Can someone please check if I tackled this question correctly? I don't have answers to refer to. Thank you in advance!
The lifetime of a lightbulb expressed in days is exponentially distributed with paremeter 0.004.
a) What is the probability that the lightbulb lasts at least 300 days?
So we are looking for $1-P([0,299])$.
$P([a,b]) = e^{-\lambda a}-e^{-\lambda b}$
$\lambda=0.004$
$P([0,299]) = e^{-0.004 *0}-e^{-0.004*299} = 1-0.3024=0.6975$
From here follows that: $P([300,∞)=1- 0.6975=0.3025$
b) What is the probability that the lightbulb lasts at most one year?
Here we compute probability for the interval of $[0,365]$.
Hints:
For $a)$ you have to regard that the random variable of the lifetime of a lightbulb is continuous. That means $P(X\geq t)=1-P(X\leq t)$
For b) you just insert the value for $t=365$ into the cdf. Or you integrate the pdf:
$$P(X\leq 365)=\int_0^{365} 0.004\cdot e^{-0.004x} \, dx$$