1","status":"publish","type":"post","link":"https:\/\/statorials.org\/tr\/hipergeometrik-dagilim-1\/","title":{"rendered":"Hipergeometrik da\u011f\u0131l\u0131m"},"content":{"rendered":"

Bu yaz\u0131da hipergeometrik da\u011f\u0131l\u0131m\u0131n ne oldu\u011funu ve bu t\u00fcr da\u011f\u0131l\u0131mla olas\u0131l\u0131\u011f\u0131n nas\u0131l hesapland\u0131\u011f\u0131n\u0131 a\u00e7\u0131klayaca\u011f\u0131z. \u00c7evrimi\u00e7i olarak hipergeometrik da\u011f\u0131l\u0131m\u0131n form\u00fcl\u00fcn\u00fc, \u00f6zelliklerinin neler oldu\u011funu ve ayr\u0131ca hipergeometrik da\u011f\u0131l\u0131m\u0131n olas\u0131l\u0131\u011f\u0131n\u0131 hesaplamak i\u00e7in bir hesap makinesini bulacaks\u0131n\u0131z. <\/p>\n

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

Hipergeometrik da\u011f\u0131l\u0131m,<\/strong> bir pop\u00fclasyondan n<\/em> \u00f6\u011fenin de\u011fi\u015ftirilmesi gerekmeden rastgele bir \u00e7\u0131karma i\u015flemindeki ba\u015far\u0131l\u0131 vakalar\u0131n say\u0131s\u0131n\u0131 tan\u0131mlayan bir olas\u0131l\u0131k da\u011f\u0131l\u0131m\u0131d\u0131r.<\/p>\n

Yani hipergeometrik da\u011f\u0131l\u0131m, bir pop\u00fclasyondan herhangi birini de\u011fi\u015ftirmeden n<\/em> \u00f6\u011fe \u00e7\u0131kar\u0131rken x<\/em> ba\u015far\u0131 elde etme olas\u0131l\u0131\u011f\u0131n\u0131 hesaplamak i\u00e7in kullan\u0131l\u0131r.<\/p>\n

Hipergeometrik da\u011f\u0131l\u0131m\u0131n \u00fc\u00e7 parametresi vard\u0131r:<\/p>\n

    \n
  • N<\/em><\/strong> : pop\u00fclasyondaki elementlerin say\u0131s\u0131d\u0131r (N = 0, 1, 2,\u2026).<\/li>\n
  • K<\/em><\/strong> : Maksimum ba\u015far\u0131 durumu say\u0131s\u0131d\u0131r (K = 0, 1, 2,\u2026,N). Hipergeometrik bir da\u011f\u0131l\u0131mda bir \u00f6\u011fe yaln\u0131zca “ba\u015far\u0131l\u0131” veya “ba\u015far\u0131s\u0131zl\u0131k” olarak de\u011ferlendirilebilece\u011finden, NK<\/em> maksimum ba\u015far\u0131s\u0131zl\u0131k durumu say\u0131s\u0131d\u0131r.<\/li>\n
  • n<\/em><\/strong> : ger\u00e7ekle\u015ftirilen de\u011fi\u015ftirilmeden getirme say\u0131s\u0131d\u0131r.<\/li>\n<\/ul>\n

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

    \u00d6rne\u011fin, N=8, K=5 ve n=3 parametreleriyle hipergeometrik da\u011f\u0131l\u0131ma sahip ayr\u0131k bir rastgele de\u011fi\u015fken X a\u015fa\u011f\u0131daki \u015fekilde tan\u0131mlan\u0131r: <\/p>\n<\/p>\n

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

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

    Hipergeometrik da\u011f\u0131l\u0131m form\u00fcl\u00fc,<\/strong> K<\/em> b\u00f6l\u00fc x’in<\/em> kombinatoryal say\u0131s\u0131n\u0131n, NK<\/em> b\u00f6l\u00fc nx’in<\/em> kombinatoryal say\u0131s\u0131 ile N<\/em> b\u00f6l\u00fc n’nin<\/em> kombinatoryal say\u0131s\u0131na b\u00f6l\u00fcnmesinin \u00e7arp\u0131m\u0131d\u0131r. <\/p>\n

