(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)
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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e?
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
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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