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The lifespan, in years, of a certain computer is exponentially distributed. The probability that its lifespan exceeds four years is 0.30. Let f(x) represent the density function of the computer's lifespan, in years, for x>0. Determine an expression for f(x).

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Cricetus

Answer:

The correct answer is "[tex]0.300993e^{-0.300993x}[/tex]".

Step-by-step explanation:

According to the question,

⇒ [tex]P(x>4)=0.3[/tex]

We know that,

⇒ [tex]P(X > x) = e^{(-\lambda\times x)}[/tex]

⇒     [tex]e^{(-\lambda\times 4)} = 0.3[/tex]

∵ [tex]\lambda = 0.300993[/tex]

Now,

⇒ [tex]f(x) = \lambda e^{-\lambda x}[/tex]

By putting the value, we get

           [tex]=0.300993e^{-0.300993x}[/tex]

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