    \n
    \"hipergeometrik<\/figure>\n<\/div>\n

    N’nin<\/em> pop\u00fclasyon b\u00fcy\u00fckl\u00fc\u011f\u00fc, K’n\u0131n<\/em> toplam olumlu durum say\u0131s\u0131, n’nin<\/em> yerine yenisi konulmayan \u00e7\u0131karmalar\u0131n say\u0131s\u0131 ve x’in<\/em> ise ger\u00e7ekle\u015fme olas\u0131l\u0131\u011f\u0131n\u0131n hesaplanmas\u0131 gereken olumlu durumlar\u0131n say\u0131s\u0131 oldu\u011fu durumlarda.<\/p>\n

    \ud83d\udc49Hipergeometrik da\u011f\u0131l\u0131ma uyan bir de\u011fi\u015fkenin olay olas\u0131l\u0131\u011f\u0131n\u0131 hesaplamak i\u00e7in a\u015fa\u011f\u0131daki hesaplay\u0131c\u0131y\u0131 kullanabilirsiniz.<\/u> <\/p>\n

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

    Hipergeometrik da\u011f\u0131l\u0131m\u0131n tan\u0131m\u0131n\u0131 ve form\u00fcl\u00fcn\u00fc g\u00f6rd\u00fckten sonra, \u015fimdi hipergeometrik da\u011f\u0131l\u0131m\u0131n olas\u0131l\u0131\u011f\u0131n\u0131 nas\u0131l hesaplayaca\u011f\u0131n\u0131z\u0131 bilmeniz i\u00e7in ad\u0131m ad\u0131m bir \u00f6rnek \u00e7\u00f6zece\u011fiz.<\/p>\n

      \n
    • Bir torbaya 20 adet mavi, 30 adet k\u0131rm\u0131z\u0131 top koyuyoruz yani torban\u0131n i\u00e7erisinde toplam 50 adet top bulunmaktad\u0131r. Hi\u00e7birini de\u011fi\u015ftirmeden 12 top \u00e7ekersek 4 mavi top \u00e7ekme olas\u0131l\u0131\u011f\u0131n\u0131 bulun.<\/li>\n<\/ul>\n

      Al\u0131\u015ft\u0131rmay\u0131 \u00e7\u00f6zmek i\u00e7in yapmam\u0131z gereken ilk \u015fey hipergeometrik da\u011f\u0131l\u0131m\u0131n parametrelerini belirlemektir. Bu durumda pop\u00fclasyondaki toplam eleman say\u0131s\u0131 50 ( N<\/em> =50), maksimum uygun durum say\u0131s\u0131 20 ( K<\/em> =20) ve 12 top \u00e7ekiliyor ( n<\/em> =12).<\/p>\n<\/p>\n

      \"\\left.\\begin{array}{c}N=50\\\\[2ex]K=20\\\\[2ex]n=12\\end{array}\\right\\}<\/p>\n<\/p>\n

      4 mavi top \u00e7ekme olas\u0131l\u0131\u011f\u0131n\u0131 hesaplamak istiyoruz ( x<\/em> =4), bu nedenle hipergeometrik da\u011f\u0131l\u0131m form\u00fcl\u00fcn\u00fc uyguluyoruz, de\u011fi\u015fkenleri kar\u015f\u0131l\u0131k gelen de\u011ferlerle de\u011fi\u015ftiriyoruz ve hesaplamay\u0131 ger\u00e7ekle\u015ftiriyoruz: <\/p>\n<\/p>\n

      \"P\\bigl[X=x\\bigr]=\\cfrac{\\begin{pmatrix}K\\\\x\\end{pmatrix}\\begin{pmatrix}N-K\\\\n-x\\end{pmatrix}}{\\begin{pmatrix}N\\\\n\\end{pmatrix}}\"<\/p>\n<\/p>\n

