{"id":2031,"date":"2023-07-24T00:29:32","date_gmt":"2023-07-24T00:29:32","guid":{"rendered":"https:\/\/statorials.org\/tr\/ustel-dagilim-gercek-ornekleri\/"},"modified":"2023-07-24T00:29:32","modified_gmt":"2023-07-24T00:29:32","slug":"ustel-dagilim-gercek-ornekleri","status":"publish","type":"post","link":"https:\/\/statorials.org\/tr\/ustel-dagilim-gercek-ornekleri\/","title":{"rendered":"\u00dcstel da\u011f\u0131l\u0131ma 4 ger\u00e7ek \u00f6rnek"},"content":{"rendered":"<p><\/p>\n<hr>\n<p><span style=\"color: #000000;\"><a href=\"https:\/\/statorials.org\/tr\/ustel-dagilim-1\/\" target=\"_blank\" rel=\"noopener\">\u00dcstel da\u011f\u0131l\u0131m,<\/a> belirli bir olay ger\u00e7ekle\u015fene kadar beklememiz gereken s\u00fcreyi modellemek i\u00e7in kullan\u0131lan bir olas\u0131l\u0131k da\u011f\u0131l\u0131m\u0131d\u0131r.<\/span><\/p>\n<p> <span style=\"color: #000000;\">E\u011fer bir <em>X<\/em> <a href=\"https:\/\/statorials.org\/tr\/rastgele-degiskenler\/\" target=\"_blank\" rel=\"noopener\">rastgele de\u011fi\u015fkeni<\/a> \u00fcstel bir da\u011f\u0131l\u0131m izliyorsa, <em>X&#8217;in<\/em> k\u00fcm\u00fclatif yo\u011funluk fonksiyonu \u015fu \u015fekilde yaz\u0131labilir:<\/span><\/p>\n<p> <span style=\"color: #000000;\"><strong><em>F<\/em> (x; \u03bb) = 1 \u2013 e <sup>-\u03bbx<\/sup><\/strong><\/span><\/p>\n<p> <span style=\"color: #000000;\">Alt\u0131n:<\/span><\/p>\n<ul>\n<li> <span style=\"color: #000000;\"><strong>\u03bb:<\/strong> oran parametresi (\u03bb = 1\/\u03bc olarak hesaplan\u0131r)<\/span><\/li>\n<li> <span style=\"color: #000000;\"><strong>e:<\/strong> Yakla\u015f\u0131k olarak 2,718&#8217;e e\u015fit bir sabit<\/span><\/li>\n<\/ul>\n<p> <span style=\"color: #000000;\">Bu yaz\u0131m\u0131zda ger\u00e7ek hayattan 5 \u00fcstel da\u011f\u0131l\u0131m \u00f6rne\u011fini payla\u015f\u0131yoruz.<\/span><\/p>\n<h3> <span style=\"color: #000000;\"><strong>\u00d6rnek 1: \u015eofben patlamalar\u0131 aras\u0131ndaki s\u00fcre<\/strong><\/span><\/h3>\n<p> <span style=\"color: #000000;\">Belirli bir gayzerin patlamalar\u0131 aras\u0131ndaki dakika say\u0131s\u0131 \u00fcstel da\u011f\u0131l\u0131mla modellenebilir.<\/span><\/p>\n<p> <span style=\"color: #000000;\">\u00d6rne\u011fin,<\/span> <span style=\"color: #000000;\">belirli bir gayzerin patlamalar\u0131 aras\u0131ndaki ortalama dakika say\u0131s\u0131n\u0131n 40 dakika oldu\u011funu varsayal\u0131m. E\u011fer bir gayzer patlarsa, bir sonraki patlama i\u00e7in 50 dakikadan daha az beklemek zorunda kalma olas\u0131l\u0131\u011f\u0131m\u0131z nedir?