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Answer:
[tex]P(50 < x < 150) =0.3834[/tex]
[tex]P(x = 100) =0.0074[/tex]
Step-by-step explanation:
Given
[tex]f(x) = \left \{ {{\frac{1}{100}e^{-x/100}\ x\ge 0} \atop {0\ x<0}} \right.[/tex]
Solving (a): Probability that it will function between 50 and 150 hr before it breaks down
This is represented as:
[tex]P(50 < x < 150) = \int\limits^{150}_{50} {f(x)} \, dx[/tex]
So, we have:
[tex]P(50 < x < 150) = \int\limits^{150}_{50} {\frac{1}{100}e^{-x/100}} \, dx[/tex]
Integrate:
[tex]P(50 < x < 150) =- e^{-x/100}|\limits^{150}_{50}[/tex]
This gives:
[tex]P(50 < x < 150) =- e^{-150/100} - - e^{-50/100}[/tex]
[tex]P(50 < x < 150) =- e^{-150/100} + e^{-50/100}[/tex]
[tex]P(50 < x < 150) =- e^{-1.5} + e^{-0.5}[/tex]
[tex]P(50 < x < 150) =- 0.2231 + 0.6065[/tex]
[tex]P(50 < x < 150) =0.3834[/tex]
Solving (a): Probability that it will function exactly 100 hr before it breaks down
This is represented as:
[tex]P(x= 100)[/tex]
This can be rewritten as:
[tex]P(x= 100) = P(99<x<101)[/tex]
So, we have:
[tex]P(99 < x < 101) = \int\limits^{101}_{99} {f(x)} \, dx[/tex]
So, we have:
[tex]P(99 < x < 101) = \int\limits^{101}_{99} {\frac{1}{100}e^{-x/100}} \, dx[/tex]
Integrate:
[tex]P(99 < x < 101) =- e^{-x/100}|\limits^{101}_{99}[/tex]
This gives:
[tex]P(99 < x < 101) =- e^{-101/100} - - e^{-99/100}[/tex]
[tex]P(99 < x < 101) =- e^{-101/100} + e^{-99/100}[/tex]
[tex]P(99 < x < 101) =- e^{-1.01} + e^{-0.99}[/tex]
[tex]P(99 < x < 101) =- 0.3642 + 0.3716[/tex]
[tex]P(99 < x < 101) =0.0074[/tex]
Hence:
[tex]P(x = 100) =P(99 < x < 101) =0.0074[/tex]