\u0130statistikte, merkezi limit teoremi<\/strong> olarak da adland\u0131r\u0131lan merkezi limit teoremi<\/strong> , pop\u00fclasyonun olas\u0131l\u0131k da\u011f\u0131l\u0131m\u0131na<\/a> bak\u0131lmaks\u0131z\u0131n , numune ortalamalar\u0131n\u0131n da\u011f\u0131l\u0131m\u0131n\u0131n<\/a> , numune boyutu artt\u0131k\u00e7a normal bir da\u011f\u0131l\u0131ma yakla\u015ft\u0131\u011f\u0131n\u0131 belirtir.<\/p>\n Yani, merkezi limit teoremi, yeterince fazla say\u0131da \u00f6rnek al\u0131rsak, bu \u00f6rneklerin ortalamas\u0131n\u0131n normal da\u011f\u0131l\u0131ma yakla\u015fabilece\u011fini s\u00f6yl\u00fcyor.<\/p>\n Ek olarak merkezi limit teoremi, \u00f6rneklem b\u00fcy\u00fckl\u00fc\u011f\u00fc artt\u0131k\u00e7a \u00f6rneklem ortalamas\u0131n\u0131n pop\u00fclasyon ortalamas\u0131n\u0131n de\u011ferine yakla\u015faca\u011f\u0131n\u0131 belirtir. Bu, istatistiksel pop\u00fclasyonun parametrelerini yakla\u015f\u0131k olarak tahmin etmemizi sa\u011flar. A\u015fa\u011f\u0131da bunun nas\u0131l yap\u0131ld\u0131\u011f\u0131n\u0131 g\u00f6rece\u011fiz. <\/p>\n Genel olarak merkezi limit teoreminin uygulanabilmesi i\u00e7in \u00f6rneklem b\u00fcy\u00fckl\u00fc\u011f\u00fcn\u00fcn en az 30 g\u00f6zlem olmas\u0131 gerekti\u011fi d\u00fc\u015f\u00fcn\u00fclmektedir, ancak bu durum \u00e7al\u0131\u015f\u0131lan de\u011fi\u015fkenin \u00f6zelliklerine ba\u011fl\u0131d\u0131r.<\/p>\n Merkezi limit teoreminin bir\u00e7ok uygulamas\u0131 vard\u0131r, \u00e7\u00fcnk\u00fc normal da\u011f\u0131l\u0131m hipotez testi veya g\u00fcven aral\u0131klar\u0131 gibi \u00e7\u0131kar\u0131msal istatistiksel hesaplamalara izin verir. \u00d6rne\u011fin finansta, bir yat\u0131r\u0131m\u0131n getirisini ve riskini analiz etmek i\u00e7in merkezi limit teoremi kullan\u0131l\u0131r. <\/p>\n Merkezi limit teoreminin tan\u0131m\u0131n\u0131 g\u00f6rd\u00fckten sonra anlam\u0131n\u0131 tam olarak anlamak i\u00e7in bir \u00f6rne\u011fe bakal\u0131m.<\/p>\n Merkezi limit teoreminin bir \u00f6rne\u011fi, bir zar\u0131n yuvarlanmas\u0131d\u0131r. T\u00fcm sonu\u00e7lar e\u015fit olas\u0131l\u0131kl\u0131 oldu\u011fundan, kal\u0131p rulosu ayr\u0131k bir d\u00fczg\u00fcn da\u011f\u0131l\u0131m<\/a> izler. Ancak birka\u00e7 sonucun toplam\u0131n\u0131n da\u011f\u0131l\u0131m\u0131 normal da\u011f\u0131l\u0131ma yakla\u015fmaktad\u0131r. <\/p>\n Dolay\u0131s\u0131yla, ne kadar \u00e7ok at\u0131\u015f varsa, ortalamalar\u0131n da\u011f\u0131l\u0131m \u015feklinin normal da\u011f\u0131l\u0131m grafi\u011fine benzeme olas\u0131l\u0131\u011f\u0131 da o kadar artar. <\/p>\n Merkezi limit teoremi, e\u011fer bir pop\u00fclasyon ortalama \u03bc ve standart sapma \u03c3’ya sahipse ve yeterince fazla say\u0131da \u00f6rnek al\u0131rsak (n\u226530), \u00f6rnek ortalamalar\u0131 k\u00fcmesinin ortalama \u03bc ve standart sapma \u03c3 ile normal bir da\u011f\u0131l\u0131ma yakla\u015ft\u0131r\u0131labilece\u011fini belirtir. \/\u221an.