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The probabilities for the given hours are,
[tex](a) Pr(100 < x < 250) = 0.302440605\\(b) Pr(\frac{x > 300}{x > 200})=0.670320046\\ (c) Pr(x > 100) = 0.670320046[/tex]
What is probability?
The area of mathematics known as probability deals with numerical descriptions of how likely it is for an event to happen or for a proposition to be true. A number between 0 and 1 is the probability of an event, where, roughly speaking, 0 denotes the event's impossibility and 1 denotes its certainty.
P.D.F , [tex]f(x) = \frac{1}{250}e^\frac{-x}{250} ; x \geq 0\\[/tex]
C.D.F , [tex]F(X) = 1-e^\frac{-x}{250} ;x \geq 0[/tex]
(a)
[tex]Pr(100 < x < 250) = F(250) - F(100)\\Pr(100 < x < 250) = (1-e^\frac{-250}{250})-(1-e^\frac{100}{250} )\\ Pr(100 < x < 250) = (e^\frac{-2}{5}-e^-^1) \\ Pr(100 < x < 250) = 0.670320046 - 0.367079441\\ Pr(100 < x < 250) = 0.302440605[/tex]
(b)
[tex]Pr(\frac{x > 300}{x > 200}) = \frac{Pr[(x > 300) intersection(x > 200)]}{Pr(x > 200)} \\ Pr(\frac{x > 300}{x > 200}) = \frac{Pr(x > 300)}{Pr(x > 200)} \\Pr(\frac{x > 300}{x > 200}) = \frac{1 - F(300)}{1-F(200)} \\Pr(\frac{x > 300}{x > 200}) = \frac{e^\frac{-300}{250} }{e^\frac{-200}{250} } \\Pr(\frac{x > 300}{x > 200}) = e^\frac{-2}{5} \\Pr(\frac{x > 300}{x > 200})= 0.670320046[/tex]
(c)
[tex]Pr(x > 100) = 1-Pr(x < 100) = 1- F(100) = e^\frac{-100}{250} = e^\frac{-2}{5} = 0.670320046[/tex]
Hence, the probabilities for the given hours are,
[tex](a) Pr(100 < x < 250) = 0.302440605\\(b) Pr(\frac{x > 300}{x > 200})=0.670320046\\ (c) Pr(x > 100) = 0.670320046[/tex]
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Complete question:
Complete question is attached below,
