Question

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?

Answer 1

Solution:

We are given that: 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. The probability the engineer will spend less than 20 minutes at the computer.

P( X < 20) = ........?

Put x = 20 in F(X) to get P( X < 20 )

( Use excel to 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 approximately 0

thus P( X= 60) = 0

Part c. The probability the engineer will spend more than 1 hour at the computer.

For 1 hour = Number of minutes = 60

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