Question

Lead times have been found to be exponentially distributed with a mean of 3.7 days. Generate three random lead times from this distribution using the inverse-transform technique. Please show how to do it on pen and paper. 

          Lead times have been found to be exponentially distributed with a mean of 3.7 days. Generate three random lead times from this distribution using the inverse-transform technique. Please show how to do it on pen and paper.

Added by Tracy S.

Elementary Statistics a Step by Step Approach
Elementary Statistics a Step by Step Approach
Allan G. Bluman 9th Edition
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Lead times have been found to be exponentially distributed with a mean of 3.7 days. Generate three random lead times from this distribution using the inverse-transform technique. Please show how to do it on pen and paper.
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Transcript

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00:01 Question this is that the late times has been found to be exponential to it's true with me in 3 .7 days we have to generate 5 random daytime vary from x1 to x5 from this distribution for random numbers i 1 r2 r3 r 4 and r5 whose values are provided to us now here we consider that x is a variable that denotes the lead time and this is exponentially distributed with mean it 3 .7 days right so here the probability function of x is 1 4 upon 3 .7 into x divided by 3 .7 for x squared 0 and 0 elsewhere right so probability integral transformation the distribution function as it follows the uniform distribution that is right so in this case f x is equal to 1 minus e to require minus x divided by 3 .7 x7 x .0 so we can consider that the random variables from r to r5 they denote the observations from the uniform distribution right so here r i is equal to 1 minus 8 to the bar minus x i divided by 3 .7 that is e to devere minus x i by 3 .70 to 1 minus r i that is x i .0 to minus 3 .7 log natural 1 minus ri so this is the formula we will be using to determine the value of x so first r1 is given to a 0 .94737 so for this x 1 will be minus 3 .7 into log natural 1 minus 0 .94737 which will be equal to 10 .89454.
01:59 Next r2 is 0 .08225 so for that the value x x2 is 3 .7 to log natural 1 minus 0 .0225 so this is equal to 0 .31757 r3 is 0 .35619 so x3 is equal to n7 1 minus 0 .3 569 and this is will be equal to 1 .62901.
02:35 R4 is equal to 0 .24 -24826...
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