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(e) P(0 < X < 3) = (i) 0 (i) 을 (ii) 유 (iv) 1 Why integrate from 2 to 3 and not 0 to 3?

(e) P(0 < X < 3) = (i) 0 (i) 을 (ii) 유 (iv) 1 Why integrate from 2 to 3 and not 0 to 3?

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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(a) Verify function f(x) satisfies the second property of pdfs, (i) 0 (ii) 0.15 (ii) 0.5 (iv) 1 (b) Р(2 < X < 3) (i) 0 (ii) (ii) (iv) 1 (e) P(0 < X < 3) = () 0 (i) 를 (ii1) 을 (iv) 1 Why integrate from 2 to 3 and not 0 to 3? (f) Р(X < 3) — (g) Determine cdf (not pdf) F(3). (i) 0 (ii) 음 (ii)을 (iv) 1 (h) Determine F(3) – F(2). (i) 0 (ii) (iii) (iv) 1 12
Transcribed Image Text:(a) Verify function f(x) satisfies the second property of pdfs, (i) 0 (ii) 0.15 (ii) 0.5 (iv) 1 (b) Р(2 < X < 3) (i) 0 (ii) (ii) (iv) 1 (e) P(0 < X < 3) = () 0 (i) 를 (ii1) 을 (iv) 1 Why integrate from 2 to 3 and not 0 to 3? (f) Р(X < 3) — (g) Determine cdf (not pdf) F(3). (i) 0 (ii) 음 (ii)을 (iv) 1 (h) Determine F(3) – F(2). (i) 0 (ii) (iii) (iv) 1 12
Let the time waiting in line, in minutes, be described by the random variable X which has the following pdf, ) = { i, S x, 2< x < 4, 0, f(x) = otherwise. density, pdf f(x) probability, cdf F(x) 1 probability less than 3 - area under curve, P(X < 3) = 5/12 probability = value of function, F(3) = P(X < 3) = 5/12 2/3 0.75 · 1/2 probability at 3, P(X - 3) = 0 1/3 0.25+ 4
Transcribed Image Text:Let the time waiting in line, in minutes, be described by the random variable X which has the following pdf, ) = { i, S x, 2< x < 4, 0, f(x) = otherwise. density, pdf f(x) probability, cdf F(x) 1 probability less than 3 - area under curve, P(X < 3) = 5/12 probability = value of function, F(3) = P(X < 3) = 5/12 2/3 0.75 · 1/2 probability at 3, P(X - 3) = 0 1/3 0.25+ 4
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