Customers arrive at a checkout counter in a department score according to a Poisson distribution at an average of seven per hour. During a given hour, what are the probabilities that:
a.) no more than three customers arrive?
b.) at least two customers arrive?
c.) exactly five customers arrive?
Poisons Formula: $\frac{X^y}{y!}e^{-X}$
My answers:
a.) $1 – (P(Y=3) + P(Y=2)+P(Y=1)+P(Y=0))$
Basically I just use the Poisons formula and substitution $y$ and the $X$ as $7$.
b.) $1 – (P(Y=1) + P(Y=0))$
c.) $P(Y=0)+P(Y=1)+P(Y=2)+P(Y=3)+P(Y=4)+P(Y=5)$
Is this correct? I am not really good at this stuff.Thanks.
Best Answer
Poisson Distribution is given by $$f(y) = \frac{\lambda^ye^{-\lambda}}{y!}$$
For part A the Question asks for the probability that NO MORE THAN THREE arrive. So that is simply $$P(0 ~~arrive) + P(1 ~~arrives) + P(2 ~~arrive) + P(3 ~~arrive)$$ What you've solved for instead is $$P(more ~~than ~~3 ~~customers ~~arrive)$$ Get rid of the 1 minus and you're good to go. Just plug 7 hours in for $\lambda$ and each of your $Y$ values in for $y$
Part B looks good to me.
For part C, all you want to solve for is $P(Y=5)$, you don't care about the other terms. So again, just solve using the above distribution function (plug $5$ in for $y$ and $7$ in for $\lambda$)