Gambaran kongkrit teorema central limit

Jejer ieu ngagambarkeun tiori central limit ngaliwatan conto keur penghitungan bisa diitung sacara gancang ku leungeun dina kertas, teu saperti conto-intesip dina jejer gambaran dina teorema central limit. Anggap sebaran probabiliti variabel random X beuratna sarua dina 1, 2, jeung 3:

$X=\{\begin{array}{cc}1& \text{with}\text{probability}1/3,\\ 2& \text{with}\text{probability}1/3,\\ 3& \text{with}\text{probability}1/3.\end{array}$

Fungsi probabiliti massa tina variabel random X digambarkeun ku:

    o    o    o
   -------------
    1    2    3

Katembong jelas teu siga kurva bentuk-bel.

Ayeuna tempo jumlah dua kopi-an X bébas:

$\left\{\begin{array}{ccc}1+1& =& 2\\ 1+2& =& 3\\ 1+3& =& 4\\ 2+1& =& 3\\ 2+2& =& 4\\ 2+3& =& 5\\ 3+1& =& 4\\ 3+2& =& 5\\ 3+3& =& 6\end{array}\right\}=\left\{\begin{array}{cc}2& \text{with}\text{probability}1/9\\ 3& \text{with}\text{probability}2/9\\ 4& \text{with}\text{probability}3/9\\ 5& \text{with}\text{probability}2/9\\ 6& \text{with}\text{probability}1/9\end{array}\right\}$

Fungsi probabiliti massa tina jumlah ieu digambarkeun ku:

              o
         o    o    o
    o    o    o    o    o
   ----------------------------
    2    3    4    5    6

Ieu ogé can katembong leuwih siga tina kurva bentuk-bell, tapi, saperti bentuk-bel sarta teu saperti fungsi probabiliti massa X éta sorangan, leuwih luhur dibagian tengah tinimbang di dua sisina.

Ayeuna tempo jumlah tilu kopian bébas ieu random variabel:

$\left\{\begin{array}{ccc}1+1+1& =& 3\\ 1+1+2& =& 4\\ 1+1+3& =& 5\\ 1+2+1& =& 4\\ 1+2+2& =& 5\\ 1+2+3& =& 6\\ 1+3+1& =& 5\\ 1+3+2& =& 6\\ 1+3+3& =& 7\\ 2+1+1& =& 4\\ 2+1+2& =& 5\\ 2+1+3& =& 6\\ 2+2+1& =& 5\\ 2+2+2& =& 6\\ 2+2+3& =& 7\\ 2+3+1& =& 6\\ 2+3+2& =& 7\\ 2+3+3& =& 8\\ 3+1+1& =& 5\\ 3+1+2& =& 6\\ 3+1+3& =& 7\\ 3+2+1& =& 6\\ 3+2+2& =& 7\\ 3+2+3& =& 8\\ 3+3+1& =& 7\\ 3+3+2& =& 8\\ 3+3+3& =& 9\end{array}\right\}=\left\{\begin{array}{cc}3& \text{with}\text{probability}1/27\\ 4& \text{with}\text{probability}3/27\\ 5& \text{with}\text{probability}6/27\\ 6& \text{with}\text{probability}7/27\\ 7& \text{with}\text{probability}6/27\\ 8& \text{with}\text{probability}3/27\\ 9& \text{with}\text{probability}1/27\end{array}\right\}$

Fungsi probabiliti tina jumlah ieu digambarkeun ku:

                   o
              o    o    o
              o    o    o
              o    o    o
         o    o    o    o    o
         o    o    o    o    o
    o    o    o    o    o    o    o
   ---------------------------------
    3    4    5    6    7    8    9

Ieu heunteu ngan leuwih gedé di tengah tinimbang dua sisina, tapi pindah ka arah tengah ti sisi nu séjén, miring nu mimiti naek sarta saterusna turun, siga kurva bentuk-bel.

Urang bisa ngitung tingkatna tina susnan kana kurva bentuk-bel siga di handap ieu. Tempo

Pr(X₁ + X₂ + X₃ ≤ 7) = 1/27 + 3/27 + 6/27 + 7/27 + 6/27 = 23/27 = 0.851 851 851 ... .

Sakumaha raket hal ieu ngadeukeutan kana normal? Ieu bisa ditempo tina nilai ekspektasi Y = X₁ + X₂ + X₃ nyaéta 6 sarta simpangan baku Y mangrupa akar kuadrat 2. Saprak Y ≤ 7 (kateusaruaan lemah) lamun jeung lamun Y < 8 (kateusaruaan kuat), bisa maké koreksi kontinyu sarta ditembongkeun ku

$\text{Pr}(Y\le 7.5)=\text{P}(\frac{Y-6}{\sqrt{2}}\le \frac{7.5-6}{\sqrt{2}})=\text{Pr}(Z\le 1.606602\dots )\approx 0.8555778$

nu mana Z mangrupa standar normal sebaran. béda antara 0.85185... sarta 0.8556... katempo beuki ngaleutikan waktu éta ditempo salaku wilangan variabel random bébas nu ditambahkeun ngan tilu.

Conto dina simulasi

Gambar di handap nembongkeun hasil simulasi dumasar kana conto di luhur. Data dicokot tina sebaran seragam ku cara "pengulangan" 1'000 kali tur hasilna dijumlahkeun.

Conto ieu dumasar kana [Monte Carlo method], prosés "pengulanganna" 10'000 kali. Hasilna nembongkeun yén sebaran jumlah 1'000 nu dicokot sacara saragam nembongkeung bentuk kurva nu siga bel kacida alusna.

[[Image:Gambar:Central theorem 2.png]]

Rujukan

Wikipedia Basa Inggris, disunting panungtung, 24 Juli 2006