Suppose a certain type of bulb has an average lifetime 500 hours. Suppose the pr

Question

Suppose a certain type of bulb has an average lifetime 500 hours. Suppose the probability distribution of failure of this type of bulb after time t is

f(t) = µe^(µt) , t 0

f(t)= 0 , t < 0 .

(a) Find µ.

(b) Suppose a new lamp comes with one bulb and a replacement. Suppose this lamp uses one bulb at a time. If the failure of one bulb is independent of the other, find the probability that this lamp will be in use for at least 800 hours before one needs to buy a third bulb. Leave your answer in the calculator ready form.

Explanation / Answer

Here average life time of the bulb = 500 hours

(a) Here  µ = 1/500 = 0.002 hours-1

(b) Here there are two lamps which will be used one by one. That means we have to solve this problem with the help of poisson distribution.

We have to find here the probability that these two bulbs will survive for next 800 hours.

Expected number of bulbs that will fail in 800 hours = 800 * 0.002 = 1.6

so, we have to find the probability of number of bulb failure that would be lesser than 1 that means that there should be one or less bulb should be failed so we dont need to buy a third bulb.

so Pr(X < 2) = POISSON(X < 2 ; 1.6) = Pr (X = 0) + Pr(X = 1)

= e-1.6 1.60/0! + e-1.6 1.61/1!

= 0.2019 + 0.3230 = 0.525

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