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) 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,and0.6 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.

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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) 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,and0.6 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.

Clelioo · Accepted Answer

Given  The time between successive customers coming to the market X is assumed to have Exponential distribution with parameter I.  X≈Exp(I)  a)  The mean of exponential distribution with parameter I is I.  As the sample mean is unbiased estimate of population mean we say the following  1I^=∑xin  This is the estimate for the parameter of the distribution.  b)  The data given 12 randomly selected times between successive customers as shown below  1.8,1.2,0.8,1.4,1.2,0.9,0.6,1.2,1.2,0.8,1.5,0.6  The parameter is estimated as shown below  1I^=∑xin  1I^=1.8,1.2,0.8,1.4,1.2,0.9,0.6,1.2,1.2,0.8,1.5,0.612  I^=1011=0.9091  So the exponential distribution is shown below  P(X=x)=IeIxx≥0  c)  Given  The margin of error E=0.3  Level of significance =4%  If X is exponential distribution with parameter I then X is approximately normal distribution with mean and variance 1Iand(1I)2 respectively.  We can approximate the distribution is normal with mean and variance as 1011,(1011)2 respectively.  Thus the Z critical score for 4% level of significance or 96% confidence interval is  Z1−0.04/2=Z0.98=2.05  The sample size required is calculated as shown below  E=Zσn  n=(ZσE)2  n=(10112.050.3)2  n=42.45  Thus we can say the minimum sample size required is 43. (rounding to next integer).

The time between successive customers coming to the market is assumed to have Exponential distribution with parameter l. a) If displaystyle{X}_{{1}},{

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