A.2) The time between successive customers coming to the market is assumed to have Exponential distribution with parameter lambda. a) If X1, X2, ..., Xp are the times, in minutes, between Successive customers selected randomly, estimate the parameter of the distribution. b) The randomly selected 15 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, 1.8, 0.9, 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 0.4 error and 4% risk, find the minimum sample size.

A.2) The time between successive customers coming to the market is assumed to have Exponential distribution with parameter lambda. a) If X1, X2, ..., Xp are the times, in minutes, between Successive customers selected randomly, estimate the parameter of the distribution. b) The randomly selected 15 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, 1.8, 0.9, 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 0.4 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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Estimation

A new framework of data analysis where the new statistical methods are used for interpreting the results and analyzing the data is known as estimation in statistics.

Question

A.2) The time between successive customers coming to
the market is assumed to have Exponential distribution
with parameter lambda.
a) If X1, X2. ..., Xp are the times, in minutes, between
Successive customers selected randomly, estimate the
parameter of the distribution.
b) The randomly selected 15 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, 1.8, 0.9, 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
0.4 error and 4% risk, find the minimum sample size.

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