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 $1-P([0,299])$.

$P([a,b]) = e^{-\lambda a}-e^{-\lambda b}$


$P([0,299]) = e^{-0.004 *0}-e^{-0.004*299} = 1-0.3024=0.6975$

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 $[0,365]$.



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:

$$P(X\leq 365)=\int_0^{365} 0.004\cdot e^{-0.004x} \, dx$$

  • $\begingroup$ So my solution to a) should involve 300 as well and not take into consideration only the interval of [0,299]? $\endgroup$ – VRT Feb 3 at 19:19
  • $\begingroup$ @VRT That´s true. $\endgroup$ – callculus Feb 3 at 21:12

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