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# gamexamp - Gamma& Other Continuous Distributions...

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Unformatted text preview: Gamma & Other Continuous Distributions Examples Example 1 Suppose the time spent by a randomly selected student who uses a terminal connected to a local time-sharing computer facility has a gamma distribution with mean 20 minutes and variance 80 minutes2 . a. What are the values of and ? Solution If = 20 and 2 = 80, then = 20 and 2 = 80. Therefore = 20 = 20 = 5. 4 2 == 80 20 = 4. Hence b. What is the probability that a student uses the terminal for at most 24 minutes? Solution P (X 24) = F ( 24 ; 5) = F (6; 5) = 0:7149: 4 c. What is the probability that a student spends between 20 and 40 minutes using the terminal? Solution P (20 X 40) = F ( 40 ; 5) F ( 20 ; 5) = F (10; 5) F (5; 5) = 0:4112: 4 4 Example 2 Let X be the time between two successive arrivals at the drive-up window of a local bank. If X has an exponential distribution with = 1 (which is identical to a standard gamma distribution with = 1), compute the following: a. The expected time between two successive arrivals. Solution E(X) = q 1 = 1 1 = 1: b. The standard deviation of the time between successive arrivals. Solution X = c. P (X 4) 1 2 = 1 = 1 1 = 1: Solution P (X 4) = 1 exp(4) = 0:9817: d. P (2 X 5) Solution P (2 X 5) = [1 exp(5)] [1 exp(2)] = 0:1286: Example 3 The paper "Determination of the MTF of Positive Photoresists Using the Monte Carlo Method" (Photographic Sci. and Engr., 1983, pp. 254-260) proposes the exponential distribution with parameter = 0:93 as a model for the distribution of a photon's free path length (m) under certain circumstances. Suppose this is the correct model. a. What is the expected path length? Solution E(X) = 1 = 1 0:93 = 1:0753: b. What is the standard deviation of path length? Solution X = 1 = 1 0:93 = 1:0753: 1 c. What is the probability that the path length exceeds 3.0? Solution P (X > 3:0) = 1 P (X 3) = 1 F (3) = exp [3(0:93)] = 0:0614: d. What is the probability that the path length is between 1.0 and 3.0? Solution P (1:0 X 3:0) = F (3) F (1) = exp [3(0:93)] exp(0:93) = 0:3331: e. What value is exceeded by only 10% of all path lengths? Solution Let c be the 90th percentile. Then 0:9 = F (c) = 1 exp [c(0:93)]. Hence c = 2:4759: ln(0:1) 0:93 = Example 4 The lifetime X (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters = 2 and = 3. Compute the following: a. E(X) Solution E(X) = 1 + b. V (X) Solution V (X) = c. P (X 6) 2 1 = 3 3 2 = 2:6587: o2 n o2 3 2 1+ 2 1+ n 1 = 9 (2) = 1:9314: Solution P (X 6) = F (6; 2; 3) = 1 exp d. P (1:5 X 6) 2 6 3 = 0:9817: Solution P (1:5 X 6) = F (6; 2; 3) F (1:5; 2; 3) = 1 exp 0:7605: 2 6 3 1 exp 1:5 3 2 = Example 5 Suppose the proportion X of surface area in a randomly selected quadrant that is covered by a certain plant has a standard beta distribution with = 5 and = 2. a. Compute E(X). Solution E(X) = A + (B A) b. Compute V (X). Solution V (X) = (BA)2 (+)2 (++1) + = 0 + (1 0) 5 5+2 5 = 7: = (10)2 (5)(2) (5+2)2 (5+2+1) = 5 : 196 c. Compute P (X 0:2). Solution d. Compute P (0:2 X 0:4). Solution R 0:4 (+) 1 (1 0:2 ()() x R 0:2 0 (+) 1 x (1 ()() x)1 dx = x)1 dx = R 0:2 0 (7) (x4 (5)(2) x5 )dx = 0:0016: x5 )dx = 0:0394: R 0:4 (7) 4 0:2 (5)(2) (x 2 e. What is the expected proportion of the sampling region not covered by the plant? Solution 1 E(X) = 1 5 7 = 2: 7 Example 6 Calls are received at a 24-hour "suicide hotline" randomly and uniformly at a rate of 0.5 calls per day. a. What is the probability that more than two days elapse between calls? Solution Let X be the number of days between successive calls. Then X has an exponential 1 1 distribution with mean = 0:5 = 2 days. Therefore: P (X > 2) = 1 P (X 2) = 1 [1 exp(1)] = exp(1) = 0:3679: b. What is the probability that the elapsed time before the ...fth call is more than ten days? Solution Let X be the number of calls that occur in ten days. Then the elapsed time for the ...fth call exceeds ten days if and only if fewer than ...ve calls occur in ten days. Hence X is a Poisson random variable with mean 10 = 5. So: P (X < 5) = P (X 4) = 4 X 5i exp(5) i=0 i! = 0:4405: 3 ...
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## This note was uploaded on 01/08/2012 for the course EXST 4050 taught by Professor Staff during the Fall '10 term at LSU.

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