      \"\\begin{aligned}P\\bigl[X=4\\bigr]&=\\cfrac{\\begin{pmatrix}20\\\\4\\end{pmatrix}\\begin{pmatrix}50-20\\\\12-4\\end{pmatrix}}{\\begin{pmatrix}50\\\\12\\end{pmatrix}}<\/p>\n<\/p>\n

      <\/span> Hipergeometrik Da\u011f\u0131l\u0131m Hesaplay\u0131c\u0131<\/span><\/h2>\n

      \u0130stenilen olay\u0131n meydana gelme olas\u0131l\u0131\u011f\u0131n\u0131 hesaplamak i\u00e7in hipergeometrik da\u011f\u0131l\u0131m\u0131n parametrelerini a\u015fa\u011f\u0131daki \u00e7evrimi\u00e7i hesaplay\u0131c\u0131ya girin.<\/p>\n

      N’nin<\/em> pop\u00fclasyon b\u00fcy\u00fckl\u00fc\u011f\u00fc, K’n\u0131n<\/em> toplam olumlu vaka say\u0131s\u0131, n’nin<\/em> \u00f6rneklem b\u00fcy\u00fckl\u00fc\u011f\u00fc ve x’in<\/em> bunun ger\u00e7ekle\u015fme olas\u0131l\u0131\u011f\u0131n\u0131 bulmak istedi\u011fimiz de\u011fer oldu\u011funu unutmay\u0131n. <\/p>\n

      \n
        \n
      • \n

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

        = <\/span><\/li>\n

      • \n

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

        = <\/span><\/li>\n

      • \n

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

        = <\/span><\/li>\n

      • \n

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

        = <\/span><\/li>\n<\/ul>\n

        <\/div>\n<\/form>\n

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

        Hipergeometrik da\u011f\u0131l\u0131m a\u015fa\u011f\u0131daki \u00f6zelliklere sahiptir:<\/p>\n

          \n
        • Bir hipergeometrik da\u011f\u0131l\u0131m\u0131n beklenen de\u011feri,<\/strong> \u00f6rnekteki element say\u0131s\u0131 ile olumlu durumlar\u0131n toplam say\u0131s\u0131 \u00e7arp\u0131m\u0131n\u0131n pop\u00fclasyondaki element say\u0131s\u0131na b\u00f6l\u00fcnmesine e\u015fittir.<\/li>\n<\/ul>\n

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

            \n
          • Hipergeometrik da\u011f\u0131l\u0131m\u0131n modu,<\/strong> n+1<\/em> \u00e7arp\u0131 K+1<\/em> b\u00f6l\u00fc N+2’nin<\/em> \u00e7arp\u0131m\u0131ndan a\u015fa\u011f\u0131 yuvarlanan de\u011ferdir.<\/li>\n<\/ul>\n

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

              \n
            • Bir hipergeometrik da\u011f\u0131l\u0131m\u0131n varyans\u0131<\/strong> a\u015fa\u011f\u0131daki ifade kullan\u0131larak elde edilebilir:<\/li>\n<\/ul>\n

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

                \n
              • Hipergeometrik bir da\u011f\u0131l\u0131m\u0131n moment \u00fcreten fonksiyonu a\u015fa\u011f\u0131daki gibidir:<\/li>\n<\/ul>\n

                \"\\cfrac{{N-K<\/p>\n<\/p>\n

                  \n
                • Hipergeometrik da\u011f\u0131l\u0131m\u0131n karakteristik fonksiyonu a\u015fa\u011f\u0131daki gibidir:<\/li>\n<\/ul>\n

