In a recent year, an author wrote 169 checks. Use the Poisson distribution to find the probability that, on a randomly selected day, he wrote at least one check.
We should first calculate the average number of checks he wrote per day. To do that, divide 169 by 365 (the number of days in a year) and you get (rounded) 0.463. This will be λ in our Poisson distribution. Our formula is [tex]P(X=k)= \frac{ \lambda ^{k}-e^{-\lambda} }{k!} [/tex]. We want to evaluate this formula for X≥1, so first we must evaluate our case at k=0. [tex]P(X=0)= \frac{0.463 ^{0}-e ^{-0.463} }{0!} \\ = \frac{1-e ^{-0.463} }{1} =0.3706[/tex] To find P(X≥1), we find 1-P(X<1). Since the author cannot write a negative number of checks, this means we are finding 1-P(X=0). Therefore we have 1-0.3706=0.6294. There is a 63% chance that the author will write a check on any given day in the year.