<\/p>\n<\/p>\n Ayr\u0131ca<\/em> X<\/em> 1<\/sub> ise<\/sub> a\u015fa\u011f\u0131daki form\u00fclle tan\u0131mlanan normal da\u011f\u0131l\u0131ma g\u00f6re<\/sub> : <\/p>\n<\/p>\n Kavram\u0131 tamamen \u00f6z\u00fcmseyebilmeniz i\u00e7in burada merkezi limit teoreminin \u00e7\u00f6z\u00fclm\u00fc\u015f bir al\u0131\u015ft\u0131rmas\u0131n\u0131 bulacaks\u0131n\u0131z.<\/p>\n Parti boyutu b\u00fcy\u00fck oldu\u011fundan (n=100), sorunu \u00e7\u00f6zmek i\u00e7in merkezi limit teoremini uygulayabiliriz.<\/p>\n B\u00f6ylece, merkezi limit teoremi form\u00fcl\u00fcn\u00fc kullanarak, \u00f6rnek ortalamalar\u0131n da\u011f\u0131l\u0131m\u0131 a\u015fa\u011f\u0131daki parametrelerle normal da\u011f\u0131l\u0131ma yakla\u015ft\u0131r\u0131labilir: <\/p>\n<\/p>\n \u015eimdi al\u0131\u015ft\u0131rman\u0131n bizden istedi\u011fi olas\u0131l\u0131\u011f\u0131 bulabilmek i\u00e7in yazma i\u015flemini ger\u00e7ekle\u015ftiriyoruz. Bunu yapmak i\u00e7in ortalamay\u0131 da\u011f\u0131l\u0131mdan \u00e7\u0131karmam\u0131z ve ard\u0131ndan standart sapmaya b\u00f6lmemiz gerekir:<\/p>\n<\/p>\n Dolay\u0131s\u0131yla partideki par\u00e7alar\u0131n ortalama a\u011f\u0131rl\u0131\u011f\u0131n\u0131n 305 g’dan b\u00fcy\u00fck olma olas\u0131l\u0131\u011f\u0131n\u0131 bulmak i\u00e7in normal da\u011f\u0131l\u0131m tablosunda<\/a> Z>1 de\u011ferinin hangi de\u011fere kar\u015f\u0131l\u0131k geldi\u011fine bakmam\u0131z gerekir:<\/p>\n<\/p>\n \u00d6te yandan, merkezi limit teoremi sayesinde 100 adetlik bir partinin normal da\u011f\u0131l\u0131ma yakla\u015fabilece\u011fini biliyoruz \u00e7\u00fcnk\u00fc t\u00fcm madeni paralar ayn\u0131 da\u011f\u0131l\u0131m\u0131 takip ediyor. Bu nedenle, 100 madeni paradan olu\u015fan bir partinin 31 kg’dan daha a\u011f\u0131r olma olas\u0131l\u0131\u011f\u0131n\u0131 belirlemek i\u00e7in merkezi limit teoreminin di\u011fer form\u00fcl\u00fcn\u00fc uygulamam\u0131z gerekir:<\/p>\n<\/p>\n B\u00f6ylece yazma i\u015flemini yeniden yap\u0131yoruz ve ard\u0131ndan sorunun bize sordu\u011fu ikinci olas\u0131l\u0131\u011f\u0131 buluyoruz:<\/p>\n<\/p>\n Son olarak, normal da\u011f\u0131l\u0131m tablosunu kullanarak 100 par\u00e7al\u0131k bir partinin 31 kg’dan fazla a\u011f\u0131rl\u0131\u011fa sahip olma olas\u0131l\u0131\u011f\u0131n\u0131 belirleyebiliriz: <\/p>\n<\/p>\n![\"merkezi](\"https:\/\/statorials.org\/wp-content\/uploads\/2023\/08\/theoreme-central-limite.png\")
<\/figure>\n<\/span> Merkezi limit teoremine \u00f6rnek<\/span><\/h2>\n
