Answer to Question #169479 in Statistics and Probability for Sunera

Question

The maintenance department of a city’s electric power company finds that it is cost-effective to replace all street bulbs at once, rather than to replace the bulbs individually as they burn out. Assume that the lifetime of a bulb is normally distributed, with a mean of 3000 hours and a standard deviation of 200 hours.

a) If the department wants to more than 1% of the bulbs to burn out before they are replaced, after how many hours should all of the bulbs be replaced?

b) If two bulbs are selected at random from among those that have been replaced, what is the probability that at least one of them has burned out?

Answer 1

a) According to distribution function of a normal random variable X with mean of 3000 hours and standart deviation of 200 hours:

$Pr(X < X_0) = \Phi(\frac{X_0 - 3000}{200}) = 0.01$ , where $\Phi(x) = \int_{-\infty}^{x}e^{\frac{-t^2}{2}}\frac{dt}{\sqrt{2\pi}}$ is distribution function of a normal random variable with mean = 0 and standard deviation = 1, hence:$X_0 = 3000 + 200\cdot \Phi^{-1}(0.01) = 2535$

So, department should replace all bulbs after 2535 hours 

b) We want to estimate the probability that at least one of two bulbs (with lifetimes $X_1$ , $X_2$ ) have lifetime less than $X_0$ :

$Pr(min(X_1, X_2) < X_0) = 1 - Pr(min(X_1, X_2) \geq X_0) = 1 - (Pr(X_1 \geq X_0))^2 = 1 - (1 - Pr(X_1 < X_0))^2 = 1 - (1 - 0.01)^2 = 0.02$

So, the probability that at least one of these two bulbs has burned out is approximately 2%.