利用R軟體模擬投擲一顆公正骰子2次，並計算平均點數，共模擬10000次，繪製模擬之抽樣分配圖如下：

par(mfrow = c(1,2))
x <- c()
for (i in 1:10000) {x <- rbind(x,(sample(1:6,1)+sample(1:6,1))/2)}
hist(x, seq(0,7,.5), main = "size=10,000")
hist(x, seq(0,7,.01), main = "size=10,000")

某不透明箱中內裝有五顆球。五顆球中有兩顆標示2號，一顆標示3號，兩顆標示9號。利用R軟體模擬從箱中，分別以 1.取後放回 2.取後不放回 方式抽球2次，並計算平均，將模擬10000次結果繪圖，與理論抽樣分配做比較。

par(mfrow = c(1,2))

x <- c()
for (i in 1:10000) {x <- rbind(x, sum(sample(c(2,2,3,9,9),2,rep=T))/2)}
hist(x, seq(0,10,.01), main = "size=10,000") 

x <- c()
for (i in 1:10000) {x <- rbind(x, sum(sample(c(2,2,3,9,9),2,rep=F))/2)}
hist(x, seq(0,10,.01), main = "size=10,000")

投擲一顆公正骰子(1)100次 (2)400次 (3)10000次，求平均點數之期望值與變異數? 利用R軟體各模擬10000次

par(mfrow = c(1,3))

x <- c()
for (i in 1:10000) {x <- rbind(x, sum(sample(1:6,100,rep=T))/100)}
hist(x, seq(0,7,.01), main = "size=100") 

for (i in 1:10000) {x <- rbind(x, sum(sample(1:6,400,rep=T))/400)}
hist(x, seq(0,7,.01), main = "size=400")

for (i in 1:10000) {x <- rbind(x, sum(sample(1:6,10000,rep=T))/10000)}
hist(x, seq(0,7,.01), main = "size=10,000")

調整程式，讓比較的基準更一致

par(mfrow = c(1,3))

x <- c()
for (i in 1:10000) {x <- rbind(x, sum(sample(1:6,100,rep=T))/100)}
hist(x, seq(0,7,.01), main = "size=100", freq = TRUE, ylim = c(0, 3000)) 

x <- c()
for (i in 1:10000) {x <- rbind(x, sum(sample(1:6,400,rep=T))/400)}
hist(x, seq(0,7,.01), main = "size=400", freq = TRUE, ylim = c(0, 3000))

x <- c()
for (i in 1:10000) {x <- rbind(x, sum(sample(1:6,10000,rep=T))/10000)}
hist(x, seq(0,7,.01), main = "size=10,000", freq = TRUE, ylim = c(0, 3000))

計算一下平均數介於3.4至3.6之間的比例

理論值:

\(\bar{X} \sim N(3.5, \left( \sqrt{\frac{35/12} {100}} \right)^2 ) \Rightarrow P(3.4 < \bar{X} < 3.6) = P(\frac{3.4-3.5}{\sqrt{\frac{35/12} {100}} } < Z < \frac{3.6-3.5}{\sqrt{\frac{35/12} {100}} }) \approx P(-0.58554 < Z < 0.58554) \approx 0.4418153\)

x <- c()
for (i in 1:10000) {x <- rbind(x, sum(sample(1:6,100,rep=T))/100)}
length(x[x<3.6 & x>3.4,])/length(x)
## [1] 0.4234

理論值:

\(\bar{X} \sim N(3.5, \left( \sqrt{\frac{35/12} {400}} \right)^2 ) \Rightarrow P(3.4 < \bar{X} < 3.6) = P(\frac{3.4-3.5}{\sqrt{\frac{35/12} {400}} } < Z < \frac{3.6-3.5}{\sqrt{\frac{35/12} {400}} }) \approx P(-1.17108 < Z < 1.17108) \approx 0.7584334\)

x <- c()
for (i in 1:10000) {x <- rbind(x, sum(sample(1:6,400,rep=T))/400)}
length(x[x<3.6 & x>3.4,])/length(x)
## [1] 0.7582

理論值:

\(\bar{X} \sim N(3.5, \left( \sqrt{\frac{35/12} {10000}} \right)^2 ) \Rightarrow P(3.4 < \bar{X} < 3.6) = P(\frac{3.4-3.5}{\sqrt{\frac{35/12} {10000}} } < Z < \frac{3.6-3.5}{\sqrt{\frac{35/12} {10000}} }) \approx P(-5.8554 < Z < 5.8554) \approx 1.0\)

x <- c()
for (i in 1:10000) {x <- rbind(x, sum(sample(1:6,10000,rep=T))/10000)}
length(x[x<3.6 & x>3.4,])/length(x)
## [1] 1