The Bernoulli Distribution

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# The Bernoulli Distribution - PowerPoint PPT Presentation

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

• 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:

The Bernoulli Distribution

• 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

• 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？

The Binomial Distribution二項分配

• 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。

The Binomial Distribution二項分配

• 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？

Page 236, Figure 5.4

The Binomial Distribution二項分配

• 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

The Binomial Distribution二項分配

• 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

The Binomial Distribution二項分配

• 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

Page 240, Figure 5.6

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。

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.

• 設X~b(n,p)已知E(X)=3, Var(X)=2，求P(X=7)(中山企研）

Cumulative binomial distribution function
• 累積二項分配機率函數

p =.3, n=5

• 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。

• 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.

• 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

• 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

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

• 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?

• 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.