---
tags: 機率與統計
title: 第四週活動
---
# 第四週-課堂活動

### 活動一: 常態分布的應用

某一跨國商務公司在招考員工時，必須先考初試 (滿分為200分之性向測驗)，再以其分數高低決定面試與否。過去經驗顯示測驗分數為常態分配。假設今年公司預定錄取分數最好的30人面試，而報考人數有150人，計算其平均成績為120 分和標準差分數為15 分。該公司面試前錄取分數會是多少？
(Hint: You can use Table I at page 582)
>$30/150=0.2=20\%$
>經查表得Ｚ值為0.84(0.79<X<0.8023)
>$0.84=(x-120)/15$
>$15*0.84+120=132.6$

### 活動二: 二項分佈與驗貨程序

Components arrive at a distributor in very large batches. A batch can be characterized as acceptable only if the fraction of defective components in the batch is at most .10. The distributor decides to randomly select 10 components from the batch, test each one, and accept the batch only if the sample contains at most two defective components. Assume that the condition of any particular component is independent of any other.

(a) If the actual fraction of defectives in each batch is only $π= .01$, what proportion of batches will be a accepted? Repeat this calculation for the following values of $π: .05, .10, .20$ and $.25.$
>$P(x=2,n=10,p=0.01)=0.0042$
>$P(x=1,n=10,p=0.01)=0.0914$
>$P(x=0,n=10,p=0.01)=0.9044$

>$P(x=2,n=10,p=0.05)=0.0746$
>$P(x=1,n=10,p=0.05)=0.3151$
>$P(x=0,n=10,p=0.05)=0.5987$

>$P(x=2,n=10,p=0.10)=0.1937$
>$P(x=1,n=10,p=0.10)=0.3874$
>$P(x=0,n=10,p=0.10)=0.3487$

>$P(x=2,n=10,p=0.20)=0.3020$
>$P(x=1,n=10,p=0.20)=0.2684$
>$P(x=0,n=10,p=0.20)=0.1074$

>$P(x=2,n=10,p=0.25)=0.2816$
>$P(x=1,n=10,p=0.25)=0.1877$
>$P(x=0,n=10,p=0.25)=0.0563$

(b) A graph of the proportion of batches accepted versus the actual fraction of defectives pi is called the operating characteristic curve. Use the results of part (a) to sketch this curve for $0<= π<= 1$(proportion of batches accepted is on the vertical axis and pi is on the horizontal axis).

>$\pi=0.01,sum=1.0000$
>$\pi=0.05,sum=0.9884$
>$\pi=0.10,sum=0.9298$
>$\pi=0.20,sum=0.6778$
>$\pi=0.25,sum=0.5256$
>

(c) Suppose the distributor decides to be more demanding by accepting a batch only if the sample contains at most one defective component. Repeat parts (a) and (b) with this new acceptance sampling plan.

>$P(x=1,n=10,p=0.01)=0.0914$
>$P(x=0,n=10,p=0.01)=0.9044$

>$P(x=1,n=10,p=0.05)=0.3151$
>$P(x=0,n=10,p=0.05)=0.5987$

>$P(x=1,n=10,p=0.10)=0.3874$
>$P(x=0,n=10,p=0.10)=0.3487$

>$P(x=1,n=10,p=0.20)=0.2684$
>$P(x=0,n=10,p=0.20)=0.1074$

>$P(x=1,n=10,p=0.25)=0.1877$
>$P(x=0,n=10,p=0.25)=0.0563$

>$\pi=0.01,sum=0.9958$
>$\pi=0.05,sum=0.9138$
>$\pi=0.10,sum=0.7361$
>$\pi=0.20,sum=0.3758$
>$\pi=0.25,sum=0.2440$