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

The correct answer is "
`0.300993e^(-0.300993x)`".

Explanation:

According to the question,

⇒
`P(x>4)=0.3`

We know that,

⇒
`P(X > x) = e^((-\lambda* x))`

⇒
`e^((-\lambda* 4)) = 0.3`

∵
`\lambda = 0.300993`

Now,

⇒
`f(x) = \lambda e^(-\lambda x)`

By putting the value, we get

`=0.300993e^(-0.300993x)`