                  \"\\cfrac{{N-K<\/p>\n<\/p>\n

                    \n
                  • Belirli say\u0131da olay\u0131n meydana gelme olas\u0131l\u0131\u011f\u0131, hipergeometrik da\u011f\u0131l\u0131m i\u00e7in \u00f6zyineleme kullan\u0131larak \u00f6nceki say\u0131n\u0131n olas\u0131l\u0131\u011f\u0131ndan hesaplanabilir: <\/li>\n<\/ul>\n

                    \"P[X=x+1]=\\cfrac{(K-x)(n-x)}{(x+1)(N-K-n+x-1)}\\cdot<\/p>\n<\/p>\n

                    <\/span> Hipergeometrik da\u011f\u0131l\u0131m ve binom da\u011f\u0131l\u0131m\u0131<\/span><\/h2>\n

                    Hipergeometrik da\u011f\u0131l\u0131m ile binom da\u011f\u0131l\u0131m\u0131 aras\u0131ndaki fark<\/strong> yer de\u011fi\u015ftirmedir. Hipergeometrik da\u011f\u0131l\u0131m, al\u0131mlar de\u011fi\u015ftirilmedi\u011finde kullan\u0131l\u0131r, ancak binom da\u011f\u0131l\u0131m\u0131nda al\u0131mlar de\u011fi\u015ftirilir.<\/p>\n

                    \u00d6rne\u011fin, bir desteye rastgele be\u015f kart \u00e7ekiyorsak ve belirli bir kart\u0131n \u00e7\u0131kma olas\u0131l\u0131\u011f\u0131n\u0131 hesaplamak istiyorsak, \u00e7ekti\u011fimiz her kart\u0131 de\u011fi\u015ftirmiyorsak, hesaplamay\u0131 yapmak i\u00e7in hipergeometrik da\u011f\u0131l\u0131m\u0131 kullanmam\u0131z gerekir. Ancak bir kart\u0131 \u00e7\u0131kar\u0131rken bir sonraki \u00e7\u0131karma i\u015flemini ger\u00e7ekle\u015ftirmeden \u00f6nce geri koyarsak, olas\u0131l\u0131\u011f\u0131 hesaplamak i\u00e7in binom da\u011f\u0131l\u0131m\u0131n\u0131 kullanmam\u0131z gerekir.<\/p>\n

                    N<\/em> say\u0131s\u0131 b\u00fcy\u00fck, n\/N<\/em> oran\u0131 k\u00fc\u00e7\u00fck ve istenen olumlu durumlar\u0131n say\u0131s\u0131 \u00e7ok k\u00fc\u00e7\u00fck oldu\u011funda, hipergeometrik da\u011f\u0131l\u0131m\u0131 binom da\u011f\u0131l\u0131m\u0131n\u0131n bir yakla\u015f\u0131m\u0131 olarak kullanabiliriz. Ancak sonucun o kadar g\u00fcvenilir olmayaca\u011f\u0131 ve \u00fcstelik binom yasas\u0131yla olas\u0131l\u0131klar\u0131 hesaplaman\u0131n hipergeometri yasas\u0131na g\u00f6re hesaplamaktan daha kolay olmas\u0131 nedeniyle bunu \u00f6nermiyorum.<\/p>\n","protected":false},"excerpt":{"rendered":"

