Question

2. Suppose that you are waiting for a friend to call you and that the time you wait in minutes has an exponential distribution with parameter λ = 0.1. (a) What is the expectation of your waiting time? (b) What is the probability that you will wait longer than 10 minutes? (c) What is the probability that you will wait less than 5 minutes? (d) Suppose that after 5 minutes you are still waiting for the call. What is the distribution of your additional waiting time? In this case, what is the probability that your total waiting time is longer than 15 minutes? (e) Suppose now that the time you wait in minutes for the call has a U(0, 20) distribution. What is the expectation of your waiting time? If after 5 minutes you are still waiting for the call, what is the distribution of your additional waiting time?

3. The arrival times of workers at a factory first-aid room satisfy a Poisson process with an average of 1.8 per hour. (a) What is the value of the parameter λ of the Poisson process? (b) What is the expectation of the time between two arrivals at the first-aid room? (c) What is the probability that there is at least 1 hour between two arrivals at the firstaid room? (d) What is the distribution of the number of workers visiting the first-aid room during a 4-hour period? (e) What is the probability that at least four workers visit the first-aid room during a 4- hour period?

Answer 1

2 Answer:

A)

So, the distrubtion is exponentional with parameter λ =0.1

So ,the expectation is given by,

E(X) =1/λ

Now, calculate the expectation of the waiting time

E(X) = 1/λ

         = 1/0.1

         = 10

B):

The cumulative distributive function for exponentional function given by

F(x) = 1- e^-λx

The probability   of waiting longer than 10 min is given by,

P(X≥10) =1 – F(10)

Putting values of   λ =0.1 and x=10, we get

P(X ≥10) = 1-(1-e^-0.1×10)

                = 0.3679

C:

The probability of waiting less than 5 min is given by,

P(X ≤ 5) = F(5)

So ,

Calculating theprobanility.

P(X ≤5) =F(5)

              =1-e^ -0.1×5

               = 0.3935

D)

If I am still waiting after 5 minutes for the call, the distribution of my additional waiting time is an exponentional distribution with parameter λ =0.1

5 minutes has already elapsed and hence the probability that the total waiting time is longer than 15 minutes is equal to the probability of the additional waiting time longer than 10 minutes.

So, the probability would be same for 10 minutes which is,

P(x ≥ 10) =1-(1-e ^-0.1×10)

                   = 0.3679

E)

In this case we have Uniform density function,

U(0,20)

So, the expection of the waiting time is given by,

E(X)= (a+b)/2

Calculate the expection.

E(X)= (0+20)/2

         = 10

But in this case for the waiting call after 5 minutes, the uniform density will be U(0,15)..