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The amount of time, in hours, that a computer functions before breakingdown is a continuous random variable with probability density functiongiven byfX(x) ={1100e−x/100x≥00x <0.What is the probability that1. a computer will function between 50 and 150 hours before breakingdown (3 pts)2. It will function less than 100 hours? (3 pts)3. It will function exactly 100 hours before breaking down? (3 pts)

Sagot :

MrRoyal

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]

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