{"id":766,"date":"2023-07-28T20:14:26","date_gmt":"2023-07-28T20:14:26","guid":{"rendered":"https:\/\/statorials.org\/tr\/hipergeometrik-dagilim\/"},"modified":"2023-07-28T20:14:26","modified_gmt":"2023-07-28T20:14:26","slug":"hipergeometrik-dagilim","status":"publish","type":"post","link":"https:\/\/statorials.org\/tr\/hipergeometrik-dagilim\/","title":{"rendered":"Hipergeometrik da\u011f\u0131l\u0131ma giri\u015f"},"content":{"rendered":"<p><\/p>\n<hr>\n<p><span style=\"color: #000000;\"><strong>Hipergeometrik da\u011f\u0131l\u0131m,<\/strong> bu \u00f6zelli\u011fe sahip <em>K<\/em> nesneyi i\u00e7eren <em>N<\/em> boyutlu sonlu bir pop\u00fclasyondan, <em>n<\/em> \u00e7izimde belirli bir \u00f6zelli\u011fe sahip <em>k<\/em> nesneyi de\u011fi\u015ftirmeden se\u00e7me olas\u0131l\u0131\u011f\u0131n\u0131 tan\u0131mlar.<\/span><\/p>\n<p> <span style=\"color: #000000;\">Bir <em>X<\/em> <a href=\"https:\/\/statorials.org\/tr\/rastgele-degiskenler\/\" target=\"_blank\" rel=\"noopener\">rastgele de\u011fi\u015fkeni<\/a> hipergeometrik bir da\u011f\u0131l\u0131m izliyorsa, belirli bir \u00f6zelli\u011fe sahip <em>k<\/em> nesneyi se\u00e7me olas\u0131l\u0131\u011f\u0131 a\u015fa\u011f\u0131daki form\u00fclle bulunabilir:<\/span><\/p>\n<p> <span style=\"color: #000000;\"><strong>P(X=k) = <sub>K<\/sub> C <sub>k<\/sub> ( <sub>NK<\/sub> C <sub>nk<\/sub> ) \/ <sub>N<\/sub> C <sub>n<\/sub><\/strong><\/span><\/p>\n<p> <span style=\"color: #000000;\">Alt\u0131n:<\/span><\/p>\n<ul>\n<li> <span style=\"color: #000000;\"><strong>N:<\/strong> n\u00fcfus b\u00fcy\u00fckl\u00fc\u011f\u00fc<\/span><\/li>\n<li> <span style=\"color: #000000;\"><strong>K:<\/strong> pop\u00fclasyondaki belirli bir \u00f6zelli\u011fe sahip nesnelerin say\u0131s\u0131<\/span><\/li>\n<li> <span style=\"color: #000000;\"><strong>n:<\/strong> \u00f6rneklem b\u00fcy\u00fckl\u00fc\u011f\u00fc<\/span><\/li>\n<li> <span style=\"color: #000000;\"><strong>k:<\/strong> \u00f6rnekteki belirli bir i\u015flevselli\u011fe sahip nesnelerin say\u0131s\u0131<\/span><\/li>\n<li> <span style=\"color: #000000;\"><strong><sub>K<\/sub> C <sub>k<\/sub> :<\/strong> bir seferde k al\u0131nan K \u015feyin kombinasyon say\u0131s\u0131<\/span><\/li>\n<\/ul>\n<p> <span style=\"color: #000000;\">\u00d6rne\u011fin, standart 52 kartl\u0131k bir destede 4 Krali\u00e7e vard\u0131r. Bir desteden rastgele bir kart se\u00e7ti\u011fimizi ve daha sonra de\u011fi\u015ftirmeden desteden rastgele ba\u015fka bir kart se\u00e7ti\u011fimizi varsayal\u0131m. Her iki kart\u0131n da k\u0131z olma olas\u0131l\u0131\u011f\u0131 nedir?<\/span><\/p>\n<p> <span style=\"color: #000000;\">Bunu cevaplamak i\u00e7in hipergeometrik da\u011f\u0131l\u0131m\u0131 a\u015fa\u011f\u0131daki parametrelerle kullanabiliriz:<\/span><\/p>\n<ul>\n<li> <span style=\"color: #000000;\"><strong>N:<\/strong> pop\u00fclasyon b\u00fcy\u00fckl\u00fc\u011f\u00fc = 52 kart<\/span><\/li>\n<li> <span style=\"color: #000000;\"><strong>K:<\/strong> belirli bir \u00f6zelli\u011fe sahip pop\u00fclasyondaki nesne say\u0131s\u0131 = 4 krali\u00e7e<\/span><\/li>\n<li> <span style=\"color: #000000;\"><strong>n:<\/strong> \u00f6rneklem