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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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.
Transcribed Image Text: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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