An article suggested that under some circumstances the distribution of waiting time X could be modeled with the following pdf. 9- 1 F(x; 8, r) = otherwise (a) Graph f(x; 6, 80) for the three cases 8 = 3, 1, and 0.5 and comment on their shapes. f(x) fox) f(x) 0.05 0.04 0.05 8= 0.5 0.050= 0.5 0= 0.5 0.04 0.04 0.03 0.03 0.03 e= 3 8= 3 8= 3 0.02 0.02 0.02 =1 0=1 0.01 0.01 0.01 20 40 60 80 20 40 60 20 40 60 80 f(x) 0.05 = 0.5 0.04 0.03 = 3 0.02 0.01 20 40 60 80 (b) Obtain the cumulative distribution function of X. -1)-1 -(1-주)이 F(x) %3D 0

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ISBN:9781337111348
Author:Bruce Crauder, Benny Evans, Alan Noell
Publisher:Bruce Crauder, Benny Evans, Alan Noell
Chapter3: Straight Lines And Linear Functions
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Problem 12SBE: Find the equation of the regression line for the following data set. x 1 2 3 y 0 3 4
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I need help with section B, C and D

An article suggested that under some circumstances the distribution of waiting time X could be modeled with the following pdf.
Osx<T
F(x; 8, 1) =
otherwise
(a) Graph f(x; 8, 80) for the three cases 8 = 3, 1, and 0.5 and comment on their shapes.
f(x)
f(x)
f(x)
0.05
8 = 0.5
0.05 = 0.5
0.05
8 = 0.5
0.04
0.04
0.04
0.03
0.03
0.03
8= 3
8= 3
8= 3
0.02
0.02
0.02
8-1
8=1
0.01
0.01
0.01
20
40
60
80
20
40
60
80
20
40
60
80
f(x)
0.05 0= 0.5
0.04
0.03
0.02
0.01
20
40
60
80
(b) Obtain the cumulative distribution function of X.
F(x) =
1-
0 <x<T
(c) Obtain an expression for the median of the waiting time distribution.
u= T(1- 0.5) °
(d) For the case 8 = 3, 1 = 80, calculate P(40 sXs 70) without at this point doing any additional integration. (Round your answer to four decimal places.)
0.2421
Transcribed Image Text:An article suggested that under some circumstances the distribution of waiting time X could be modeled with the following pdf. Osx<T F(x; 8, 1) = otherwise (a) Graph f(x; 8, 80) for the three cases 8 = 3, 1, and 0.5 and comment on their shapes. f(x) f(x) f(x) 0.05 8 = 0.5 0.05 = 0.5 0.05 8 = 0.5 0.04 0.04 0.04 0.03 0.03 0.03 8= 3 8= 3 8= 3 0.02 0.02 0.02 8-1 8=1 0.01 0.01 0.01 20 40 60 80 20 40 60 80 20 40 60 80 f(x) 0.05 0= 0.5 0.04 0.03 0.02 0.01 20 40 60 80 (b) Obtain the cumulative distribution function of X. F(x) = 1- 0 <x<T (c) Obtain an expression for the median of the waiting time distribution. u= T(1- 0.5) ° (d) For the case 8 = 3, 1 = 80, calculate P(40 sXs 70) without at this point doing any additional integration. (Round your answer to four decimal places.) 0.2421
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Section D 

An article suggested that under some circumstances the distribution of waiting time X could be modeled with the following pdf.
e - 1
f(x; 8, 7) =
otherwise
ULDL
(a) Graph f(x; 0, 80) for the three cases e = 3, 1, and 0.5 and comment on their shapes.
f(x)
f(x)
f(x)
f(x)
0.05 A 0 = 0.5
0.05
0 = 0.5
0.05 A 0 = 0.5
0.05
e = 0.5
0.04
0.04
0.04
0.04
0.03
0.03
0.03
0.03
0 = 3
0 = 3
0 = 3
0 = 3
0.02
0.02
0.02
0.02
0 = 1
0 = 1
0 = 1
8 = 1
0.01
0.01
0.01
0.01
20
40
60
80
20
40
60
80
20
40
60
80
20
40
60
80
(b) Obtain the cumulative distribution function of X.
XS0
-(1-)°
F(x) =
0 <x<T
1.
TSX
(c) Obtain an expression for the median of the waiting time distribution.
%3D
-0.5
(d) For the case e = 3, 1 = 80, calculate P(40 SXS 70) without at this point doing any additional integration. (Round your answer to four decimal places.)
0.1231
Transcribed Image Text:An article suggested that under some circumstances the distribution of waiting time X could be modeled with the following pdf. e - 1 f(x; 8, 7) = otherwise ULDL (a) Graph f(x; 0, 80) for the three cases e = 3, 1, and 0.5 and comment on their shapes. f(x) f(x) f(x) f(x) 0.05 A 0 = 0.5 0.05 0 = 0.5 0.05 A 0 = 0.5 0.05 e = 0.5 0.04 0.04 0.04 0.04 0.03 0.03 0.03 0.03 0 = 3 0 = 3 0 = 3 0 = 3 0.02 0.02 0.02 0.02 0 = 1 0 = 1 0 = 1 8 = 1 0.01 0.01 0.01 0.01 20 40 60 80 20 40 60 80 20 40 60 80 20 40 60 80 (b) Obtain the cumulative distribution function of X. XS0 -(1-)° F(x) = 0 <x<T 1. TSX (c) Obtain an expression for the median of the waiting time distribution. %3D -0.5 (d) For the case e = 3, 1 = 80, calculate P(40 SXS 70) without at this point doing any additional integration. (Round your answer to four decimal places.) 0.1231
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