5. Suppose light bulb lifetimes X are exponentially distributed with mean 100 ho

Question

5. Suppose light bulb lifetimes X are exponentially distributed with mean 100 hours (a) Find the probability that a light bulb burns out before 25.8 hours In the remaining parts, suppose we have two light bulbs. We install the first at time 0, and then when it burns out, immediately replace it with the second (b) Find the probability that the first light bulb lasts less than 25.8 hours and the lifetime of the second is more than 120 hours. There are also issues such as the nonzero thickness of the dart, and so on, further restricting our measurement. 158 CHAPTER 7. CONTINUOUS PROBABILITY MODELS (c) Find the probability that the second burnout occurs after time 192.5.

Explanation / Answer

Solution

Let X = lifetime (in hours) of bulbs. Then, we are given X ~ Exp(100), where 100 is the average life of bulbs.

Back-up Theory

If X ~ Exponential with parameter (average inter-event time), the pdf (probability density function) of X is given by f(x) = (1/)e-x/, 0 x < ………………………………(1)

CDF (cumulative distribution function), F(t) = P(X t) = 1- e-t/ ……………….…(2)

From (2), P(X > t) = e-t/ ……………….……………………………………………(3)

Part (a)

Probability that the bulb burns out before 25.8 hours = P(X < 25.8) = 1- e-25.8/100 [by (2)]

= 1 – 0.7726 = 0.2274 ANSWER

Part (b)

Since life of second bulb is independent of life of first bulb, joint probability = product of individual probabilities.

So, probability that the first bulb lasts less than 25.8 hours and second bulb lasts more than 120 hours = (probability that the first bulb lasts less than 25.8 hours) x (probability that the second bulb lasts more than 120 hours)

= P(X < 25.8) x P(X > 120) = 0.2274[from Part (a)] x e-120/100 [by (3)] = 0.2274 x 0.3012

= 0.0685 ANSWER

Part (c)

Second burnt-out occurs after time 192.5 => first bulb lasts less than 25.8 hours and second bulb lasts more than (192.5 – 25.8) hours = (probability that the first bulb lasts less than 25.8 hours) x (probability that the second bulb lasts more than 166.7 hours)

= P(X < 25.8) x P(X > 166.7) = 0.2274[from Part (a)] x e-166.7/100[by (3)] = 0.2274 x 0.1888

= 0.0429 ANSWER

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