{"id":1418,"date":"2023-07-26T11:32:19","date_gmt":"2023-07-26T11:32:19","guid":{"rendered":"https:\/\/statorials.org\/tr\/hata-yayilimi\/"},"modified":"2023-07-26T11:32:19","modified_gmt":"2023-07-26T11:32:19","slug":"hata-yayilimi","status":"publish","type":"post","link":"https:\/\/statorials.org\/tr\/hata-yayilimi\/","title":{"rendered":"Hata yay\u0131l\u0131m\u0131 nedir? (tan\u0131m & #038; \u00f6rnek)"},"content":{"rendered":"

<\/p>\n

\n

Hata yay\u0131l\u0131m\u0131,<\/strong> baz\u0131 a<\/em> , b<\/em> , c<\/em> , \u2026 niceliklerini \u03b4a<\/em> , \u03b4b<\/em> , \u03b4c<\/em> \u2026 belirsizlikleriyle \u00f6l\u00e7t\u00fc\u011f\u00fcn\u00fczde ve ard\u0131ndan a<\/em> , b<\/em> , c<\/em> , vb. \u00f6l\u00e7\u00fcmlerini kullanarak ba\u015fka bir Q<\/em> niceli\u011fini hesaplamak istedi\u011finizde meydana gelir.<\/span><\/p>\n

\u03b4 a<\/em> , \u03b4 b<\/em> , \u03b4c<\/em> belirsizliklerinin Q<\/em> belirsizli\u011fine kadar yay\u0131laca\u011f\u0131<\/strong> (yani “yay\u0131laca\u011f\u0131”) ortaya \u00e7\u0131kt\u0131.<\/span><\/p>\n

\u03b4Q<\/em> ile g\u00f6sterilen Q<\/em> belirsizli\u011fini hesaplamak i\u00e7in a\u015fa\u011f\u0131daki form\u00fclleri kullanabiliriz.<\/span><\/p>\n

Not:<\/strong> A\u015fa\u011f\u0131daki form\u00fcllerin her biri i\u00e7in a<\/em> , b<\/em> , c<\/em> vb. b\u00fcy\u00fckl\u00fcklerin oldu\u011fu varsay\u0131lmaktad\u0131r. rastgele<\/em> ve ili\u015fkisiz<\/em> hatalar i\u00e7erir.<\/span><\/p>\n

Toplama veya \u00e7\u0131karma<\/strong><\/span><\/h3>\n

E\u011fer Q<\/em> = a + b + \u2026 + c \u2013 (x + y + \u2026 + z)<\/span><\/p>\n

O halde \u03b4 Q<\/em> = \u221a (\u03b4a) 2<\/sup> + (\u03b4b) 2<\/sup> + \u2026 + (\u03b4c) 2<\/sup> + (\u03b4x) 2<\/sup> + (\u03b4y) 2<\/sup> + \u2026 + (\u03b4z) 2<\/sup><\/span><\/span><\/p>\n

\u00d6rnek:<\/strong> Bir ki\u015finin yerden beline kadar olan uzunlu\u011funu 40 in\u00e7 \u00b1 0,18 in\u00e7 olarak \u00f6l\u00e7t\u00fc\u011f\u00fcn\u00fcz\u00fc varsayal\u0131m. Daha sonra bir ki\u015finin belinden ba\u015f\u0131n\u0131n tepesine kadar olan uzunlu\u011funu 30 in\u00e7 \u00b1 0,06 in\u00e7 olacak \u015fekilde \u00f6l\u00e7ersiniz.<\/span><\/p>\n

Daha sonra ki\u015finin toplam boyunu hesaplamak i\u00e7in bu iki \u00f6l\u00e7\u00fcm\u00fc kulland\u0131\u011f\u0131n\u0131z\u0131 varsayal\u0131m. Y\u00fckseklik \u015fu \u015fekilde hesaplan\u0131r: 40 in\u00e7 + 30 in\u00e7 = 70<\/strong> in\u00e7. Bu tahminin belirsizli\u011fi \u015fu \u015fekilde hesaplanacakt\u0131r:<\/span><\/p>\n