b\u00fcy\u00fckl\u00fc\u011f\u00fc = 2 \u00e7ekili\u015f<\/span><\/li>\n<li> <span style=\"color: #000000;\"><strong>k:<\/strong> \u00f6rnekteki belirli bir \u00f6zelli\u011fe sahip nesnelerin say\u0131s\u0131 = 2 krali\u00e7e<\/span><\/li>\n<\/ul>\n<p> <span style=\"color: #000000;\">Bu say\u0131lar\u0131 form\u00fcle yerle\u015ftirdi\u011fimizde olas\u0131l\u0131\u011f\u0131n \u015f\u00f6yle oldu\u011funu buluruz:<\/span><\/p>\n<p> <span style=\"color: #000000;\"><strong>P(X=2)<\/strong> = <sub>K<\/sub> C <sub>k<\/sub> ( <sub>NK<\/sub> C <sub>nk<\/sub> ) \/ <sub>N<\/sub> C <sub>n<\/sub> = <sub>4<\/sub> C <sub>2<\/sub> ( <sub>52-4<\/sub> C <sub>2-2<\/sub> ) \/ <sub>52<\/sub> C <sub>2<\/sub> = 6*1\/ 1326 = <strong>0,00452<\/strong> .<\/span><\/p>\n<p> <span style=\"color: #000000;\">Bu sezgisel olarak anlaml\u0131 olmal\u0131d\u0131r. Bir desteden arka arkaya iki kart \u00e7ekti\u011finizi hayal ederseniz, <em>her iki<\/em> kart\u0131n da k\u0131z olma olas\u0131l\u0131\u011f\u0131 \u00e7ok d\u00fc\u015f\u00fck olmal\u0131d\u0131r.<\/span><\/p>\n<h3> <strong>Hipergeometrik da\u011f\u0131l\u0131m\u0131n \u00f6zellikleri<\/strong><\/h3>\n<p> <span style=\"color: #000000;\">Hipergeometrik da\u011f\u0131l\u0131m a\u015fa\u011f\u0131daki \u00f6zelliklere sahiptir:<\/span><\/p>\n<p> <span style=\"color: #000000;\">Da\u011f\u0131l\u0131m\u0131n ortalamas\u0131 <b>(nK) \/ N&#8217;dir<\/b><\/span><\/p>\n<p> <span style=\"color: #000000;\">Da\u011f\u0131l\u0131m\u0131n varyans\u0131 <strong>(nK)(NK)(Nn) \/ (N <sup>2<\/sup> (n-1))<\/strong> \u015feklindedir.<\/span><\/p>\n<h3> <strong>Hipergeometrik Da\u011f\u0131t\u0131m Uygulama Problemleri<\/strong><\/h3>\n<p> <span style=\"color: #000000;\">Hipergeometrik da\u011f\u0131l\u0131m bilginizi test etmek i\u00e7in a\u015fa\u011f\u0131daki al\u0131\u015ft\u0131rma problemlerini kullan\u0131n.<\/span><\/p>\n<p> <span style=\"color: #000000;\"><em><strong>Not:<\/strong> Bu sorular\u0131n cevaplar\u0131n\u0131 hesaplamak i\u00e7in <a href=\"https:\/\/statorials.org\/tr\/hipergeometrik-dagilim-hesaplayicisi\/\" target=\"_blank\" rel=\"noopener\">Hipergeometrik Da\u011f\u0131l\u0131m Hesaplay\u0131c\u0131s\u0131n\u0131<\/a> kullanaca\u011f\u0131z.<\/em><\/span><\/p>\n<p> <span style=\"color: #000000;\"><strong>Sorun 1<\/strong><\/span><\/p>\n<p> <span style=\"color: #000000;\"><strong>Soru:<\/strong><\/span> <span style=\"color: #000000;\">Bir desteden, yerine koymadan rastgele d\u00f6rt kart se\u00e7ti\u011fimizi varsayal\u0131m. Kartlardan ikisinin k\u0131z olma olas\u0131l\u0131\u011f\u0131 nedir?<\/span><\/p>\n<p> <span style=\"color: #000000;\">Bunu cevaplamak i\u00e7in hipergeometrik da\u011f\u0131l\u0131m\u0131 a\u015fa\u011f\u0131daki parametrelerle kullanabiliriz:<\/span><\/p>\n<ul>\n<li> <span style=\"color: #000000;\"><strong>N:<\/strong> pop\u00fclasyon b\u00fcy\u00fckl\u00fc\u011f\u00fc = 52 kart<\/span><\/li>\n<li> <span style=\"color: #000000;\"><strong>K:<\/strong> belirli bir \u00f6zelli\u011fe sahip pop\u00fclasyondaki nesne say\u0131s\u0131 = 4 krali\u00e7e<\/span><\/li>\n<li> <span style=\"color: #000000;\"><strong>n:<\/strong> \u00f6rneklem