The distance between defects in an automated weaving process at Craft Mills, Inc. is exponentially distributed. On average there are 0.025 defects per foot. Use the random number 0.749 to simulate the distance between two defects. Give your answer to 3 decimal places. (Note: For this problem, the average (represented by tau) would be the average distance (in feet) between defects.)

55.292 feets

Explanation:

Given that :

Average defect per foot, λ = 0.025

Random number generated = 0.791

Distance between two defects :

b(x) = 1 - e^-λx = random number

1 - e^-λx = 0.749

e^-λx = 0.749 - 1

λ = 0.025

e^-0.025x = - 0.251

Take the In of both sides ;

-0.025x = - ln(0.251)

0.025x = In(0.251)

x = In(0.251) / 0.025

x = 1.382302 / 0.025

x = 55.29209

x = 55.292 feets

Hence, distance between two defects is 55.292 feets

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