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In the cumulative distribution example below, P(X > 1/4) is equivalent to, 1 – P(X < 1/4). 3-52. + The thickness of wood paneling (in inches) that a customer orders is a random variable with the following cumu- lative distribution function: x<1/8 |0.2 F(x)= 0.9 1/4sx<3/8 1/81/4) (e) P(x<1/2) f1'18) = P(x=V8) >0.20 (14) - Plx= V4)• 0.70 4//) - P(x: %) . 0.10 O True O False

In the cumulative distribution example below, P(X > 1/4) is equivalent to, 1 – P(X < 1/4). 3-52. + The thickness of wood paneling (in inches) that a customer orders is a random variable with the following cumu- lative distribution function: x<1/8 |0.2 F(x)= 0.9 1/4sx<3/8 1/81/4) (e) P(x<1/2) f1'18) = P(x=V8) >0.20 (14) - Plx= V4)• 0.70 4//) - P(x: %) . 0.10 O True O False

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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In the cumulative distribution example below, P(X > 1/4) is equivalent to, 1 – P(X < 1/4). 3-52. + The thickness of wood paneling (in inches) that a customer orders is a random variable with the following cumu- lative distribution function: x<1/8 |0.2 F(x)= 0.9 1/4sx<3/8 1/8<x<1/4 1 3/85 x Determine the following probabilities: (a) P(X<1/18) (b) P(x<1/4) (c) P(X<5/16) (d) P(X >1/4) (e) P(x<1/2) e) - P(x=/e) - 0:20 (14) - Plx= V4)• 0.70 f/0/1) - P(x. R) : 0. 10 O True O False
Transcribed Image Text:In the cumulative distribution example below, P(X > 1/4) is equivalent to, 1 – P(X < 1/4). 3-52. + The thickness of wood paneling (in inches) that a customer orders is a random variable with the following cumu- lative distribution function: x<1/8 |0.2 F(x)= 0.9 1/4sx<3/8 1/8<x<1/4 1 3/85 x Determine the following probabilities: (a) P(X<1/18) (b) P(x<1/4) (c) P(X<5/16) (d) P(X >1/4) (e) P(x<1/2) e) - P(x=/e) - 0:20 (14) - Plx= V4)• 0.70 f/0/1) - P(x. R) : 0. 10 O True O False
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