b\u00fcy\u00fckl\u00fc\u011f\u00fc = 4 \u00e7ekili\u015f<\/span><\/li>\n<li> <span style=\"color: #000000;\"><strong>k:<\/strong> \u00f6rnekteki belirli bir \u00f6zelli\u011fe sahip nesnelerin say\u0131s\u0131 = 2 krali\u00e7e<\/span><\/li>\n<\/ul>\n<p> <span style=\"color: #000000;\">Bu say\u0131lar\u0131 hipergeometrik da\u011f\u0131l\u0131m hesaplay\u0131c\u0131s\u0131na yerle\u015ftirdi\u011fimizde olas\u0131l\u0131\u011f\u0131n <strong>0,025<\/strong> oldu\u011funu g\u00f6r\u00fcyoruz.<\/span><\/p>\n<p> <span style=\"color: #000000;\"><strong>Sorun 2<\/strong><\/span><\/p>\n<p> <span style=\"color: #000000;\"><strong>Soru:<\/strong> Bir kavanozda 3 k\u0131rm\u0131z\u0131 ve 5 ye\u015fil top bulunmaktad\u0131r. Rastgele 4 top se\u00e7iyorsunuz. Tam olarak 2 k\u0131rm\u0131z\u0131 top se\u00e7me olas\u0131l\u0131\u011f\u0131n\u0131z nedir?<\/span><\/p>\n<p> <span style=\"color: #000000;\">Bunu cevaplamak i\u00e7in hipergeometrik da\u011f\u0131l\u0131m\u0131 a\u015fa\u011f\u0131daki parametrelerle kullanabiliriz:<\/span><\/p>\n<ul>\n<li> <span style=\"color: #000000;\"><strong>N:<\/strong> pop\u00fclasyon b\u00fcy\u00fckl\u00fc\u011f\u00fc = 8 top<\/span><\/li>\n<li> <span style=\"color: #000000;\"><strong>K:<\/strong> pop\u00fclasyondaki belirli bir \u00f6zelli\u011fe sahip nesnelerin say\u0131s\u0131 = 3 k\u0131rm\u0131z\u0131 top<\/span><\/li>\n<li> <span style=\"color: #000000;\"><strong>n:<\/strong> \u00f6rneklem b\u00fcy\u00fckl\u00fc\u011f\u00fc = 4 \u00e7ekili\u015f<\/span><\/li>\n<li> <span style=\"color: #000000;\"><strong>k:<\/strong> \u00f6rnekteki belirli bir \u00f6zelli\u011fe sahip nesnelerin say\u0131s\u0131 = 2 k\u0131rm\u0131z\u0131 top<\/span><\/li>\n<\/ul>\n<p> <span style=\"color: #000000;\">Bu say\u0131lar\u0131 hipergeometrik da\u011f\u0131l\u0131m hesaplay\u0131c\u0131s\u0131na yerle\u015ftirdi\u011fimizde olas\u0131l\u0131\u011f\u0131n <strong>0,42857<\/strong> oldu\u011funu g\u00f6r\u00fcyoruz.<\/span><\/p>\n<p> <span style=\"color: #000000;\"><strong>Sorun 3<\/strong><\/span><\/p>\n<p> <span style=\"color: #000000;\"><strong>Soru:<\/strong> Bir sepette 7 mor ve 3 pembe bilye vard\u0131r. Rastgele 6 bilye se\u00e7iyorsunuz. Tam olarak 3 pembe bilye se\u00e7me olas\u0131l\u0131\u011f\u0131n\u0131z nedir?<\/span><\/p>\n<p> <span style=\"color: #000000;\">Bunu cevaplamak i\u00e7in hipergeometrik da\u011f\u0131l\u0131m\u0131 a\u015fa\u011f\u0131daki parametrelerle kullanabiliriz:<\/span><\/p>\n<ul>\n<li> <span style=\"color: #000000;\"><strong>N:<\/strong> pop\u00fclasyon b\u00fcy\u00fckl\u00fc\u011f\u00fc = 10 bilye<\/span><\/li>\n<li> <span style=\"color: #000000;\"><strong>K:<\/strong> pop\u00fclasyondaki belirli bir \u00f6zelli\u011fe sahip nesnelerin say\u0131s\u0131 = 3 pembe top<\/span><\/li>\n<li> <span style=\"color: #000000;\"><strong>n:<\/strong> \u00f6rneklem b\u00fcy\u00fckl\u00fc\u011f\u00fc = 6 \u00e7ekili\u015f<\/span><\/li>\n<li> <span style=\"color: #000000;\"><strong>k:<\/strong> \u00f6rnekteki belirli bir \u00f6zelli\u011fe sahip nesnelerin say\u0131s\u0131 = 3 pembe top<\/span><\/li>\n<\/ul>\n<p> <span style=\"color: #000000;\">Bu say\u0131lar\u0131 hipergeometrik da\u011f\u0131l\u0131m hesaplay\u0131c\u0131s\u0131na yerle\u015ftirdi\u011fimizde olas\u0131l\u0131\u011f\u0131n <strong>0,16667<\/strong> oldu\u011funu g\u00f6r\u00fcyoruz.