In: Statistics and Probability
2. An engineering office has a specialized computer for performing a particular analysis. If we assume that the time an engineer spends at this computer in one sitting is describable with an exponential random variable and that the average time spent is 40 minutes, determine the following:
a. The probability the engineer will spend less than 20 minutes at the computer
.b. The probability the engineer will spend 60 minutes at the computer.
c. The probability the engineer will spend more than 1 hour at the computer.
d. Suppose the engineer has already been using the computer for 50 minutes. What is the probability they will use it for more than one additional hour?
Solution:
Given: An engineering office has a specialized computer for performing a particular analysis.
The time an engineer spends at this computer in one sitting is describable with an exponential random variable and that the average time spent is 40 minutes.
That is: X ~ Exp
Cumulative distribution function of Exponential distribution is given by:
Part a. We have to find the probability the engineer will spend less than 20 minutes at the computer.
that is: P( X < 20) = ........?
Use F(X) to find this probability , put x = 20 in F(X) and simplify:
Use excel commandto find e-0.5 ,
=EXP(-0.5)
or use scientific calculator
thus we get:
That is: P( X < 20) = 0.393469
Part b. The probability the engineer will spend 60 minutes at the computer.
Since X is continuous random variable and probability of continuous random variable at any exact value is 0
thus P( X= 60) = 0
Part c. The probability the engineer will spend more than 1 hour at the computer.
We know, 1 hour = 60 minutes
thus we have to find:
P( X > 60) = ......?
thus we use following steps:
P( X > 60) = 1 - P( X < 60)
P( X > 60) = 1 - F(X)
P( X > 60) = 1 - ( 1 - )
P( X > 60) = 1 - 1 +
P( X > 60) =
P( X > 60) =
P( X > 60) =
P( X > 60) = 0.223130
Part d. Suppose the engineer has already been using the computer for 50 minutes. What is the probability they will use it for more than one additional hour ( 60 minutes)?
That is we have to find:
P( X > 60 + 50 | X > 50 ) =...........?
Here we use forgetfullness property of Exponential distribution:
P( X > a+b | X > a) = P( X > b)
So here we have a = 50 , b = 60
thus we get:
P( X > 60 + 50 | X > 50 ) = P ( X > 60)
from part c) we have P( X > 60) = 0.223130
thus
P( X > 60 + 50 | X > 50 ) = 0.223130