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Continuous random variable (stats)

by francisg3
Tags: continuous, random, stats, variable
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francisg3
#1
Jun7-11, 01:13 PM
P: 32
The probability density function of the time customers arrive at a terminal (in minutes after 8:00 A.M) is
f(x)= (e^(-x/10))/10 for 0 < x


c) Determine the probability that:

two or more customers arrive before 8:40 A.M among five that arrive at the terminal. Assume arrivals are independent




my logic is the following:


Probability= 1-Probability 0 or 1 customers arrive before 8:40 A.M

the answer is the following:

P(X1>40)+ P(X1<40 and X2>40)= e-4+(1- e-4) e-4= 0.0363

from what is written above, it seems to be the probability that no one arrives before 8:40 P(X1>40) and the probability that one arrives before 8:40 (X1<40) and another arrives after 8:40 (X2>40).

i tihnk i just need some help on understanding why X2 is brought in.



Thanks!
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jashua
#2
Jun8-11, 01:11 PM
P: 43
Your logic is correct. But the solution is wrong.

Let p=P(a customer arrives at the terminal before 08:40)=P(X<40)=1-e^(-4)

Then, P(0 or 1 customers arrive at the terminal before 08:40) = (1-p)^5 + nchoosek(5, 1)*p*(1-p)^4 = 5.5255*10^(-7) (approximately zero).

Hence, P(2 or more customers arrive at the terminal before 08:40) = 1-5.5255*10^(-7) (approximately 1)


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