Let X and Y be independent random variables representing the lifetime (in 100 ho
ID: 3182755 • Letter: L
Question
Let X and Y be independent random variables representing the lifetime (in 100 hours) of Type A and Type B light bulbs, respectively. Both variables have exponential distributions, and the mean of X is 2 and the mean of Y is 3. (a) Find the joint pdf f(x|y) of X and Y. (b) Find the conditional pdf f_2(y|x) of Y given X = x. (c) Find the probability that a Type A bulb lasts at least 300 hours and a Type B bulb lasts at least 400 hours. (d) Given that a Type B bulb foils at 300 hours, find the probability that a Type A bulb lasts longer than 300 hours. (e) What is the expected total lifetime of two Type A bulbs and one Type B bulb? (f) What is the variance of the total lifetime of two Type A bulbs and one Type B bulb?Explanation / Answer
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)
If two random variables, X and Y are independent, P(X = x, Y = y) = P(X = x)xP(Y = y)..(4)
P(X = x/Y) = P(X)…………………………………………………………………...(5)
Now, to work out solution,
X = life time of Type A bulb in hours and Y = life time of Type B bulb in hours.Then, we are given, X ~ Exp(200), Y ~ Exp(300) [note: question says life in 100 hours and mean of A and B as 2 and 3. But, we are expressing everything in terms of hours.]
Also given is that X and Y are independent.
Part (a)
Joint pdf of X and Y = P(X = x, Y = y) = P(X = x) x P(Y = y) [vide (4) under Back-up Theory]
= (1/200)e-x/200 x (1/300)e-y/300 = (1/60000)e-{(x/200) + (y/300)} ANSWER
Part (b)
[vide (5) under Back-up Theory], conditional pdf of Y/X = pdf of Y = (1/300)e-y/300 ANSWER
Part (c)
Probability Type A bulbs lasts at least 300 hours and Type B lasts at least 400 hours
= (Probability Type A bulbs lasts at least 300 hours) x (Probability Type B bulbs lasts at least 400 hours) [vide (4) under Back-up Theory]
= P(X 300) x P(Y 400) = e-300/200 x e-400/300 [vide (3) under Back-up Theory]
= e-17/6
= 0.0588 [using Excel Function] ANSWER
Part (d)
Given Type B bulb fails at 300 hours, probability Type A bulb lasts longer than 300 hours
= Probability Type A bulb lasts longer than 300 hours because of independence [vide (5) under Back-up Theory]
= P(X 300) = e-300/200 [vide (3) under Back-up Theory]
= e-1/5 = 0.2231 ANSWER
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