2) The time between successive customers coming to the market is assumed to have Exponential distribution with parameter l. a) If X1, X2, . . . , Xn are the times, in minutes, between successive customers selected randomly, estimate the parameter of the distribution. b) b) The randomly selected 12 times between successive customers are found as 1.8, 1.2, 0.8, 1.4, 1.2, 0.9, 0.6, 1.2, 1.2, 0.8, 1.5, and 0.6 mins. Estimate the mean time between successive customers, and write down the distribution function. c) In order to estimate the distribution parameter with 0.3 error and 4% risk, find the minimum sample size
2) The time between successive customers coming to the market is assumed to have Exponential distribution with parameter l. a) If X1, X2, . . . , Xn are the times, in minutes, between successive customers selected randomly, estimate the parameter of the distribution. b) b) The randomly selected 12 times between successive customers are found as 1.8, 1.2, 0.8, 1.4, 1.2, 0.9, 0.6, 1.2, 1.2, 0.8, 1.5, and 0.6 mins. Estimate the mean time between successive customers, and write down the distribution function. c) In order to estimate the distribution parameter with 0.3 error and 4% risk, find the minimum sample size
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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2) The time between successive customers coming to the market is assumed to have Exponential distribution with parameter l.
a) If X1, X2, . . . , Xn are the times, in minutes, between successive customers selected randomly, estimate the parameter of the distribution.
b) b) The randomly selected 12 times between successive customers are found as 1.8, 1.2, 0.8, 1.4, 1.2, 0.9, 0.6, 1.2, 1.2, 0.8, 1.5, and 0.6 mins. Estimate the mean time between successive customers, and write down the distribution
c) In order to estimate the distribution parameter with 0.3 error and 4% risk, find the minimum
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