<\/span><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Hipergeometrik da\u011f\u0131l\u0131m, bu \u00f6zelli\u011fe sahip K nesneyi i\u00e7eren N boyutlu sonlu bir pop\u00fclasyondan, n \u00e7izimde belirli bir \u00f6zelli\u011fe sahip k nesneyi de\u011fi\u015ftirmeden se\u00e7me olas\u0131l\u0131\u011f\u0131n\u0131 tan\u0131mlar. Bir X rastgele de\u011fi\u015fkeni hipergeometrik bir da\u011f\u0131l\u0131m izliyorsa, belirli bir \u00f6zelli\u011fe sahip k nesneyi se\u00e7me olas\u0131l\u0131\u011f\u0131 a\u015fa\u011f\u0131daki form\u00fclle bulunabilir: P(X=k) = K C k ( NK C nk ) \/ [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[11],"tags":[],"class_list":["post-766","post","type-post","status-publish","format-standard","hentry","category-rehber"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v21.3 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Hipergeometrik Da\u011f\u0131l\u0131ma Giri\u015f - Statorials<\/title>\n<meta name=\"description\" content=\"Resmi bir tan\u0131m ve birka\u00e7 \u00f6rnek i\u00e7eren, hipergeometrik da\u011f\u0131l\u0131ma basit bir giri\u015f.\" \/>\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\/\" \/>\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\u0131ma Giri\u015f - Statorials\" \/>\n<meta property=\"og:description\" content=\"Resmi bir tan\u0131m ve birka\u00e7 \u00f6rnek i\u00e7eren, hipergeometrik da\u011f\u0131l\u0131ma basit bir giri\u015f.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/statorials.org\/tr\/hipergeometrik-dagilim\/\" \/>\n<meta property=\"og:site_name\" content=\"Statorials\" \/>\n<meta property=\"article:published_time\" content=\"2023-07-28T20:14:26+00:00\" \/>\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=\"3 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\/\",\"url\":\"https:\/\/statorials.org\/tr\/hipergeometrik-dagilim\/\",\"name\":\"Hipergeometrik Da\u011f\u0131l\u0131ma Giri\u015f - Statorials\",\"isPartOf\":{\"@id\":\"https:\/\/statorials.org\/tr\/#website\"},\"datePublished\":\"2023-07-28T20:14:26+00:00\",\"dateModified\":\"2023-07-28T20:14:26+00:00\",\"author\":{\"@id\":\"https:\/\/statorials.org\/tr\/#\/schema\/person\/365dc158a39a7c8ae256355451e3de48\"},\"description\":\"Resmi bir tan\u0131m ve birka\u00e7 \u00f6rnek i\u00e7eren, hipergeometrik da\u011f\u0131l\u0131ma basit bir giri\u015f.\",\"breadcrumb\":{\"@id\":\"https:\/\/statorials.org\/tr\/hipergeometrik-dagilim\/#breadcrumb\"},\"inLanguage\":\"tr\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/statorials.org\/tr\/hipergeometrik-dagilim\/\"]}]},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/statorials.org\/tr\/hipergeometrik-dagilim\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Ev\",\"item\":\"https:\/\/statorials.org\/tr\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"Hipergeometrik da\u011f\u0131l\u0131ma giri\u015f\"}]},{\"@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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