確率分布について

正規分布

x <- seq(-4, 4, 0.1)
 plot(x = x,y = dnorm(x, -2, 1),type = "b",ylim = c(0,1))
 par(new = T)
 plot(x = x,y = dnorm(x, 0, 1),type = "b",ylim = c(0,1),col = 2)
 par(new = T)
 plot(x = x,y = dnorm(x, 0, 2),type = "b",ylim = c(0,1),col = 3)
 par(new = T)
 plot(x = x,y = dnorm(x, 0, 0.5),type = "b",ylim = c(0,1),,col = 4)

二項分布

i <- seq(1,30)
 plot(x = i,dbinom(x = i,size = 30,prob = 0.2),type = "b",ylim = c(0,0.5))
 par(new = T)
 plot(x = i,dbinom(x = i,size = 30,prob = 0.5),type = "b",col = 2,ylim = c(0,0.5))
 par(new = T)
 plot(x = i,dbinom(x = i,size = 30,prob = 0.1),type = "b",col = 3,ylim = c(0,0.5))
 par(new = T)
 plot(x = i,dbinom(x = i,size = 100,prob = 0.2),type = "b",col = 4,ylim = c(0,0.5))
 legend_name <- c("size = 30,prob = 0.2","size = 30,prob = 0.5","size = 30,prob = 0.1","size = 100,prob = 0.2")
  legend(x = 15,y = 0.5,legend = legend_name,col = c(1:4),lwd=3)

par(mfrow = c(2,2))
 plot(x = i,dbinom(x = i,size = 30,prob = 0.2),type = "b",ylim = c(0,0.3))
 plot(x = i,dbinom(x = i,size = 30,prob = 0.5),type = "b",col = 2,ylim = c(0,0.3))
 plot(x = i,dbinom(x = i,size = 30,prob = 0.1),type = "b",col = 3,ylim = c(0,0.3))
 plot(x = i,dbinom(x = i,size = 100,prob = 0.2),type = "b",col = 4,ylim = c(0,0.3))

ポアソン分布

• 決まった試行回数k回の中でlambdaは決まった期間や区間の間で起きる平均発生数。
• ポイントはこの平均発生数なので、
• 例
  • 1日に平均5回成約する営業員が60件訪問した場合の成功率は0.2程度のような見方

ポアソン分布
 rpois(n = 30,lambda = 5)
 x <- seq(1,60)
 par(mfrow = c(2,2))
 plot(x, y = dpois(x = x, lambda = 1), type = "b")
 plot(x, y = dpois(x = x, lambda = 5), type = "b",col = 2)
 plot(x, y = dpois(x = x, lambda = 10), type = "b",col = 3)
 plot(x, y = dpois(x = x, lambda = 30), type = "b",col = 4)
 par(mfrow = c(1,1))

負の二項分布

1. 統計的に独立なベルヌーイ試行を続けて行ったときに、r 回の失敗をする前に成功した試行回数の分布。成功と失敗の定義は逆になることもある。
2. 同様に、統計的に独立なベルヌーイ試行を続けて行ったとき、r 回の成功を得るのに必要な試行回数の分布。
3. 数学的に、1番目の意味でのベルヌーイ試行の r を整数から実数に拡張して考えるもの。

x <- seq(1,60)
 par(mfrow = c(2,2))
 plot(x, y = dnbinom(x = x,size = 1,prob = 0.1), type = "b")
 plot(x, y = dnbinom(x = x,size = 1,prob = 0.3), type = "b",col = 2)
 plot(x, y = dnbinom(x = x,size = 1,prob = 0.5), type = "b",col = 3)
 plot(x, y = dnbinom(x = x,size = 1,prob = 0.7), type = "b",col = 4)
 par(mfrow = c(1,1))

• 以下は事象rが10回起きたパターン

par(mfrow = c(2,2))
 plot(x, y = dnbinom(x = x,size = 10,prob = 0.1), type = "b")
 plot(x, y = dnbinom(x = x,size = 10,prob = 0.3), type = "b",col = 2)
 plot(x, y = dnbinom(x = x,size = 10,prob = 0.5), type = "b",col = 3)
 plot(x, y = dnbinom(x = x,size = 10,prob = 0.7), type = "b",col = 4)
 par(mfrow = c(1,1))

カイ2乗分布

例題１）ここで、分布を利用した統計推理をしてみることにします。いまあるコーヒー喫茶店にコーヒーを飲みにやって来るお客さんの人数は１日に平均16人であることがわかっている場合、32人以上のお客さんがこの店にやって来る日数は１ヶ月のうちに何日あるかを推定してみて下さい。また、１日に8人以下しかお客さんが来ない日数は１ヶ月のうち何日あるでしょうか。

（解答）分布の平均は自由度に等しいから、この場合となります。

par(mfrow = c(2,2)) px <- seq(1,60)
 x
 par(mfrow = c(2,2))
 plot(x, y = dchisq(x = x,df = 1,ncp = 0), type = "b")
 plot(x, y = dchisq(x = x,df = 5,ncp = 0), type = "b",col = 2)
 plot(x, y = dchisq(x = x,df = 10,ncp = 0), type = "b",col = 3)
 plot(x, y = dchisq(x = x,df = 30,ncp = 0), type = "b",col = 4)
 par(mfrow = c(1,1))lot(x, y = dchisq(x = x,df = 1,ncp = 0), type = "b") plot(x, y = dchisq(x = x,df = 5,ncp = 0), type = "b",col = 2) plot(x, y = dchisq(x = x,df = 10,ncp = 0), type = "b",col = 3) plot(x, y = dchisq(x = x,df = 30,ncp = 0), type = "b",col = 4)

様々な確率密度関数、乱数発生関数