                    Bu yaz\u0131da hipergeometrik da\u011f\u0131l\u0131m\u0131n ne oldu\u011funu ve bu t\u00fcr da\u011f\u0131l\u0131mla olas\u0131l\u0131\u011f\u0131n nas\u0131l hesapland\u0131\u011f\u0131n\u0131 a\u00e7\u0131klayaca\u011f\u0131z. \u00c7evrimi\u00e7i olarak hipergeometrik da\u011f\u0131l\u0131m\u0131n form\u00fcl\u00fcn\u00fc, \u00f6zelliklerinin neler oldu\u011funu ve ayr\u0131ca hipergeometrik da\u011f\u0131l\u0131m\u0131n olas\u0131l\u0131\u011f\u0131n\u0131 hesaplamak i\u00e7in bir hesap makinesini bulacaks\u0131n\u0131z. Hipergeometrik da\u011f\u0131l\u0131m nedir? Hipergeometrik da\u011f\u0131l\u0131m, bir pop\u00fclasyondan n \u00f6\u011fenin de\u011fi\u015ftirilmesi gerekmeden rastgele bir \u00e7\u0131karma i\u015flemindeki ba\u015far\u0131l\u0131 vakalar\u0131n say\u0131s\u0131n\u0131 tan\u0131mlayan bir olas\u0131l\u0131k da\u011f\u0131l\u0131m\u0131d\u0131r. […]<\/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-42","post","type-post","status-publish","format-standard","hentry","category-olasilik"],"yoast_head":"\nHipergeometrik da\u011f\u0131l\u0131m: form\u00fcl, \u00f6rnek, hesap makinesi,...<\/title>\n<meta name=\"description\" content=\"Hipergeometrik da\u011f\u0131l\u0131m\u0131n ne oldu\u011funu ve t\u00fcm \u00f6zelliklerini a\u00e7\u0131kl\u0131yoruz. \u00d6rnekler ve hipergeometrik da\u011f\u0131l\u0131m hesaplay\u0131c\u0131s\u0131 ile.\" \/>\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\/hipergeometrik-dagilim-1\/\" \/>\n<meta property=\"og:locale\" content=\"tr_TR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Hipergeometrik da\u011f\u0131l\u0131m: form\u00fcl, \u00f6rnek, hesap makinesi,...\" \/>\n<meta property=\"og:description\" content=\"Hipergeometrik da\u011f\u0131l\u0131m\u0131n ne oldu\u011funu ve t\u00fcm \u00f6zelliklerini a\u00e7\u0131kl\u0131yoruz. \u00d6rnekler ve hipergeometrik da\u011f\u0131l\u0131m hesaplay\u0131c\u0131s\u0131 ile.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/statorials.org\/tr\/hipergeometrik-dagilim-1\/\" \/>\n<meta property=\"og:site_name\" content=\"Statorials\" \/>\n<meta property=\"article:published_time\" content=\"2023-08-06T06:39:02+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/statorials.org\/wp-content\/ql-cache\/quicklatex.com-bd43d7c14739c66e63b224abf6cc20b3_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=\"4 dakika\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"WebPage\",\"@id\":\"https:\/\/statorials.org\/tr\/hipergeometrik-dagilim-1\/\",\"url\":\"https:\/\/statorials.org\/tr\/hipergeometrik-dagilim-1\/\",\"name\":\"Hipergeometrik da\u011f\u0131l\u0131m: form\u00fcl, \u00f6rnek, hesap makinesi,...\",\"isPartOf\":{\"@id\":\"https:\/\/statorials.org\/tr\/#website\"},\"datePublished\":\"2023-08-06T06:39:02+00:00\",\"dateModified\":\"2023-08-06T06:39:02+00:00\",\"author\":{\"@id\":\"https:\/\/statorials.org\/tr\/#\/schema\/person\/365dc158a39a7c8ae256355451e3de48\"},\"description\":\"Hipergeometrik da\u011f\u0131l\u0131m\u0131n ne oldu\u011funu ve t\u00fcm \u00f6zelliklerini a\u00e7\u0131kl\u0131yoruz. \u00d6rnekler ve hipergeometrik da\u011f\u0131l\u0131m hesaplay\u0131c\u0131s\u0131 ile.\",\"breadcrumb\":{\"@id\":\"https:\/\/statorials.org\/tr\/hipergeometrik-dagilim-1\/#breadcrumb\"},\"inLanguage\":\"tr\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/statorials.org\/tr\/hipergeometrik-dagilim-1\/\"]}]},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/statorials.org\/tr\/hipergeometrik-dagilim-1\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Ev\",\"item\":\"https:\/\/statorials.org\/tr\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"Hipergeometrik da\u011f\u0131l\u0131m\"}]},{\"@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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