Ch 8 實習

1. Ch 8 實習

2. 0 1/3 1/2 1 2/3 Continuous Probability Distributions • The probability that a continuous variable X will assume any particular value is zero. Why? • A continuous random variable has an uncountably infinite number of values in the interval (a,b). The probability of each value 1/4 + 1/4 + 1/4 + 1/4 = 1 1/3 + 1/3 + 1/3 = 1 1/2 + 1/2 = 1 Jia-Ying Chen

3. 0 1/3 1/2 1 2/3 Continuous Probability Distributions As the number of values increases the probability of each value decreases. This is so because the sum of all the probabilities remains 1. When the number of values approaches infinity (because X is continuous) the probability of each value approaches 0. The probability of each value 1/4 + 1/4 + 1/4 + 1/4 = 1 1/3 + 1/3 + 1/3 = 1 1/2 + 1/2 = 1 Jia-Ying Chen

4. Area = 1 x1 x2 Probability Density Function • To calculate probabilities we define a probability density function f(x). • The density function satisfies the following conditions • f(x) is non-negative, • The total area under the curve representing f(x) equals 1. P(x1<=X<=x2) • The probability that X falls between x1 and x2 is found by calculating the area under the graph of f(x) between x1 and x2. Jia-Ying Chen

5. Uniform Distribution • A random variable X is said to be uniformly distributed if its density function is • The expected value and the variance are Jia-Ying Chen

6. Example 1 • The weekly output of a steel mill is a uniformly distributed random variable that lies between 110 and 175 metric tons. a. Compute the probability that the steel mill will produce more than 150 metric tons next week. b. Deter the probability that the steel mill will produce between 120 and 160 metric tons next week. Jia-Ying Chen

7. Solution • f(x) = ,110 ≦ x ≦ 175 • a. P(X ≧ 150) = = 0.3846 • b. P(120 ≦ X ≦ 160) = = 0.6154 Jia-Ying Chen

8. Example 2 • The following function is the density function for the random variable X： f(x)=(x-1)/8, 1≦x ≦ 5 a. Graph the density function b. Find the probability that X lies between 2 and 4 c. What is the probability that X is less than 3? Jia-Ying Chen

9. f(x) 4/8 0 x 1 5 Solution • a. • b. P(2 < X < 4) = P(X < 4) – P(X < 2) = (.5)(3/8)(4–1) – (.5)(1/8)(2–1) = .5625 – .0625 = .5 • c P(X < 3) = (.5)(2/8)(3–1) = .25 Jia-Ying Chen

10. Normal Distribution • A random variable X with mean m and variance s2is normally distributed if its probability density function is given by Jia-Ying Chen

11. Finding Normal Probabilities • Two facts help calculate normal probabilities: • The normal distribution is symmetrical. • Any normal distribution can be transformed into a specific normal distribution called… • “Standard Normal Distribution” • Example: • The amount of time it takes to assemble a computer is normally distributed, with a mean of 50 minutes and a standard deviation of 10 minutes. What is the probability that a computer is assembled in a time between 45 and 60 minutes? Jia-Ying Chen

12. Finding Normal Probabilities • Solution • If X denotes the assembly time of a computer, we seek the probability P(45 ≦ X ≦ 60). • This probability can be calculated by creating a new normal variable the standard normal variable. Every normal variable with some m and s, can be transformed into this Z. Therefore, once probabilities for Z are calculated, probabilities of any normal variable can be found. V(Z) = s2 = 1 E(Z) = m = 0 Jia-Ying Chen

13. Finding Normal Probabilities • Example - continued - m 45 - 50 X 60 - 50 P(45 ≦X ≦60) = P( ≦≦ ) s 10 10 = P(-0.5 ≦ Z ≦ 1) To complete the calculation we need to compute the probability under the standard normal distribution Jia-Ying Chen

14. Jia-Ying Chen

15. z0 = 1 z0 = -.5 Finding Normal Probabilities • Example - continued - m 45 - 50 X 60 - 50 P(45 ≦X ≦60) = P( ≦≦ ) s 10 10 = P(-.5 ≦Z ≦ 1)=(0.8413-0.5)+(0.6915-0.5)=0.5328 We need to find the shaded area Jia-Ying Chen

16. Example 3 • X is normally distributed with mean 300 and standard deviation 40. What value of X does only the top 15 % exceed? • Solution • P(0 < Z < ) = 1-0.15 = 0.85 • = 1.04; Jia-Ying Chen

