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The Bernoulli Distribution

The Bernoulli Distribution. 定義. 若一隨機試驗只有兩種課能的結果（正面反面、成功失敗），則此試驗稱之為伯弩利試驗。 A random variable X has a Bernoulli distribution with parameter p (0  p  1) if X can take only the values 0 and 1 and the probabilities are P(X=1) = p P(X=0) = (1-p)

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The Bernoulli Distribution

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  1. The Bernoulli Distribution 定義 • 若一隨機試驗只有兩種課能的結果（正面反面、成功失敗），則此試驗稱之為伯弩利試驗。 • A random variable X has a Bernoulli distribution with parameter p (0  p  1) if X can take only the values 0 and 1 and the probabilities are • P(X=1) = p • P(X=0) = (1-p) • If we let q = 1- p, then the p.f of X can be written as follows: 社會統計（上）

  2. The Bernoulli Distribution 定義 • E(X) = 1·p +0·q = p • E(X2) =X2 f(x)=12·p+02·q = p • Var(X) = E(X2) –[E(X)]2 =p-p2 • =p(1-p) = p·q 社會統計（上）

  3. 例題 • 執銅板一次，X為出現正面的數目，其分配為何？其期待值及變異數為何？ 社會統計（上）

  4. The Binomial Distribution二項分配 定義 • 若間斷r.v X的機率分配函數為： • n為完全相同且獨立之試驗的次數。 • 每次試驗只有「成功」「失敗」兩種戶斥可能 • p為每次試行成功之機率，失敗的機率為q = 1 – p, 其中 0<p<1。 • 隨機變數X表示n次獨立試驗中成功之次數。 社會統計（上）

  5. The Binomial Distribution二項分配 定義 • 一個正常20歲的成年人活至65歲的機率為80%，請問三個三個年輕人中有兩人活到65歲的機率為？ 社會統計（上）

  6. Page 236, Figure 5.4 社會統計（上）

  7. The Binomial Distribution二項分配 定義 • 每一個人存活至65的機率為85%，三人中有兩人可以存活至65: • (0.8)2 • 一個死亡的機率： • (0.2)1 • 根據上圖我們知道這種情形共有(ssf)(sfs)(fss)三種: • C3,2 = 3!/(2!(3-2)!)=3 社會統計（上）

  8. The Binomial Distribution二項分配 定義 • 鑽油井的成功機率為.30，某公司找到五處有可能蘊藏石油的地點。求正好兩處挖到石油的機率？ • p=.3, q=1-.3=.7 • n=5 • 五次中有兩次成功， • P(正好挖到兩處）=P(X=2) = (.3)2(.7)3+ (.3)2(.7)3+…(.3)2(.7)3=10 (.3)2(.7)3=.3087 社會統計（上）

  9. The Binomial Distribution二項分配 定義 • X為五次獨立的試驗成功的次數，列出X的機率分配： Binomial distribution, n=5 p=.3 社會統計（上）

  10. Page 240, Figure 5.6 社會統計（上）

  11. The Binomial Distribution二項分配 • 如果p=.5, 則成功失敗的機率各半，此機率分配為對稱(symmetric)。 • 若p>.5，表示「成功」的機率大於「失敗」，圖形右方的機率會大於左方。 • n愈大，機率分配愈接近鐘型(bell shaped) • 如果p很接近.5，既使n很小，機率分配也會呈現鐘型狀態。 • 圖5.6顯示，隨著p增加，圖形的高峰愈往右邊偏移，且愈接近.5，愈呈現鐘型。 社會統計（上）

  12. The Binomial Distribution二項分配 • If the random variable X1, X2,…Xn form n Bernoulli trials with parameter p and if X =X1+X2…+Xn, then X has a binomial distribution with parameter n and p. 社會統計（上）

  13. 例題 • 設X~b(n,p)已知E(X)=3, Var(X)=2，求P(X=7)(中山企研） 社會統計（上）

  14. Cumulative binomial distribution function • 累積二項分配機率函數 p =.3, n=5 社會統計（上）

  15. 在EXCEL中求解 • 語法： • BINOMDIST(成功次數number_s,實驗次數trials, 成功機率probability_s, 求累積函數cumulative) • Number_s   為欲求解的實驗成功次數。 • Trials   為獨立實驗的次數。 • Probability_s    為每一次實驗的成功機率。 • Cumulative   為一邏輯值，主要用來決定函數的型態。如果 cumulative 為 TRUE，則 BINOMDIST 傳回累加分配函數值，其代表最多有 number_s 次成功的機率；如果其值為 FALSE，則傳回機率密度函數的機率值，代表有 number_s 次成功的機率。 社會統計（上）

  16. 在EXCEL中求解 社會統計（上）

  17. 在EXCEL中求解 社會統計（上）

  18. 例題 • According to IRS, approximately 20% of all income tax returns contain mathematical errors. • (a) find the probability that 3 or fewer returns out of a sample of 10 contain mathematical errors. • (b) Find the probability that fewer than three of the returns contain errors. • (c) Find the probability that exactly three of the returns contains errors. • (d) find the probability that three or more of the returns contain errors. 社會統計（上）

  19. 例題 • Let X denote the number of errors, then X follows the binomial distribution with n=10 and p=.20 • (a) P(X3) = P(X=0) +P(X=1)+P(X=2)+P(X=3) • 查表可之n=10, p=.2, c=3 P(X 3) = .879 • (b) P(X<3) • N=10, p=.2, c=2, P(X<3)= P(X 2)=.678 • (c) P(X=3)=P(X 3) – P(X 2)=.879-.678=.201 • (d) P(X3) =1- P(X 2) = 1-.678=.322 社會統計（上）

  20. 例題 • 生壞血病復原之機率為40%，現有15人患此病，求 • （一）至少10人存活的機率 • P(X10)=1-P(X9) =.0338 • （二）3-8人存活的機率 • P(3X 8)=P(X 8)-P(X 2)=.8779 • （三）恰巧5人存活的機率 • P(X=5) • （四）期望值及變異數 • E(X)=15(.4)=6 Var(X)=15(.4)(.6)=3.6 社會統計（上）

  21. Sample proportion of successes • Statisticians frequently are more interested in the proportion of successes in a sample than in the number of successes. • If we obtain X successes in n trials, then the sample proportion ^p = X/n • P(X=x) = P(^p=x/n) • E(^p)=E(X/n)=np/n=p • The sample proportion ^p is an unbiased estimator of population proportion p. • Var(^p)=Var(X/n)=(1/n)2Var(X)= npq/n2 = pq/n 社會統計（上）

  22. 例題 • A councilman claims that at least 30% of the voters of a large city are in favor of increasing taxes on alcoholic beverages. To test this claim, a polling agent obtains a random sample of 500 voters. Suppose that X=100 voters in the sample say they favor the tax. Thus, the sample proportion is ^p=100/500=.2. Is it reasonable to reject the claim that, in the population, p is at least .3? 社會統計（上）

  23. 例題 • If the claim is true, the the sample proportion ^p has expected value E(^p)=p=.3 • Var(^p)=pq/n=(.3)(.7)/500=.00042 • S^p=sqrt(.00042)=.02 • Empirical rule more than 99.7% of the value of ^p should fall within 3 standard deviation of the mean: • (.3-.06, .3+.06) = (.24, .36) • .02 lies outside this interval, we have strong evidence that, in the population, p does not equal to .3. If p were .3, it would be quite unusual to observe a value as extreme as ^p=.2 in a sample of n=500. 社會統計（上）

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