Is the answer to this Poisson process question correct?

Question

A factory has two production lines which work independently and deliver products to a central packing area. Products arrive from each of the lines at a rate of 1 every two minutes according to a Poisson process. Mary has just started work at the packing are.

i. How long should she expect to wait before the first product arrives?

This is simply the mean of the exponential r.v. X with parameter $\lambda$ = 0.5

E{X} = 2

ii. What is the probability she will receive more than two products in the next minute?

Now we nee to use the Poisson r.v. Y with parameter $\lambda$ = 0.5

Pr{Y>2} = 1 - Pr{Y=<2} = 1-e^-0.5[1+0.5+0.25] = 0.0144

iii. How long should she expect to wait before the 10th product arrives?

This is the mean of the Gamma distributed r.v. Z with parameters $\alpha$ = 10 and $\beta$ = o.5

E{Z} = 20

iv. The time Mary takes to pack each item may be assumed to follow an exponential distribution with mean 30 seconds. What percentage of each hour can Mary expect to be actively packing products.

By X, with $\lambda$ = 0.5, we should expect 30 products to arrive in the full hour. She will only be able to pack the first 29. This should take around 30 seconds per product, giving a total time of 14.5 minutes, which is 24.167% of an hour.

Answer 1

You are on the right track! First off, you should recognize that the sum of two independent poisson random variables (with parameters $ \lambda_1 $ and $ \lambda_2 $) is also poisson random variable (with parameter $ \lambda_1 + \lambda_2 $).

Let $ X_1 $ be the number of products delivered in 2 minutes from production line 1, and let $ X_2 $ be the number of products delivered in 2 minutes from production line 2. Since $ X_1 $ ~ $ Poisson(\lambda_1=1) $ and $ X_2 $ ~ $ Poisson(\lambda_2=1) $, then $ Y $ ~ $ Poisson(\lambda=\lambda_1+\lambda_2=2) $, where $ Y $ is the number of products delivered in 2 minutes from both production lines.

Finally, you can re-scale the poisson parameter to get "the number of products delivered in time t" as opposed to "the number of products delivered in 2 minutes." $ Z $ ~ $ Poisson(\lambda t) $, where $ t $ is given in increments of 2 minutes. It is often easier to use $ 1 $ minute as a unit of time; we can do this be letting $ t = \frac{1}{2} $, which gives us: $ Z $ ~ $ Poisson \big( \lambda t = (\lambda_1+\lambda_2)\frac{1}{2} = 1 \big) $, where $ Z $ is the number of products delivered in $ 1 $ minute from both production lines.

Part 1:

The "waiting time" of a Poisson distribution follows an exponential distribution with parameter $ \lambda $. The mean of an exponential distribution with parameter $ \lambda$ is equal to $ \frac{1}{\lambda} = \frac{1}{1} = 1 $.

Part 2:

$ P(Z > 2) = 1 - P(Z=0) - P(Z=1) - P(Z=2) = 1 - e^{-1} - e^{-1} - \frac{e^{-1}}{2} = 0.0803 $

Part 3:

The "waiting time of the 10th event" of a Poisson distribution is simply the sum of 10 independent exponential distributions with parameter $ \lambda $. Since the mean of an exponential distribution with parameter $ \lambda $ is $ \frac{1}{\lambda} = \frac{1}{1} = 1 $, then the mean of the sum of 10 independent exponential distributions (each with parameter $ \lambda $) is equal to $ 10 \times \frac{1}{\lambda} = 10 $.

Part 4:

Let $ t = 30 $ so that the unit of time in the Poisson distribution is $ 60 $ minutes = 1 $ hour $ (i.e. $ 30 \times 2 = 60 $ minutes). This means that we have: $ Z_1 $ ~ $ Poisson \big( \lambda t = (\lambda_1+\lambda_2)30 = 60 \big) $.

Since $ E[Z] = 60 $, Mary will receive (on average) $ 60 $ products in $ 1 $ hour. On average, Mary will not be able to wrap the last gift (as you pointed out in your question), so she will only wrap for $ 1,770 $ seconds (i.e. $ 59 $ gifts $ \times $ $ 30 $ seconds per gift $ = 1,770 $ seconds).

$ Percentage = \frac{1,770}{3,600} \times 100 \approx 48.36 \% $.