Poisson Distribution Lambda or Mu?

Question

The number of goals in a hockey game is modelled using a Poisson process with rate 1.5 per minute.

(a) What is the distribution for the number of goals in the first quarter (12 minutes)? Write down the pmf.

My question is with regard to the rate parameter. Should I leave lambda calculated as 1.5 and continue the calculation like so:

$$ \begin{equation} f_{X}(x) = P_{X}\, ({\{X = x}\} = \frac{e ^{-\lambda} \lambda ^ {x}} {x!} = \frac{e ^{-1.5} 1.5 ^ {x}} {x!}\qquad \qquad x = 0,1,2 , \ldots \,. \end{equation} $$

Or instead should I take into account the information concerning the 1st quarter and assign a value to $\mu $ as $\mu = 1.5 \cdot 12 = 18$?

$$ \begin{equation} f_{X}(x) = P_{X}\, ({\{X = x}\} = \frac{\mu ^ {\kern 0.08 em x} e ^ {-\mu}} {x!} = \frac{18 ^ {\kern 0.08 em x} e ^ {-18}} {x!} \qquad \qquad x = 0,1,2 , \ldots \,. \end{equation} $$

Or are these the same thing?

Answer 1

The rate tells you the number of goals per minute, so to find the average number of goals, you need to multiply the number of goals per minute by the number of minutes. So the second thing you do is correct.

They aren't the same thing... they are Poisson distributions with different means. The first one has mean 1.5 and the second has the correct mean of 18.

Answer 2

Remember that the mean value of a Poisson distribution with parameter $\lambda$ is $\lambda$. The average number of goals in the first period is $18$, so you should be using the second distribution.