{"id":215,"date":"2023-08-04T02:51:27","date_gmt":"2023-08-04T02:51:27","guid":{"rendered":"https:\/\/statorials.org\/tr\/cok-terimli-dagilim-1\/"},"modified":"2023-08-04T02:51:27","modified_gmt":"2023-08-04T02:51:27","slug":"cok-terimli-dagilim-1","status":"publish","type":"post","link":"https:\/\/statorials.org\/tr\/cok-terimli-dagilim-1\/","title":{"rendered":"\u00c7ok terimli da\u011f\u0131l\u0131m"},"content":{"rendered":"

Bu makale istatistikte \u00e7ok terimli da\u011f\u0131l\u0131m\u0131n ne oldu\u011funu a\u00e7\u0131klamaktad\u0131r. B\u00f6ylece, \u00e7ok terimli bir da\u011f\u0131l\u0131m\u0131n tan\u0131m\u0131n\u0131, form\u00fcl\u00fcn\u00fcn ne oldu\u011funu, \u00e7\u00f6z\u00fclm\u00fc\u015f bir al\u0131\u015ft\u0131rmay\u0131 ve bu t\u00fcr olas\u0131l\u0131k da\u011f\u0131l\u0131m\u0131n\u0131n \u00f6zelliklerinin neler oldu\u011funu bulacaks\u0131n\u0131z. Ek olarak, \u00e7evrimi\u00e7i bir hesap makinesiyle \u00e7ok terimli bir da\u011f\u0131l\u0131m\u0131n olas\u0131l\u0131\u011f\u0131n\u0131 hesaplayabileceksiniz. <\/p>\n

<\/span> \u00c7ok terimli da\u011f\u0131l\u0131m nedir?<\/span><\/h2>\n

\u00c7ok terimli da\u011f\u0131l\u0131m<\/strong> (veya \u00e7ok terimli da\u011f\u0131l\u0131m<\/strong> ), birbirini d\u0131\u015flayan birka\u00e7 olay\u0131n birka\u00e7 denemeden sonra belirli say\u0131da meydana gelme olas\u0131l\u0131\u011f\u0131n\u0131 tan\u0131mlayan bir olas\u0131l\u0131k da\u011f\u0131l\u0131m\u0131d\u0131r.<\/p>\n

Yani, rastgele bir deney \u00fc\u00e7 veya daha fazla \u00f6zel olayla sonu\u00e7lanabiliyorsa ve her olay\u0131n ayr\u0131 ayr\u0131 meydana gelme olas\u0131l\u0131\u011f\u0131 biliniyorsa, \u00e7ok terimli da\u011f\u0131l\u0131m, birden fazla deney ger\u00e7ekle\u015ftirildi\u011finde belirli say\u0131da olay\u0131n meydana gelme olas\u0131l\u0131\u011f\u0131n\u0131 hesaplamak i\u00e7in kullan\u0131l\u0131r. her zaman.<\/p>\n

Dolay\u0131s\u0131yla \u00e7ok terimli da\u011f\u0131l\u0131m, binom da\u011f\u0131l\u0131m\u0131n\u0131n bir genellemesidir. <\/p>\n

<\/span> \u00c7ok terimli da\u011f\u0131l\u0131m form\u00fcl\u00fc<\/span><\/h2>\n

\u00c7ok terimli bir da\u011f\u0131l\u0131m olas\u0131l\u0131\u011f\u0131n\u0131 hesaplamak i\u00e7in,<\/strong> \u00f6ncelikle toplam veri say\u0131s\u0131n\u0131n fakt\u00f6riyeli ile her bir olay\u0131n meydana gelme say\u0131s\u0131n\u0131n fakt\u00f6riyelleri aras\u0131ndaki b\u00f6l\u00fcm\u00fc belirlemeniz gerekir ve sonu\u00e7, her olay\u0131n olas\u0131l\u0131\u011f\u0131n\u0131n \u00e7arp\u0131m\u0131 ile \u00e7arp\u0131l\u0131r. s\u00f6z konusu olay\u0131n meydana gelme say\u0131s\u0131na getirildi.<\/p>\n

Ba\u015fka bir deyi\u015fle multinomial da\u011f\u0131l\u0131m\u0131n form\u00fcl\u00fc<\/strong> \u015fu \u015fekildedir: <\/p>\n

\"\u00e7ok<\/figure>\n

Alt\u0131n:<\/p>\n

    \n
  • \n

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

    hesaplanan \u00e7ok terimli da\u011f\u0131l\u0131m\u0131n olas\u0131l\u0131\u011f\u0131d\u0131r.<\/li>\n