![\"merkezi](\"https:\/\/statorials.org\/wp-content\/uploads\/2023\/08\/exemple-du-theoreme-central-limite.png\")
<\/figure>\n<\/span> Merkezi Limit Teoremi Form\u00fcl\u00fc<\/span><\/h2>\n
![\"\\displaystyle](\"https:\/\/statorials.org\/wp-content\/ql-cache\/quicklatex.com-9bc00b8814e098df988d7c695c66506a_l3.png\" "\\"Rendered")
<\/p>\n<\/p>\n![\"\\displaystyle](\"https:\/\/statorials.org\/wp-content\/ql-cache\/quicklatex.com-7c518ada589db37029302048824df42a_l3.png\" "\\"Rendered")
<\/p>\n<\/p>\n<\/span> Merkezi limit teoreminin \u00e7\u00f6z\u00fclm\u00fc\u015f al\u0131\u015ft\u0131rmas\u0131<\/span><\/h2>\n
\n
<\/p>\n<\/p>\n![\"\\mu_{\\overline{X}}=\\mu=300\"](\"https:\/\/statorials.org\/wp-content\/ql-cache\/quicklatex.com-c62f30e3319566c578be2b03a2d983a2_l3.png\" "\\"Rendered")
<\/p>\n<\/p>\n![\"\\sigma_{\\overline{X}}=\\cfrac{\\sigma}{\\sqrt{n}}=\\cfrac{50}{\\sqrt{100}}=5\"](\"https:\/\/statorials.org\/wp-content\/ql-cache\/quicklatex.com-3d9f3d2d9a4272dca9d9e34626ff8290_l3.png\" "\\"Rendered")
<\/p>\n<\/p>\n![\"\\displaystyle](\"https:\/\/statorials.org\/wp-content\/ql-cache\/quicklatex.com-a2ccad42b29091909256610fdd711e39_l3.png\" "\\"Rendered")
<\/p>\n<\/p>\n![\"\\displaystyle](\"https:\/\/statorials.org\/wp-content\/ql-cache\/quicklatex.com-aabd7d1605e6518f17d8fc6199becc44_l3.png\")
305\\right)=P\\left(Z>\\frac{305-300}{5}\\right)=P\\left(Z>1\\right)” title=”Rendered by QuickLaTeX.com” height=”43″ width=”381″ style=”vertical-align: -17px;”><\/p>\n<\/p>\n![\"\\displaystyle](\"https:\/\/statorials.org\/wp-content\/ql-cache\/quicklatex.com-05ad215d26ded143bc689eb37c00f30e_l3.png\")
305\\right)=P\\left(Z>1\\right)=0,1587″ title=”Rendered by QuickLaTeX.com” height=”22″ width=”276″ style=”vertical-align: -7px;”><\/p>\n<\/p>\n<\/p>\n<\/p>\n![\"\\displaystyle](\"https:\/\/statorials.org\/wp-content\/ql-cache\/quicklatex.com-22186501d443fa0104f80ab2210937ec_l3.png\" "\\"Rendered")
<\/p>\n<\/p>\n![\"\\displaystyle](\"https:\/\/statorials.org\/wp-content\/ql-cache\/quicklatex.com-f2c36d2fd557394ea1f4f80d887d1625_l3.png\" "\\"Rendered")
<\/p>\n<\/p>\n![\"\\displaystyle](\"https:\/\/statorials.org\/wp-content\/ql-cache\/quicklatex.com-4e3dfdc2ecc330ef981fba7476b5e13e_l3.png\")
31000\\right)=P\\left(Z>\\frac{31000-30000}{500}\\right)=P\\left(Z>2\\right)” title=”Rendered by QuickLaTeX.com” height=”43″ width=”430″ style=”vertical-align: -17px;”><\/p>\n<\/p>\n![\"\\displaystyle](\"https:\/\/statorials.org\/wp-content\/ql-cache\/quicklatex.com-01523dbe603d64f1a7ca3187d96b062c_l3.png\")
31000\\right)=P\\left(Z>2\\right)=0,0228″ title=”Rendered by QuickLaTeX.com” height=”19″ width=”289″ style=”vertical-align: -5px;”><\/p>\n<\/p>\n