---
tags: 機率與統計
title: 第七週活動
---
# 第 7 週-課堂活動

### 活動一 常態分佈的抽樣分布應用

Lifetime of a certain battery is normally distributed with a mean value of 8 hours and a standard deviation of 1 hour. There are four such batteries in a package.

(a) What is the probability that the average lifetime of the four batteries exceeds 9 hours?
>$\frac{X-u}{\sqrt{\frac{\sigma}{n}}}=Z$ n個數 s標準差
>$\frac{9-8}{\frac{1}{2}}=2\ 標準差 =>0.9772$
>$1-0.9772=0.0228$

(b) What is the probability that the total lifetime of the batteries will exceed 36 hours?
>$\frac{36-32}{\sqrt{1\times{4}}}=2\ 標準差＝>0.9772$
>1-0.9772=0.0228

( c ) If T denotes the total lifetime of the four batteries in a randomly selected package, find the numerical value of T0 for which $P(T≥T0)= 0.95$.
>$\frac{T0-32}{\sqrt{1\times{4}}}=z \ =>0.95$
>$1-0.95=0.05$
>$\because0.95=>1.64<z<1.65$
>$\therefore0.05=>-1.64>z>-1.65$
>$-(1.645)\times2+32=28.71$

(d) Referred to your answer to part ( c ), suppose the batteries manufacturer guarantees that any package of batteries that does not yield a total lifetime of T0 hours will be replaced free of charge to the customer. If it costs the manufacturer $3 dollars to replace a package of batteries (materials plus mailing to customer), calculate the expected replacement cost per package associated with a large shipment of batteries.
>$0.05*3+0.95*0=0.15$

### 活動二 中央極限定理應用

Only 2% of a large population of 100-ohm gold-band resistors have resistances that exceed 105 ohms.

(a) For samples of size 100 from this population, describe the sampling distribution of the simple proportion of resistors that have resistances in excess of 100 ohms.
>$\mathbb{E(x)}=100\times0.02=2$
>$Var(x)=2\times(1-0.02)=1.96$
>$\sigma = \sqrt{\frac{1.96}{100}}=0.014$

(b) What is the probability that the proportion of resisters with resistances exceeding 105 ohms in a random sample of 100 will be less than 3%?
>$P(x<0.03)=P(\frac{x-0.02}{0.014}<=\frac{0.03-0.02}{0.014})=p(z<=0.714)=0.7611$

(c) What if there were 10 resisters with resistances exceeding 105 ohms in a random sample of 100? Please describe your opinion using probability evidence you can calculate.
>$\frac{10}{100}=10%$
>$P(x>0.1)=P(\frac{x-0.02}{0.014}>\frac{0.1-0.02}{0.014})=p(z>5.714)$
>$P(x>0.08)=P(\frac{x-0.02}{0.014}>\frac{0.8-0.02}{0.014})=p(z>4.28)$
>$P(x>0.07)=P(\frac{x-0.02}{0.014}>\frac{0.7-0.02}{0.014})=p(z>3.57)=0.9998$
>每100顆電阻中超過105歐姆的電阻如果為7顆的情況下，機率只有0.0002，超過7顆以上的機率近乎為零