<\/span><\/p>\n<p> <span style=\"color: #000000;\">Bu sorunu \u00e7\u00f6zmek i\u00e7in \u00f6ncelikle oran parametresini hesaplamam\u0131z gerekir:<\/span><\/p>\n<ul>\n<li> <span style=\"color: #000000;\">\u03bb = 1\/\u03bc<\/span><\/li>\n<li> <span style=\"color: #000000;\">\u03bb = 1\/40<\/span><\/li>\n<li> <span style=\"color: #000000;\">\u03bb = 0,025<\/span><\/li>\n<\/ul>\n<p> <span style=\"color: #000000;\">\u03bb = 0,025 ve x = 50&#8217;yi CDF form\u00fcl\u00fcne koyabiliriz:<\/span><\/p>\n<ul>\n<li> <span style=\"color: #000000;\">P(X \u2264 x) = 1 \u2013 e <sup>-\u03bbx<\/sup><\/span><\/li>\n<li> <span style=\"color: #000000;\">P(X \u2264 50) = 1 \u2013 e <sup>-0,025(50)<\/sup><\/span><\/li>\n<li> <span style=\"color: #000000;\">P(X \u2264 50) = 0,7135<\/span><\/li>\n<\/ul>\n<p> <span style=\"color: #000000;\">Bir sonraki patlama i\u00e7in 50 dakikadan az beklemek zorunda kalma olas\u0131l\u0131\u011f\u0131m\u0131z <strong>0,7135&#8217;tir<\/strong> .<\/span><\/p>\n<h3> <span style=\"color: #000000;\"><strong>\u00d6rnek 2: M\u00fc\u015fteriler Aras\u0131ndaki S\u00fcre<\/strong><\/span><\/h3>\n<p> <span style=\"color: #000000;\">Belirli bir ma\u011fazaya giren m\u00fc\u015fterilerin aras\u0131ndaki dakika say\u0131s\u0131 \u00fcstel da\u011f\u0131l\u0131mla modellenebilir.<\/span><\/p>\n<p> <span style=\"color: #000000;\">\u00d6rne\u011fin ortalama her iki dakikada bir yeni bir m\u00fc\u015fterinin ma\u011fazaya girdi\u011fini varsayal\u0131m. Bir m\u00fc\u015fteri geldikten sonra, yeni bir m\u00fc\u015fterinin bir dakikadan daha k\u0131sa s\u00fcrede gelme olas\u0131l\u0131\u011f\u0131n\u0131 belirleyin.<\/span><\/p>\n<p> <span style=\"color: #000000;\">Bunu \u00e7\u00f6zmek i\u00e7in istemciler aras\u0131ndaki ortalama s\u00fcrenin iki dakika oldu\u011funu bilerek ba\u015flayabiliriz. B\u00f6ylece oran \u015fu \u015fekilde hesaplanabilir:<\/span><\/p>\n<ul>\n<li> <span style=\"color: #000000;\">\u03bb = 1\/\u03bc<\/span><\/li>\n<li> <span style=\"color: #000000;\">\u03bb = 1\/2<\/span><\/li>\n<li> <span style=\"color: #000000;\">\u03bb = 0,5<\/span><\/li>\n<\/ul>\n<p> <span style=\"color: #000000;\">\u03bb = 0,5 ve x = 1&#8217;i CDF form\u00fcl\u00fcne koyabiliriz:<\/span><\/p>\n<ul>\n<li> <span style=\"color: #000000;\">P(X \u2264 x) = 1 \u2013 e <sup>-\u03bbx<\/sup><\/span><\/li>\n<li> <span style=\"color: #000000;\">P(X \u2264 1) = 1 \u2013 e <sup>-0,5(1)<\/sup><\/span><\/li>\n<li> <span style=\"color: #000000;\">P(X \u2264 1) = 0,3935<\/span><\/li>\n<\/ul>\n<p> <span style=\"color: #000000;\">Bir sonraki m\u00fc\u015fterinin gelmesi i\u00e7in bir dakikadan az beklemek zorunda kalma olas\u0131l\u0131\u011f\u0131m\u0131z <strong>0,3935&#8217;tir<\/strong> .<\/span><\/p>\n<h3> <span style=\"color: #000000;\"><strong>\u00d6rnek 3: Depremler aras\u0131ndaki s\u00fcre<\/strong><\/span> <span style=\"color: #000000;\"><strong>&nbsp;<\/strong><\/span><\/h3>\n<p> <span style=\"color: #000000;\">Depremler aras\u0131ndaki s\u00fcre \u00fcstel da\u011f\u0131l\u0131m kullan\u0131larak modellenebilir.<\/span><\/p>\n<p> <span style=\"color: #000000;\">\u00d6rne\u011fin belli bir b\u00f6lgede ortalama 400 g\u00fcnde bir deprem meydana geldi\u011fini varsayal\u0131m. Bir depremden sonra bir sonraki depremin olu\u015fmas\u0131na 500 g\u00fcnden fazla zaman ge\u00e7mesi olas\u0131l\u0131\u011f\u0131n\u0131 belirleyin.