恩儀 asked in 科學數學 · 7 years ago

統計問題~連續校正因子 二項分配逼近常態分配

請問一下

當二項分配近似常態分配

要怎麼判斷要+0.5或-0.5?

P( x-0.5 < x < x+0.5 )

並請用下題解釋

An airline has determined that 20% of its international flights are not on time. Use the normal approximation to the binomial distribution to answer the following questions. What is the probability that of the next 80 international flights

a. Fifteen or less will not be on time?

b. Eighteen or more will not be on time?

c. Exactly 17 will not be on time?

謝謝^^

2 Answers

Rating
  • 7 years ago
    Favorite Answer

    首先注意大小寫...P[x-0.5 寫 X 是隨機變數, 兩端的小寫 x 是某個 x 值.

    "連續性校正" 是在用連續型分布近似離散型(整數值隨機變數)

    分布時, 把每個整數值攤開在寬度 1 的區間.

    以常態分布近似二項分布為例, P[X=x] 攤開成連續型, 就是把

    P[X=x] 寫成 P[x-0.5

    例如:

    c. Exactly 17 will not be on time?

    用二項分布表示, 是 P[X=17]

    要用常態近似時, 把它攤開成 P[16.5<17.5].

    多值範圍, 就把每個值攤開再連接在一起, 例如

    P[16≦X≦17] = P[X=16 or X=17],

    而用常態近似時, 攤開成

    P[15.5<16.5 16.5<17.5]="P[15.5<17.5].

    以所舉練習題而言

    a. Fifteen or less will not be on time?

    在二項分布計算時是 P[X≦15], 用常態分布近似時, 變成

    P[X<15.5], 因 [X=15] 被展開成 [14.5<15.5].

    b. Eighteen or more will not be on time?

    在二項分布計算時, 是 P[X≧18], 其中 [X=18] 在常態近似

    中被展開成 [17.5<18.5]. 而 P[X 就變成 P[X≧18]>17.5].

    An airline has determined that 20% of its international flights are not on time. Use the normal approximation to the binomial distribution to answer the following questions. What is the probability that of the next 80 international flights.

    由題目設定知以 X 代表 80 航次未準時的航次數. 因不準時的

    比率是 20%, 所以 X~bin(80,0.20). 故 E[X] = 16, 而

    Var(X) = 12.8, 所以 sd(X) = √12.8 = 3.5777

    (a) 二項分布: P[X≦15]

    常態近似 P[X≦15.5] = P[Z≦(15.5-16)/3.5777],

    其中 Z 服從標準常態分布.

    (b) 二項分布 P[X≧18]

    常態近似 P[X>17.5] = P[Z>(17.5-16)/3.5777].

    (c) P[X=17] 用常態近似得

    P[16.5<X≦17.5] = P[(16.5-16)/3.5777< Z≦(17.5-16)/3.5777]

    <17.5].< body>

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  • 麻辣
    Lv 7
    7 years ago

    (1) 請問一下: 當二項分配近似常態分配.要怎麼判斷要+0.5或-0.5?

    P(-0.5<x<+0.5)=?Sol: 當二項分配近似常態分配.就以常態分配來操作:P(-0.5<x<+0.5)=90% => α=10%

    (2) 並請用下題解釋An airline has determined that 20% of its international flights are not on time. Use the normal approximation to the binomial distribution to answer the following questions. What is the probability that of the next 80 international flightsa. Fifteen or less will not be on time?b. Eighteen or more will not be on time?c. Exactly 17 will not be on time?翻譯: 某一國際航線的航班擁有誤點機率p=20%.針對下次n=80個國際航班的機率當中.使用二項式分配來回答下列問題:a.誤點: P(X<=15次)=?誤點=p=0.2準時=q=1-p=0.8二項式係數: c(x)=80!/[x!(80-x)!]設定函數: f(x)=c(x)*p^x*q^(80-x)函數可以透過電腦輔助運算:P(X<=15次)=f(1)+f(2)+......+f(15)=0.000,000,3534+0.000,003,4895+0.0000,2268+0.000,109+0.000,415+0.001296+0.003426+0.007815+0.01563+0.02774+0.04414+0.0634+0.08297+0.09927+0.0109=0.455475.......ans

    b.誤點: P(X>=18次)=?=P(X<=15次)+f(16)+f(17)+f(18)=0.455475+0.1109+0.1044+0.09133=0.76208........ans

    c.誤點P(X=17次)=?=f(17)=0.104376........ans

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