17. Example 4 • The long-distance calls made by the employees of a company are normally distributed with a mean of 7.2 minutes and a standard deviation of 1.9 minutes. Find the probability that a call • a. Last between 5 and 10 minutes • b. Last more than 7 minutes • c. Last less than 4 minutes Jia-Ying Chen

18. Solution • a. P(5 < X < 10) = = P(–1.16 < Z < 1.47) = 0.9292-(1-0.8770) = .8062 • b. P(X > 7) = = P(Z > –.11) = 0.5438 • c. P(X < 4) = =1-0.9535= .0465 Jia-Ying Chen

19. Exponential Distribution • The exponential distribution can be used to model • the length of time between telephone calls • the length of time between arrivals at a service station • the life-time of electronic components. • When the number of occurrences of an event follows the Poisson distribution, the time between occurrences follows the exponential distribution. Jia-Ying Chen

20. Exponential Distribution • A random variable is exponentially distributed if its probability density function is given by • E(X) = 1/l; V(X) = (1/l)2 • Finding exponential probabilities is relatively easy: • P(X > a) = e–la. • P(X < a) = 1 – e –la • P(a1 < X < a2) = e –l(a1)– e –l(a2) Jia-Ying Chen

21. f(x) = 2e-2x f(x) = 1e-1x f(x) = .5e-.5x 0 1 2 3 4 5 Exponential Distribution Exponential distribution for l = .5, 1, 2 Jia-Ying Chen

22. Exponential Distribution • Example • The service rate at a supermarket checkout is 6 customers per hour. • If the service time is exponential, find the following probabilities: • A service is completed in 5 minutes, • A customer leaves the counter more than 10 minutes after arriving • A service is completed between 5 and 8 minutes. Jia-Ying Chen

23. Exponential Distribution • Solution • A service rate of 6 per hour = A service rate of .1 per minute (l = .1/minute). • P(X < 5) = 1-e-lx = 1 – e-.1(5) = .3935 • P(X >10) = e-lx = e-.1(10) = .3679 • P(5 < X < 8) = e-.1(5)– e-.1(8) = .1572 Jia-Ying Chen

24. Example 5 • Cars arrive randomly and independently to a tollbooth at an average of 360 cars per hour. • Use the exponential distribution to find the probability that the next car will not arrive within half a minute. • What is the probability that no car will arrive within the next half minute? Jia-Ying Chen

25. Solution • Let X denote the time (in minutes) that elapses before the next car arrives. • X is exponentially distributed with l = 360/60 = 6 cars per minute.P(X>.5) = e-6(.5) = .0498. Jia-Ying Chen

26. Solution • If Y counts the number of cars that will arrive in the next half minute, then Y is a Poisson variable with m = (.5)(6) = 3 cars per half a minute.P(Y = 0) = e-3(30)/0! = .0498. • Comment: If the first car will not arrive within the next half a minute then no car will arrives within the next half minute. Therefore, not surprisingly, the probability found here is the exact same probability found in the previous question. Jia-Ying Chen

27. t分配 • t分配是一個理論上的機率分配，他是對稱的並且為一個鍾型分配（bell-shaped），並且相似於常態分配的曲線；不同的是他是依著自由度(df)來改變分配的形狀 • t分配和常態分配看起來非常的相似 • 當df越大時，t分配會越接近常態分配（μ=0 σ=1） • 常態分配其實是t分配的的一個特例，當df=∞，t分配就是常態分配 • 實際的例子上，只要df=30，t分配就已經很接近常態分配。 Jia-Ying Chen

28. t分配 Jia-Ying Chen

29. Jia-Ying Chen

30. 分配 • 卡方分配的性質 • 卡方分配為一定義在大於等於0(正數)範圍的右偏分配，不同的自由度決定不同的卡方分配。 • 卡方分配只有一個參數即自由度，表為v。卡方分配的平均數與變異數為： • 卡方分配當自由度增加而逐漸對稱，當自由度趨近於無窮大時，卡方分配會趨近於常態分配。 Jia-Ying Chen

31. 卡方分配 Jia-Ying Chen

32. 卡方值 Jia-Ying Chen

33. F分配 • Ｆ有兩個自由度(v1,v2) • Ｆ之倒數1/F亦為Ｆ分佈：分子與分母自由度互換 Jia-Ying Chen

34. F 分配表（α=0.05） Jia-Ying Chen