RE: Q6.10

Kambiz Asrar Haghighi (kambiz@interchange.ubc.ca) Thu, 2 Mar 2000 21:09:19 -0800 

From: "Kambiz Asrar Haghighi" <kambiz@interchange.ubc.ca>
Subject: RE: Q6.10
Date: Thu, 2 Mar 2000 21:09:19 -0800



You have to use enough precision. I couldn't get the same result even with
my HP, I had to resort to using Matlab to get enough precision for the
expression "e^(-0.01)[1.01005...]"

Kambiz

-----Original Message-----
From: owner-eece563@casas2.ece.ubc.ca
[mailto:owner-eece563@casas2.ece.ubc.ca]On Behalf Of Maria Lei [4681]
Sent: Thursday, March 02, 2000 7:24 PM
To: eece563@casas2.ece.ubc.ca
Subject: Q6.10


Hi,

In Q6.10 when evaluating eqn 6.69, I didn't get the same answer as
provided in the solution.

My answer:
Pr = 1 - e^(-0.01) [0.01^0/0! + 0.01^1/1! + 0.01^2/2! + ... + 0.01^5/5!]
   = 1 - e^(-0.01)[1.01005]
   = 1.65 * 10^-7

Regards,
Maria