An article suggested that under some circumstances the distribution of waiting time X could be modeled with the following pdf. F(x; 8, T) = 6 Osx

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An article suggested that under some circumstances the distribution of waiting time X could be modeled with the following pdf.
e - 1
1 -
F(x; 8, T) =
Osx<T
otherwise
(a) Graph f(x; 8, 80) for the three cases e = 3, 1, and 0.5 and comment on their shapes.
f(x)
f(x)
0 = 0.5
0.050 = 0.5
0.05
0.04
0.04
0.03
0.03
8 = 3
8 = 3
0.02
0.02
e = 1
0.01
0.01
20
40
60
80
20
40
60
80
f(x)
f(x)
0 = 0.5
0 = 0.5
0.05
0.05
0.04
0.04
0.03
0.03
8 = 3
e = 3
0.02
0.02
e = 1
0.01
0.01
20
40
60
80
20
40
60
80
(b) Obtain the cumulative distribution function of X.
xs 0
F(x) =
0 <x<T
(c) Obtain an expression for the median of the waiting time distribution.
(d) For the case 0 = 3, T = 80, calculate P(40 s Xs 70) without at this point doing any additional integration. (Round your answer to four decimal places.)
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 1 - F(x; 8, T) = Osx<T otherwise (a) Graph f(x; 8, 80) for the three cases e = 3, 1, and 0.5 and comment on their shapes. f(x) f(x) 0 = 0.5 0.050 = 0.5 0.05 0.04 0.04 0.03 0.03 8 = 3 8 = 3 0.02 0.02 e = 1 0.01 0.01 20 40 60 80 20 40 60 80 f(x) f(x) 0 = 0.5 0 = 0.5 0.05 0.05 0.04 0.04 0.03 0.03 8 = 3 e = 3 0.02 0.02 e = 1 0.01 0.01 20 40 60 80 20 40 60 80 (b) Obtain the cumulative distribution function of X. xs 0 F(x) = 0 <x<T (c) Obtain an expression for the median of the waiting time distribution. (d) For the case 0 = 3, T = 80, calculate P(40 s Xs 70) without at this point doing any additional integration. (Round your answer to four decimal places.)
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