Expected number of a Poisson-distributed variable

Question

Fifty spotlights have just been installed in an outdoor security system. According to the manufacturer’s specifications, these particular lights are expected to burn out at the rate of 1.1 per one hundred hours. What is the expected number of bulbs that will fail to last for at least seventy-five hours?

Here $\lambda=1.1/100$ hours. My first idea was to find the average burnout rate for $75$ hour interval, which equals $3/4 \cdot \lambda=0.825/75$ hours. However, I do not understand how it can help to find the expected number of bulbs that will burn out within $75$ hours.

Answer 1

The expected value of a Poisson distribution is equal to $\lambda$ by definition. For your particular example, $E(x)=\lambda=\frac{1.1*75}{100}=0.825$, so you have already found the expected number of bulbs that will burn out. A common misconception is the the expected value must be an integer but that is not actually the case.

Answer 2

Think better of the exponential distribution, which defines the probability of the time that elapses until the occurrence of a particular event (or between events) in Poisson processes.

The cumulative density function for such distribution is, for positive time $t$, as follows: $$F(t,\lambda)=1-e^{-\lambda\,t}$$

Our unit measure of time is 100 hours. $\lambda=1.1$ and the failure threshold we are interested in is $t=0.75$. The probability that a spotlight will fail before $t=0.75$ is $$1-e^{-1.1*0.75}$$ which is, approximately, $0.561765$. Now, since we have 50 spotlights, $50*0.561765$ is the desired output, which is approximately 28.