  • \n

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

    ger\u00e7ekle\u015ftirilen toplam test say\u0131s\u0131d\u0131r.<\/li>\n

  • \n

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

    olay\u0131n meydana gelme say\u0131s\u0131d\u0131r<\/p>\n

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

    .<\/li>\n

  • \n

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

    olay\u0131n meydana gelme olas\u0131l\u0131\u011f\u0131<\/p>\n

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

    .<\/li>\n<\/ul>\n

    \ud83d\udc49 Multinomial da\u011f\u0131l\u0131ma uyan bir de\u011fi\u015fkenin olas\u0131l\u0131\u011f\u0131n\u0131 hesaplamak i\u00e7in a\u015fa\u011f\u0131daki hesaplay\u0131c\u0131y\u0131 kullanabilirsiniz.<\/u> <\/p>\n

    <\/span> \u00c7ok Terimli Da\u011f\u0131l\u0131m \u00d6rne\u011fi<\/span><\/h2>\n

    \u00c7ok terimli da\u011f\u0131l\u0131m kavram\u0131n\u0131 anlamay\u0131 tamamlamak i\u00e7in a\u015fa\u011f\u0131da \u00e7ok terimli bir da\u011f\u0131l\u0131m\u0131n olas\u0131l\u0131\u011f\u0131n\u0131 hesaplamaya ili\u015fkin bir \u00f6rne\u011fi \u00e7\u00f6zd\u00fcn\u00fcz.<\/p>\n

      \n
    • Bir ma\u011fazada \u00fc\u00e7 farkl\u0131 \u00fcr\u00fcn sat\u0131l\u0131yor. Bir m\u00fc\u015fteri bir sat\u0131n alma i\u015flemi ger\u00e7ekle\u015ftirdi\u011finde bunun A \u00fcr\u00fcn\u00fc, B \u00fcr\u00fcn\u00fc veya C \u00fcr\u00fcn\u00fc olma olas\u0131l\u0131\u011f\u0131 s\u0131ras\u0131yla %30, %15 ve %55’tir. Ma\u011fazada 8 adet sat\u0131ld\u0131\u011f\u0131nda, 2 tanesinin A \u00fcr\u00fcn\u00fcnden, 1 tanesinin B \u00fcr\u00fcn\u00fcnden ve 5 tanesinin C \u00fcr\u00fcn\u00fcnden olma olas\u0131l\u0131\u011f\u0131n\u0131 bulun.<\/li>\n<\/ul>\n

      Tan\u0131mlanan problem \u00e7ok terimli bir da\u011f\u0131l\u0131m taraf\u0131ndan y\u00f6netildi\u011finden, bu t\u00fcr olas\u0131l\u0131k da\u011f\u0131l\u0131m\u0131 i\u00e7in form\u00fcl\u00fcn uygulanmas\u0131 gereklidir:<\/p>\n<\/p>\n

      \"P=\\cfrac{n!}{x_1!\\cdot<\/p>\n<\/p>\n

      B\u00f6ylece problemdeki verileri form\u00fclde yerine koyar\u0131z ve olas\u0131l\u0131k hesaplamas\u0131n\u0131 yapar\u0131z:<\/p>\n<\/p>\n

      \"P=\\cfrac{8!}{2!\\cdot<\/p>\n<\/p>\n

      Yani problem ifadesinin s\u00f6yledi\u011fi \u015feyin ger\u00e7ekle\u015fme olas\u0131l\u0131\u011f\u0131 %11,4’t\u00fcr. <\/p>\n

      <\/span> \u00c7ok Terimli Da\u011f\u0131t\u0131m Hesaplay\u0131c\u0131<\/span><\/h2>\n

      Her bir olay\u0131n ger\u00e7ekle\u015fme say\u0131s\u0131n\u0131 birinci kutuya, her olay\u0131n ger\u00e7ekle\u015fme olas\u0131l\u0131\u011f\u0131n\u0131 da ayn\u0131 s\u0131rayla ikinci kutuya yaz\u0131n. Daha sonra son bo\u015f alana yap\u0131lan toplam deneme say\u0131s\u0131n\u0131 girin.<\/p>\n

      Veriler bir bo\u015flukla ayr\u0131lmal\u0131 ve ondal\u0131k ay\u0131r\u0131c\u0131 olarak nokta kullan\u0131larak girilmelidir.<\/p>\n

      \n
        \n
      • Her bir olay\u0131n ger\u00e7ekle\u015fme say\u0131s\u0131\n

        \"(x_i)\"<\/p>\n

        : <\/li>\n<\/ul>\n