    \n
  • \u03b4 Q<\/em> = \u221a (\u03b4a) 2<\/sup> + (\u03b4b) 2<\/sup> + \u2026 + (\u03b4c) 2<\/sup> + (\u03b4x) 2<\/sup> + (\u03b4y) 2<\/sup> + \u2026 + (\u03b4z) 2<\/sup><\/span><\/span><\/li>\n
  • \u03b4 Q<\/em> = \u221a (0,18) 2<\/sup> + (0,06) 2<\/sup><\/span><\/span><\/li>\n
  • \u03b4Q<\/em> = 0,1897<\/strong><\/span><\/li>\n<\/ul>\n

    Bu bize 70 \u00b1 0,1897<\/strong> in\u00e7lik nihai bir \u00f6l\u00e7\u00fcm verir.<\/span><\/p>\n

    \u00c7arpma veya B\u00f6lme<\/strong><\/span><\/h3>\n

    E\u011fer Q<\/em> = (ab\u2026c) \/ (xy\u2026z)<\/span><\/p>\n

    O zaman \u03b4Q<\/em> = |Q| * \u221a (\u03b4a\/a) 2<\/sup> + (\u03b4b\/b) 2<\/sup> + \u2026 + (\u03b4c\/c) 2<\/sup> + (\u03b4x\/x) 2<\/sup> + (\u03b4y\/y) 2<\/sup> + \u2026 + (\u03b4z\/z) 2<\/sup><\/span><\/span><\/p>\n

    \u00d6rnek<\/em> :<\/strong> a \u00f6\u011fesinin uzunlu\u011funun b<\/em> \u00f6\u011fesinin uzunlu\u011funa oran\u0131n\u0131 \u00f6l\u00e7mek istedi\u011finizi varsayal\u0131m. a’n\u0131n<\/em> uzunlu\u011funu 20 in\u00e7 \u00b1 0,34 in\u00e7 ve b’nin<\/em> uzunlu\u011funu 15 in\u00e7 \u00b1 0,21 in\u00e7 olarak \u00f6l\u00e7ersiniz.<\/span><\/span><\/p>\n

    Q<\/em> = a\/b<\/em> olarak tan\u0131mlanan oran \u015fu \u015fekilde hesaplan\u0131r: 20\/15 = 1,333<\/strong> . Bu tahminin belirsizli\u011fi \u015fu \u015fekilde hesaplanacakt\u0131r:<\/span><\/p>\n

      \n
    • \u03b4Q<\/em> = |Q| * \u221a (\u03b4a\/a) 2<\/sup> + (\u03b4b\/b) 2<\/sup> + \u2026 + (\u03b4c\/c) 2<\/sup> + (\u03b4x\/x) 2<\/sup> + (\u03b4y\/y) 2<\/sup> + \u2026 + (\u03b4z\/z) 2<\/sup><\/span><\/span><\/li>\n
    • \u03b4Q<\/em> = |1,333| * \u221a (.34\/20) 2<\/sup> + (.21\/15) 2<\/sup><\/span><\/span><\/li>\n
    • \u03b4Q<\/em> = 0,0294<\/strong><\/span><\/li>\n<\/ul>\n

      Bu bize 1,333 \u00b1 0,0294<\/strong> in\u00e7lik nihai bir oran verir.<\/span><\/p>\n

      \u00d6l\u00e7\u00fclen miktar\u0131n tam say\u0131yla \u00e7arp\u0131lmas\u0131<\/strong><\/span><\/h3>\n

      A<\/em> tam olarak biliniyorsa ve Q<\/em> = A<\/em> x<\/span><\/p>\n

      O halde \u03b4 Q<\/em> = |A|\u03b4x<\/span><\/p>\n

      \u00d6rnek:<\/strong> Bir dairenin \u00e7ap\u0131n\u0131 5 metre \u00b1 0,3 metre olarak \u00f6l\u00e7t\u00fc\u011f\u00fcn\u00fcz\u00fc varsayal\u0131m. Daha sonra bu de\u011feri c = \u03c0d<\/em> \u00e7emberinin \u00e7evresini hesaplamak i\u00e7in kullan\u0131rs\u0131n\u0131z.<\/span><\/p>\n

      \u00c7evre c = \u03c0d<\/em> = \u03c0*5<\/em> = 15,708<\/strong> olarak hesaplan\u0131r. Bu tahminin belirsizli\u011fi \u015fu \u015fekilde hesaplanacakt\u0131r:<\/span><\/p>\n

