I will appreciate someone to verify my answer for exponential distribution question as I am teaching myself and don't have much confidence.
Let X, the number of years a computer works, be a random variable that follows an exponential distribution with a lambda of 3 years. You just bought a computer, what is the probability that the computer will work in 8 years?
what I have: f(x) = $\lambda \cdot e^{-\lambda \cdot x}$ given: $\lambda = 3, x = 8$ so simply $= 3 \cdot e^{-3 \cdot 8} = 3\cdot e^{-24}$?
sorry if this is too low level question. But I'm a bit confused.
If $X$ is exponentially distributed with parameter $\lambda$, the probability $P(X>x)=e^{-\lambda x}$. You can substitute $\lambda = 3$ and $x=8$ to get the probability that the computer is still working after 8 years.
By the way, some clarity is required on what $\lambda$ represents. Typically, $\lambda$ is the parameter of the distribution and $1/\lambda$ is the mean. In your example, if the mean life is 3 years (seems reasonable), $\lambda = 1/3$.
You substituted $x=8$ in the probability density function or PDF. Instead, you need to use the cumulative distribution function or CDF. For a continuous distribution (like the exponential), it is not meaningful to compute the probability $P(X=x)$. However, you can compute the probability $P(X>x)$ or $P(X \leq x)$ using the CDF (which is obtained by integrating the PDF).
Hope this helps.