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April 24th, 2017, 01:03 AM  #1 
Member Joined: Apr 2017 From: India Posts: 73 Thanks: 0  Exponential Distribution problem
Calls arrive at a switchboard following an exponential distribution with parameter $\lambda$=5 per hour. If we are at the switchboard, what is the probability that the waiting time for a call is 1) at least 15 minutes (Answer: 0.2865) 2)not more than 10 minutes (Answer: 0.6565) 3)exactly 5 minutes (Answer:0){I think this is because the function is a continuous one) My attempt: Firstly formed the Exponential Distribution Function: F(x)= 1  $e^{5x}$ Then for finding the probability : P[X$\ge$15] I am confused at this step. How to approach further? Similarly for second question I am unable to proceed further.For third question , I have given my explanation in the curly brackets above. Please help me out. 
April 24th, 2017, 01:46 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,468 Thanks: 1342 
if the call rate is exponential the distribution of the number of calls in a period of time is Poisson with the same rate parameter. 1) You want the probability of 0 calls in $\dfrac 1 4~hr$ $P[0] = \dfrac{\left(\frac 5 4\right)^0 e^{\frac 5 4}}{0!} = e^{\frac 5 4} = 0.2865$ you should be able to do (2) and (3) now 
April 27th, 2017, 03:57 AM  #3 
Member Joined: Apr 2017 From: India Posts: 73 Thanks: 0 
As per me , answer of second question must be: P [X<10] = 1 P [X>=10] =1  e^(5/6) =0.5654 ( But it is not matching as answer is 0.6565) 
April 27th, 2017, 08:34 AM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 2,468 Thanks: 1342  
April 27th, 2017, 08:50 AM  #5 
Member Joined: Apr 2017 From: India Posts: 73 Thanks: 0 
Well, the source I am looking is trustworthy, still there can be mistakes. But why the third answer is zero{(3)exactly 5 minutes}, is it because the distribution function is continuous or is there any other explanation? 
April 27th, 2017, 08:59 AM  #6 
Senior Member Joined: Sep 2015 From: USA Posts: 2,468 Thanks: 1342  you are correct about this. the probability of a single point in a continuous distribution is 0.


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