        \n
      • \u03b4Q<\/em> = |A|\u03b4x<\/span><\/li>\n
      • \u03b4Q<\/em> = | \u03c0<\/em> | * 0,3<\/span><\/li>\n
      • \u03b4Q<\/em> = 0,942<\/strong><\/span><\/li>\n<\/ul>\n

        Yani dairenin \u00e7evresi 15,708 \u00b1 0,942<\/strong> metredir.<\/span><\/p>\n

        G\u00fc\u00e7teki belirsizlik<\/strong><\/span><\/h3>\n

        E\u011fer n<\/em> tam bir say\u0131 ise ve Q<\/em> = x n<\/sup><\/em><\/span><\/p>\n

        O halde \u03b4 Q<\/em> = | Soru<\/em> | * | n<\/em> | * (\u03b4x \/x<\/em> )<\/span><\/p>\n

        \u00d6rnek:<\/strong> Bir k\u00fcp\u00fcn kenar\u0131n\u0131 s<\/em> = 2 in\u00e7 \u00b1<\/strong> 0,02 in\u00e7 olarak \u00f6l\u00e7t\u00fc\u011f\u00fcn\u00fcz\u00fc varsayal\u0131m. Daha sonra bu de\u011feri v<\/em> = s<\/em> 3<\/sup> k\u00fcp\u00fcn\u00fcn hacmini hesaplamak i\u00e7in kullan\u0131rs\u0131n\u0131z.<\/span><\/p>\n

        Hacim \u015fu \u015fekilde hesaplan\u0131r: v<\/em> = s<\/em> 3<\/sup> = 2 3<\/sup> = 8 in\u00e7 3<\/sup><\/strong> . Bu tahminin belirsizli\u011fi \u015fu \u015fekilde hesaplanacakt\u0131r:<\/span><\/p>\n

          \n
        • \u03b4Q<\/em> = | Soru<\/em> | * | n<\/em> | * (\u03b4x \/x<\/em> )<\/span><\/li>\n
        • \u03b4Q<\/em> = |8| * |3| * (.02\/2)<\/span><\/li>\n
        • \u03b4Q<\/em> = 0,24<\/strong><\/span><\/li>\n<\/ul>\n

          Yani k\u00fcp\u00fcn hacmi 8 \u00b1 0,24 in\u00e7tir. 3<\/sup><\/strong> .<\/span><\/p>\n

          Genel hata yay\u0131lma form\u00fcl\u00fc<\/strong><\/h3>\n

          E\u011fer Q<\/em> = Q(x)<\/em> , x’in<\/em> bir fonksiyonu ise, genel hata yay\u0131l\u0131m form\u00fcl\u00fc a\u015fa\u011f\u0131daki gibi tan\u0131mlanabilir:<\/span><\/p>\n

          \u03b4Q = |dQ<\/em> \/ dX<\/em> |\u03b4x<\/strong><\/span><\/p>\n

          Bu form\u00fclleri s\u0131f\u0131rdan t\u00fcretmeniz gerekmeyece\u011fini unutmay\u0131n; ancak bunlar\u0131 t\u00fcretmek i\u00e7in kullan\u0131lan genel form\u00fcl\u00fc bilmek yararl\u0131 olabilir.<\/span><\/p>\n","protected":false},"excerpt":{"rendered":"

          Hata yay\u0131l\u0131m\u0131, baz\u0131 a , b , c , \u2026 niceliklerini \u03b4a , \u03b4b , \u03b4c \u2026 belirsizlikleriyle \u00f6l\u00e7t\u00fc\u011f\u00fcn\u00fczde ve ard\u0131ndan a , b , c , vb. \u00f6l\u00e7\u00fcmlerini kullanarak ba\u015fka bir Q niceli\u011fini hesaplamak istedi\u011finizde meydana gelir. \u03b4 a , \u03b4 b , \u03b4c belirsizliklerinin Q belirsizli\u011fine kadar yay\u0131laca\u011f\u0131 (yani “yay\u0131laca\u011f\u0131”) ortaya \u00e7\u0131kt\u0131. \u03b4Q […]<\/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-1418","post","type-post","status-publish","format-standard","hentry","category-rehber"],"yoast_head":"\nHata yay\u0131l\u0131m\u0131 nedir? 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