Continuous Probability Distributions l A continuous random variable
Continuous Probability Distributions l A continuous random variable has an uncountably infinite number of values in the interval (a, b). l The probability that a continuous variable X will assume any particular value is zero. Why? 1/4 1/3 1/2 0 2 The probability of each value + 1/4 + + 1/3 1/2 2/3 1/4 = 1 1/3 = 1 1/2 = 1 1 Jia-Ying Chen
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. 1/4 1/3 1/2 0 3 The probability of each value + 1/4 + + 1/3 1/2 2/3 1/4 = 1 1/3 = 1 1/2 = 1 1 Jia-Ying Chen
Probability Density Function l l To calculate probabilities we define a probability density function f(x). The density function satisfies the following conditions l l Area = 1 f(x) is non-negative, P(x 1<=X<=x 2) The total area under the curve representing f(x) equals 1. x 1 x 2 • The probability that X falls between x 1 and x 2 is found by calculating the area under the graph of f(x) between x 1 and x 2. 4 Jia-Ying Chen
Uniform Distribution 5 l A random variable X is said to be uniformly distributed if its density function is l The expected value and the variance are Jia-Ying Chen
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. l 6 Jia-Ying Chen
Solution 7 , 110 ≦ x ≦ 175 l f(x) = l a. P(X ≧ 150) = l b. P(120 ≦ X ≦ 160) = = 0. 3846 = 0. 6154 Jia-Ying Chen
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? l 8 Jia-Ying Chen
Solution l a. f(x) 4/8 0 x 1 l l 9 5 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
Normal Distribution l 10 A random variable X with mean m and variance s 2 is normally distributed if its probability density function is given by Jia-Ying Chen
Finding Normal Probabilities l Two facts help calculate normal probabilities: l l l “Standard Normal Distribution” l Example: l 11 The normal distribution is symmetrical. Any normal distribution can be transformed into a specific normal distribution called… 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
Finding Normal Probabilities l Solution l l 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. 12 E(Z) = m = 0 Therefore, once probabilities fo are calculated, probabilities of normal variable can be found. V(Z) = s 2 = 1 Jia-Ying Chen
Finding Normal Probabilities l Example - continued 45 - 50 P(45 ≦ X ≦ 60) = P( 10 X- m ≦ s 60 - 50 ≦ 10 ) = P(-0. 5 ≦ Z ≦ 1) To complete the calculation we need to compu the probability under the standard normal distr 13 Jia-Ying Chen
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Finding Normal Probabilities l Example - continued 45 - 50 X- m 60 - 50 P(45 ≦ X ≦ 60) = P( ≦ ≦ ) s 10 10 = P(-. 5 ≦ Z ≦ 1)=(0. 8413 -0. 5)+(0. 69150. 5)=0. 5328 We need to find the shaded area z 0 = -. 5 z 0 = 1 15 Jia-Ying Chen
Example 3 l l 16 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
Example 4 l l 17 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
Solution 18 l a. P(5 < X < 10) = = P(– 1. 16 < Z < 1. 47) = 0. 9292 -(1 -0. 8770) =. 8062 l b. P(X > 7) = l c. P(X < 4) = =1 -0. 9535=. 0465 = P(Z > –. 11) = 0. 5438 Jia-Ying Chen
Exponential Distribution l The exponential distribution can be used to model l l 19 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
Exponential Distribution l A random variable is exponentially distributed if its probability density function is given by l E(X) = 1/l; V(X) = (1/l)2 Finding exponential probabilities is relatively easy: l l 20 P(X > a) = e–la. P(X < a) = 1 – e –la P(a 1 < X < a 2) = e – l(a 1) – e – l(a 2) Jia-Ying Chen
Exponential Distribution Exponential distribution for l =. 5, 1, 2 f(x) = 2 e-2 x f(x) = 1 e-1 x f(x) =. 5 e-. 5 x 0 21 1 2 3 4 5 Jia-Ying Chen
Exponential Distribution l Example l l The service rate at a supermarket checkout is 6 customers per hour. If the service time is exponential, find the following probabilities: l l l 22 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
Exponential Distribution l Solution l l 23 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
Example 5 24 l Cars arrive randomly and independently to a tollbooth at an average of 360 cars per hour. l Use the exponential distribution to find the probability that the next car will not arrive within half a minute. l What is the probability that no car will arrive within the next half minute? Jia-Ying Chen
Solution l l 25 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
Solution l 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. 26 Jia-Ying Chen
t分配 28 Jia-Ying Chen
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卡方分配 31 Jia-Ying Chen
卡方值 32 Jia-Ying Chen
F 分配表(α=0. 05) 34 Jia-Ying Chen
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