Question

Suppose your wait time for shuttle bus follows an exponential distribution with u = 5. (a) What is the probability that you have to wait longer than 10? (b) Given you already waited 10 minutes, what is the probability that you have to wait for another 10 more minutes? (c) Let X be exponentially distributed with parameter 1/u. Prove that P(X >a+b|X >a)=P(X >b)

Answer 1

Let x denotes the wait time (in minutes) for shuttle bus. Here x follows an exponential distribution with M=5.lie. mean = : The pdf of x is fin) = r é- 1 o.w . The caf of X is. Fla) = - e-x/5 no

[ we know that if a random variable X follows exponential distribution with mean M, then the pdf of X is NIM fla)=( e X > 0 , 0.w. The caf of x is F(x)=( Il-e-NM no

probability a) Required = P(x>10) a l-PlX510) - - Flio) --11015) = e- 2 = 0.135335 . by Required probability - Pl X > 10 +10 | X > 10 ) = plx>20 / x>10) Plx>20, x>10) P(x>10)

p( x>20) P ( x >10) = IP(X= 20) In PlXelo) 1 F120) - Flio) 2015 Il-é I l -p-1015 - e 0.135335

Here x distributed parameter is exponentially with The ne pdf pat :: of of X is -nlu fin)= no 0.W . The caf of X is FR)=( nco I-e-NM no

Now, / x>a) Plx> a+b PlX > atb, x> a) P(x> a) Plx> a+b) i plasa) - plxc atb) - Plxea) I Flatb) - Fla) 1- - (a+b) IM, (-é 1-11-é a

e-la-b) IM - a Ли e- Now, p( x>6) = I-Pl X=6) - 1 - Flb) = 1-l-e-blu,

- ь/ и - / x> a) .: Plx> a+b = P x >ь).