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

The time between successive customers coming to the market is assumed to have Exponential distribution with parameter l.
a) If ${X}_{1},{X}_{2},\dots ,{X}_{n}$ 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 $1.8,1.2,0.8,1.4,1.2,0.9,0.6,1.2,1.2,0.8,1.5,\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}0.6$ 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 $4\mathrm{%}$ risk, find the minimum sample size.
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Clelioo
Given
The time between successive customers coming to the market X is assumed to have Exponential distribution with parameter I.
$X\approx Exp\left(I\right)$
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
$\frac{1}{\stackrel{^}{I}}=\frac{\sum {x}_{i}}{n}$
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
$\frac{1}{\stackrel{^}{I}}=\frac{\sum {x}_{i}}{n}$
$\frac{1}{\stackrel{^}{I}}=\frac{1.8,1.2,0.8,1.4,1.2,0.9,0.6,1.2,1.2,0.8,1.5,0.6}{12}$
$\stackrel{^}{I}=\frac{10}{11}=0.9091$
So the exponential distribution is shown below
$P\left(X=x\right)=I{e}^{Ix}x\ge 0$
c)
Given
The margin of error $E=0.3$
Level of significance $=4\mathrm{%}$
If X is exponential distribution with parameter I then X is approximately normal distribution with mean and variance $\frac{1}{I}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\left(\frac{1}{I}\right)}^{2}$ respectively.
We can approximate the distribution is normal with mean and variance as $\frac{10}{11},{\left(\frac{10}{11}\right)}^{2}$ respectively.
Thus the Z critical score for $4\mathrm{%}$ level of significance or $96\mathrm{%}$ confidence interval is
${Z}_{1-0.04\text{/}2}={Z}_{0.98}=2.05$
The sample size required is calculated as shown below
$E=\frac{Z\sigma }{\sqrt{n}}$
$n={\left(\frac{Z\sigma }{E}\right)}^{2}$
$n={\left(\frac{\sqrt[2.05]{\frac{10}{11}}}{0.3}\right)}^{2}$
$n=42.45$
Thus we can say the minimum sample size required is 43. (rounding to next integer).