Question

A brand of Lightbulbs have a mean lifetime of 2000 hours.

a. Suppose we have 400 of these lightbulbs turned on. Find the probability that after 1436.2 hours, at least half of them have burned out.

b. Suppose we use 400 of these lightbulbs sequentially, what is the distrubution of the total time that all 400 lightbulbs have burned out?

Answer 1

Solution-

let X be the lifetime, then X must follow exponential distribution with mean = 2000 and parameter = 1/2000 = 0.0005

(A) We will calculate the probability that a blulb burns out before 1436.2 hours, as follows-

P(X<1436.2) = 1 - e^{-0.0005*1436.2} = 0.5123

Let Y be the number of bulbs out of 400 which burns out before 1436.2 hours, then Y must follow binomial distribution with parameters-

n = 400 and p = 0.5123

Required probability =

P(Y $\geq$ 200)

= 0.70622

(B) We know that sum of n exponential distribution s with parameters "a" follow Gamma(n,a)

So for total time of 400 lights, we will have gamma distribution with parameters as-

alhpha = n = 400 and gamma = 0.0005

So total time follows gamma(400,0.0005)

Thanks!