<\/span><\/p>\n<p> <span style=\"color: #000000;\">Bu sorunu \u00e7\u00f6zmeye depremler aras\u0131ndaki ortalama s\u00fcrenin 400 g\u00fcn oldu\u011funu bilerek ba\u015fl\u0131yoruz. B\u00f6ylece oran \u015fu \u015fekilde hesaplanabilir:<\/span><\/p>\n<ul>\n<li> <span style=\"color: #000000;\">\u03bb = 1\/\u03bc<\/span><\/li>\n<li> <span style=\"color: #000000;\">\u03bb = 1\/400<\/span><\/li>\n<li> <span style=\"color: #000000;\">\u03bb = 0,0025<\/span><\/li>\n<\/ul>\n<p> <span style=\"color: #000000;\">\u03bb = 0,0025 ve x = 500&#8217;\u00fc CDF form\u00fcl\u00fcne koyabiliriz:<\/span><\/p>\n<ul>\n<li> <span style=\"color: #000000;\">P(X \u2264 x) = 1 \u2013 e <sup>-\u03bbx<\/sup><\/span><\/li>\n<li> <span style=\"color: #000000;\">P(X \u2264 1) = 1 \u2013 e <sup>-0,0025(500)<\/sup><\/span><\/li>\n<li> <span style=\"color: #000000;\">P(X \u2264 1) = 0,7135<\/span><\/li>\n<\/ul>\n<p> <span style=\"color: #000000;\">Bir sonraki deprem i\u00e7in 500 g\u00fcnden az beklemek zorunda kalma olas\u0131l\u0131\u011f\u0131m\u0131z 0,7135&#8217;tir.<\/span><\/p>\n<p> <span style=\"color: #000000;\">Yani bir sonraki deprem i\u00e7in 500 g\u00fcnden <em>fazla<\/em> beklemek zorunda kalma ihtimalimiz 1 \u2013 0,7135 = <strong>0,2865<\/strong> .<\/span><\/p>\n<h3> <span style=\"color: #000000;\"><strong>\u00d6rnek 4: aramalar aras\u0131ndaki s\u00fcre<\/strong><\/span><\/h3>\n<p> <span style=\"color: #000000;\">Farkl\u0131 \u015firketlerdeki m\u00fc\u015fteri aramalar\u0131 aras\u0131ndaki s\u00fcre, \u00fcstel da\u011f\u0131l\u0131m kullan\u0131larak modellenebilir.<\/span><\/p>\n<p> <span style=\"color: #000000;\">\u00d6rne\u011fin bir bankan\u0131n ortalama her 10 dakikada bir yeni bir \u00e7a\u011fr\u0131 ald\u0131\u011f\u0131n\u0131 varsayal\u0131m. Bir m\u00fc\u015fteri arad\u0131ktan sonra, yeni bir m\u00fc\u015fterinin 10 ila 15 dakika i\u00e7inde arama olas\u0131l\u0131\u011f\u0131n\u0131 belirleyin.<\/span><\/p>\n<p> <span style=\"color: #000000;\">Bunu \u00e7\u00f6zmek i\u00e7in \u00e7a\u011fr\u0131lar aras\u0131ndaki ortalama s\u00fcrenin 10 dakika oldu\u011funu bilerek ba\u015fl\u0131yoruz. B\u00f6ylece oran \u015fu \u015fekilde hesaplanabilir:<\/span><\/p>\n<ul>\n<li> <span style=\"color: #000000;\">\u03bb = 1\/\u03bc<\/span><\/li>\n<li> <span style=\"color: #000000;\">\u03bb = 1\/10<\/span><\/li>\n<li> <span style=\"color: #000000;\">\u03bb = 0,1<\/span><\/li>\n<\/ul>\n<p> <span style=\"color: #000000;\">Yeni bir m\u00fc\u015fterinin 10-15 dakika i\u00e7inde arama olas\u0131l\u0131\u011f\u0131n\u0131 hesaplamak i\u00e7in a\u015fa\u011f\u0131daki form\u00fcl\u00fc kullanabiliriz:<\/span><\/p>\n<ul>\n<li> <span style=\"color: #000000;\">P(10 &lt; X \u2264 15) = (1 \u2013 e <sup>-0,1(15)<\/sup> ) \u2013 (1 \u2013 e <sup>-0,1(10)<\/sup> )<\/span><\/li>\n<li> <span style=\"color: #000000;\">P(10 &lt; X \u2264 15) = 0,7769 \u2013 0,6321<\/span><\/li>\n<li> <span style=\"color: #000000;\">P(10 &lt; X \u2264 15) = 0,1448<\/span><\/li>\n<\/ul>\n<p> <span style=\"color: #000000;\">Yeni bir m\u00fc\u015fterinin 10-15 dakika i\u00e7inde arama olas\u0131l\u0131\u011f\u0131. <strong>0,1448&#8217;dir<\/strong> .