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The article "Determination of the MTF of Positive Photoresists Using the Monte Carlo Method" (Photographic Sci.Engrg. 1983: 254-260) proposes the exponential distribution with parameter $\lambda=.93$ as a model for the distribution of a photon's free path length under certain circumstances. Suppose this is the correct model.(a) What is the expected path length, and what is the standard deviation of path length?(b) What is the probability that path length exceeds 3.0$?$ What is the probability that path length is between 1.0 and 3.0$?$(c) What value is exceeded by only 10$\%$ of all path lengths?

(a) 1.075, 1.075 (b) .0614, .333 (c) 2.476 mm

Intro Stats / AP Statistics

Chapter 3

Continuous Random Variables and Probability Distributions

Section 9

Supplementary Exercises

Continuous Random Variables

Piedmont College

Cairn University

Idaho State University

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Okay, So for a since X is exponential e of X is equal to one over Lambda, and the same is true for the standard deviation. You go to one over Lambda in the US If we plug this into a calculator given, uh, values we have will get a value of one point zero 75 for Eve X and for the standard deviation. All right, so for B, um, the probability of X being greater than three is equal to one, minus the probability that X is less than or equal to three. Could be re written, has one minus f of three. Same thing as one minus one minus. Euler is constant negative 0.93 times three. Plugging this into a calculator, we get 0.6 one four and the probability of one being between. I'm sorry. The probability of ex being between one and three inclusive is therefore f of three minus f of one. We found f of three above and f of one would be the same exact set up except you plug in one multiplied by the negative 0.93 And if we do that, we'll get a value of 0.333 and now for see, um, the 90th percentile ways requested. So 0.9 is equal to F 90th percentile and could be rewritten as one Minus Euler is constant 10.93 looking for the 90th percentile solving for are variable. We get Ln 0.1, divided by negative 0.93 Plugging this into a calculator, we'll get an answer of two 0.47 six, and that would be in units of millimeters.

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