Question

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).

Answer 1

Answer:

The correct answer is "$0.300993e^{-0.300993x}$".

Step-by-step explanation:

According to the question,

⇒ $P(x>4)=0.3$

We know that,

⇒ $P(X > x) = e^{(-\lambda\times x)}$

⇒     $e^{(-\lambda\times 4)} = 0.3$

∵ $\lambda = 0.300993$

Now,

⇒ $f(x) = \lambda e^{-\lambda x}$

By putting the value, we get

           $=0.300993e^{-0.300993x}$