<\/span><\/p>\n<h3> <span style=\"color: #000000;\"><strong>Ek kaynaklar<\/strong><\/span><\/h3>\n<p> <span style=\"color: #000000;\">A\u015fa\u011f\u0131daki makaleler di\u011fer olas\u0131l\u0131k da\u011f\u0131l\u0131mlar\u0131n\u0131n ger\u00e7ek d\u00fcnyada nas\u0131l kullan\u0131ld\u0131\u011f\u0131na dair \u00f6rnekler sunmaktad\u0131r:<\/span><\/p>\n<p> <a href=\"https:\/\/statorials.org\/tr\/normal-dagilim-ornegi\/\" target=\"_blank\" rel=\"noopener\">Normal da\u011f\u0131l\u0131m\u0131n 6 somut \u00f6rne\u011fi<\/a><br \/><a href=\"https:\/\/statorials.org\/tr\/binom-dagilimi-gercek-ornekleri\/\" target=\"_blank\" rel=\"noopener\">Binom da\u011f\u0131l\u0131m\u0131n\u0131n 5 somut \u00f6rne\u011fi<\/a><br \/><a href=\"https:\/\/statorials.org\/tr\/balik-dagitimi-gercek-ornekleri\/\" target=\"_blank\" rel=\"noopener\">Poisson da\u011f\u0131l\u0131m\u0131n\u0131n 5 somut \u00f6rne\u011fi<\/a><br \/> <a href=\"https:\/\/statorials.org\/tr\/geometrik-dagilim-gercek-ornekleri\/\" target=\"_blank\" rel=\"noopener\">Geometrik da\u011f\u0131l\u0131ma 5 somut \u00f6rnek<\/a><br \/><a href=\"https:\/\/statorials.org\/tr\/duzgun-dagilim-gercek-ornekleri\/\" target=\"_blank\" rel=\"noopener\">D\u00fczg\u00fcn da\u011f\u0131l\u0131ma 5 somut \u00f6rnek<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"<p>\u00dcstel da\u011f\u0131l\u0131m, belirli bir olay ger\u00e7ekle\u015fene kadar beklememiz gereken s\u00fcreyi modellemek i\u00e7in kullan\u0131lan bir olas\u0131l\u0131k da\u011f\u0131l\u0131m\u0131d\u0131r. E\u011fer bir X rastgele de\u011fi\u015fkeni \u00fcstel bir da\u011f\u0131l\u0131m izliyorsa, X&#8217;in k\u00fcm\u00fclatif yo\u011funluk fonksiyonu \u015fu \u015fekilde yaz\u0131labilir: F (x; \u03bb) = 1 \u2013 e -\u03bbx Alt\u0131n: \u03bb: oran parametresi (\u03bb = 1\/\u03bc olarak hesaplan\u0131r) e: Yakla\u015f\u0131k olarak 2,718&#8217;e e\u015fit bir [&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-2031","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>\u00dcstel da\u011f\u0131l\u0131ma 4 somut \u00f6rnek - Statorials<\/title>\n<meta name=\"description\" content=\"Bu e\u011fitimde \u00fcstel da\u011f\u0131l\u0131m\u0131n \u00e7e\u015fitli alanlarda nas\u0131l kullan\u0131ld\u0131\u011f\u0131 da dahil olmak \u00fczere ger\u00e7ek hayattaki \u00e7e\u015fitli \u00f6rnekler verilmektedir.\" \/>\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\/ustel-dagilim-gercek-ornekleri\/\" \/>\n<meta property=\"og:locale\" content=\"tr_TR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"\u00dcstel da\u011f\u0131l\u0131ma 4 somut \u00f6rnek - 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